applying free fall concepts to problem solving

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Application: Falling Objects (1D)

Application: projectile motion (2d), forces & newton's laws of motion, potential energy, thermal energy.

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Free Fall (Physics): Definition, Formula, Problems & Solutions (w/ Examples)

​ Free fall ​ refers to situations in physics where the only force acting on an object is gravity.

The simplest examples occur when objects fall from a given height above the surface of the Earth straight downward – a one-dimensional problem. If the object is tossed upward or forcefully thrown straight downward, the example is still one-dimensional, but with a twist.

Projectile motion is a classic category of free-fall problems. In reality, of course, these events unfold in the three-dimensional world, but for introductory physics purposes, they are treated on paper (or on your screen) as two-dimensional: ​ x ​ for right and left (with right being positive), and ​ y ​ for up and down (with up being positive).

Free-fall examples therefore often have negative values for y-displacement.

It is perhaps counterintuitive that some free-fall problems qualify as such.

Keep in mind that the only criterion is that the only force acting on the object is gravity (usually Earth's gravity). Even if an object is launched into the sky with colossal initial force, at the moment the object is released and thereafter, the only force acting on it is gravity and it is now a projectile.

• Often, high-school and many college physics problems neglect air resistance, though this always has at least a slight effect in reality; the exception is an event that unfolds in a vacuum. This is discussed in detail later.

The Unique Contribution of Gravity

A unique an interesting property of the acceleration due to gravity is that it is the same for all masses.

This was far from self-evident until the days of Galileo Galilei (1564-1642). That's because in reality gravity is not the only force acting as an object falls, and the effects of air resistance tend to cause lighter objects to accelerate more slowly – something we've all noticed when comparing the fall rate of a rock and a feather.

Galileo conducted ingenious experiments at the "leaning" Tower of Pisa, proving by dropping masses of different weights from the high top of the tower that gravitational acceleration is independent of mass.

Solving Free-Fall Problems

Usually, you are looking to determine initial velocity (v 0y ), final velocity (v y ) or how far something has fallen (y − y 0 ). Although Earth's gravitational acceleration is a constant 9.8 m/s 2 , elsewhere (such as on the moon) the constant acceleration experienced by an object in free fall has a different value.

For free fall in one dimension (for example, an apple falling straight down from a tree), use the kinematic equations in the ​ Kinematic Equations for Free-Falling Objects ​ section. For a projectile-motion problem in two dimensions, use the kinematic equations in the section ​ Projectile Motion and Coordinate Systems ​.

• You can also use the conservation of energy principle, which states that ​ the loss of potential energy (PE) ​ during the fall ​ equals the gain in kinetic energy (KE): ​ –mg(y − y 0 ) = (1/2)mv y 2 . 

Kinematic Equations for Free-Falling Objects

All of the foregoing can be reduced for present purposes to the following three equations. These are tailored for free fall, so that the "y" subscripts can be omitted. Assume that acceleration, per physics convention, equals −g (with the positive direction therefore upward).

• Note that v 0 and y 0  are initial values in any problem, not variables.

​ Example 1: ​ A strange birdlike animal is hovering in the air 10 m directly over your head, daring you to hit it with the rotten tomato you're holding. With what minimum initial velocity v 0 would you have to throw the tomato straight up in order to ensure that it reaches its squawking target?

What's happening physically is that the ball is coming to a stop owing to the force of gravity just as it reaches the required height, so here, v y = v = 0.

First, list your known quantities: ​ v = ​ 0​ , g = ​ –9.8 m/s2​ , y − y 0 = ​ 10 m

Thus you can use the third of the equations above to solve:

This is about 31 miles an hour.

Projectile Motion and Coordinate Systems

Projectile motion involves the motion of an object in (usually) two dimensions under the force of gravity. The behavior of the object in the x-direction and in the y-direction can be described separately in assembling the greater picture of the particle's motion. This means that "g" appears in most of the equations required to solve all projectile-motion problems, not merely those involving free fall.

The kinematic equations needed to solve basic projectile motion problems, which omit air resistance:

​ Example 2: ​ A daredevil decides to try to drive his "rocket car" across the gap between adjacent building rooftops. These are separated by 100 horizontal meters, and the roof of the "take-off" building is 30 m higher than the second (this almost 100 feet, or perhaps 8 to 10 "floors," i.e., levels).

Neglecting air resistance, how fast will he need to be going as he leaves the first rooftop to assure just reaching the second rooftop? Assume his vertical velocity is zero at the instant the car takes off.

Again, list your known quantities: (x – x 0 ) = 100m, (y – y 0 ) = –30m, v 0y = 0, g = –9.8 m/s 2 .

Here, you take advantage of the fact that horizontal motion and vertical motion can be assessed independently. How long the car will take to free-fall (for purposes of y-motion) 30 m? The answer is given by y – y 0 = v 0y t − (1/2)gt 2.

Filling in the known quantities and solving for t:

Now plug this value into x = x 0 + v 0x t :

​ v 0x = 40.4 m/s (about 90 miles per hour). ​

This is perhaps possible, depending on the size of the roof, but all in all not a good idea outside of action-hero movies.

Hitting it out of the Park... Far Out

Air resistance plays a major, under-appreciated role in everyday events even when free fall is only part of the physical story. In 2018, a professional baseball player named Giancarlo Stanton hit a pitched ball hard enough to blast it away from home plate at a record 121.7 miles per hour.

The equation for the maximum horizontal distance a launched projectile can attain, or ​ range equation ​ (see Resources), is:

Based on this, if Stanton had hit the ball at the theoretical ideal angle of 45 degrees (where sin 2θ is at its maximum value of 1), the ball would have traveled 978 feet! In reality, home runs almost never reach even 500 feet. Part if this is because a launch angle of 45 degrees for a batter is not ideal, as the pitch is coming in almost horizontally. But much of the difference is owed to the velocity-dampening effects of air resistance.

Air Resistance: Anything But "Negligible"

Free-fall physics problems aimed at less advanced students assume the absence of air resistance because this factor would introduce another force that can slow or decelerate objects and would need to be mathematically accounted for. This is a task best reserved for advanced courses, but it bears discussion here nonetheless.

In the real world, the Earth's atmosphere provides some resistance to an object in free fall. Particles in the air collide with the falling object, which results in transforming some of its kinetic energy into thermal energy. Since energy is conserved in general, this results in "less motion" or a more slowly increasing downward velocity.

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About the Author

Kevin Beck holds a bachelor's degree in physics with minors in math and chemistry from the University of Vermont. Formerly with ScienceBlogs.com and the editor of "Run Strong," he has written for Runner's World, Men's Fitness, Competitor, and a variety of other publications. More about Kevin and links to his professional work can be found at www.kemibe.com.

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3.5 Free Fall

Learning objectives.

By the end of this section, you will be able to:

• Use the kinematic equations with the variables y and g to analyze free-fall motion.
• Describe how the values of the position, velocity, and acceleration change during a free fall.
• Solve for the position, velocity, and acceleration as functions of time when an object is in a free fall.

An interesting application of Equation 3.4 through Equation 3.14 is called free fall , which describes the motion of an object falling in a gravitational field, such as near the surface of Earth or other celestial objects of planetary size. Let’s assume the body is falling in a straight line perpendicular to the surface, so its motion is one-dimensional. For example, we can estimate the depth of a vertical mine shaft by dropping a rock into it and listening for the rock to hit the bottom. But “falling,” in the context of free fall, does not necessarily imply the body is moving from a greater height to a lesser height. If a ball is thrown upward, the equations of free fall apply equally to its ascent as well as its descent.

The most remarkable and unexpected fact about falling objects is that if air resistance and friction are negligible, then in a given location all objects fall toward the center of Earth with the same constant acceleration , independent of their mass . This experimentally determined fact is unexpected because we are so accustomed to the effects of air resistance and friction that we expect light objects to fall slower than heavy ones. Until Galileo Galilei (1564–1642) proved otherwise, people believed that a heavier object has a greater acceleration in a free fall. We now know this is not the case. In the absence of air resistance, heavy objects arrive at the ground at the same time as lighter objects when dropped from the same height Figure 3.26 .

In the real world, air resistance can cause a lighter object to fall slower than a heavier object of the same size. A tennis ball reaches the ground after a baseball dropped at the same time. (It might be difficult to observe the difference if the height is not large.) Air resistance opposes the motion of an object through the air, and friction between objects—such as between clothes and a laundry chute or between a stone and a pool into which it is dropped—also opposes motion between them.

For the ideal situations of these first few chapters, an object falling without air resistance or friction is defined to be in free fall . The force of gravity causes objects to fall toward the center of Earth. The acceleration of free-falling objects is therefore called acceleration due to gravity . Acceleration due to gravity is constant, which means we can apply the kinematic equations to any falling object where air resistance and friction are negligible. This opens to us a broad class of interesting situations.

Acceleration due to gravity is so important that its magnitude is given its own symbol, g . It is constant at any given location on Earth and has the average value

Although g varies from 9.78 m/s 2 to 9.83 m/s 2 , depending on latitude, altitude, underlying geological formations, and local topography, let’s use an average value of 9.8 m/s 2 rounded to two significant figures in this text unless specified otherwise. Neglecting these effects on the value of g as a result of position on Earth’s surface, as well as effects resulting from Earth’s rotation, we take the direction of acceleration due to gravity to be downward (toward the center of Earth). In fact, its direction defines what we call vertical. Note that whether acceleration a in the kinematic equations has the value + g or − g depends on how we define our coordinate system. If we define the upward direction as positive, then a = − g = −9.8 m/s 2 , a = − g = −9.8 m/s 2 , and if we define the downward direction as positive, then a = g = 9.8 m/s 2 a = g = 9.8 m/s 2 .

One-Dimensional Motion Involving Gravity

The best way to see the basic features of motion involving gravity is to start with the simplest situations and then progress toward more complex ones. So, we start by considering straight up-and-down motion with no air resistance or friction. These assumptions mean the velocity (if there is any) is vertical. If an object is dropped, we know the initial velocity is zero when in free fall. When the object has left contact with whatever held or threw it, the object is in free fall. When the object is thrown, it has the same initial speed in free fall as it did before it was released. When the object comes in contact with the ground or any other object, it is no longer in free fall and its acceleration of g is no longer valid. Under these circumstances, the motion is one-dimensional and has constant acceleration of magnitude g . We represent vertical displacement with the symbol y .

Kinematic Equations for Objects in Free Fall

We assume here that acceleration equals − g (with the positive direction upward).

Problem-Solving Strategy

• Decide on the sign of the acceleration of gravity. In Equation 3.15 through Equation 3.17 , acceleration g is negative, which says the positive direction is upward and the negative direction is downward. In some problems, it may be useful to have acceleration g as positive, indicating the positive direction is downward.
• Draw a sketch of the problem. This helps visualize the physics involved.
• Record the knowns and unknowns from the problem description. This helps devise a strategy for selecting the appropriate equations to solve the problem.
• Decide which of Equation 3.15 through Equation 3.17 are to be used to solve for the unknowns.

Example 3.14

Free fall of a ball.

• Substitute the given values into the equation: y = y 0 + v 0 t − 1 2 g t 2 − 98.0 m = 0 − ( 4.9 m/s ) t − 1 2 ( 9.8 m/s 2 ) t 2 . y = y 0 + v 0 t − 1 2 g t 2 − 98.0 m = 0 − ( 4.9 m/s ) t − 1 2 ( 9.8 m/s 2 ) t 2 . This simplifies to t 2 + t − 20 = 0 . t 2 + t − 20 = 0 . This is a quadratic equation with roots t = −5.0 s and t = 4.0 s t = −5.0 s and t = 4.0 s . The positive root is the one we are interested in, since time t = 0 t = 0 is the time when the ball is released at the top of the building. (The time t = −5.0 s t = −5.0 s represents the fact that a ball thrown upward from the ground would have been in the air for 5.0 s when it passed by the top of the building moving downward at 4.9 m/s.)
• Using Equation 3.15 , we have v = v 0 − g t = −4.9 m/s − ( 9.8 m/s 2 ) ( 4.0 s ) = −44.1 m/s . v = v 0 − g t = −4.9 m/s − ( 9.8 m/s 2 ) ( 4.0 s ) = −44.1 m/s .

Significance

Example 3.15, vertical motion of a baseball.

• Equation 3.16 gives y = y 0 + v 0 t − 1 2 g t 2 y = y 0 + v 0 t − 1 2 g t 2 0 = 0 + v 0 ( 5.0 s ) − 1 2 ( 9.8 m / s 2 ) ( 5.0 s ) 2 , 0 = 0 + v 0 ( 5.0 s ) − 1 2 ( 9.8 m / s 2 ) ( 5.0 s ) 2 , which gives v 0 = 24.5 m/s v 0 = 24.5 m/s .
• At the maximum height, v = 0 v = 0 . With v 0 = 24.5 m/s v 0 = 24.5 m/s , Equation 3.17 gives v 2 = v 0 2 − 2 g ( y − y 0 ) v 2 = v 0 2 − 2 g ( y − y 0 ) 0 = ( 24.5 m/s ) 2 − 2 ( 9.8 m/s 2 ) ( y − 0 ) 0 = ( 24.5 m/s ) 2 − 2 ( 9.8 m/s 2 ) ( y − 0 ) or y = 30.6 m . y = 30.6 m .
• To find the time when v = 0 v = 0 , we use Equation 3.15 : v = v 0 − g t v = v 0 − g t 0 = 24.5 m/s − ( 9.8 m/s 2 ) t . 0 = 24.5 m/s − ( 9.8 m/s 2 ) t . This gives t = 2.5 s t = 2.5 s . Since the ball rises for 2.5 s, the time to fall is 2.5 s.
• The acceleration is 9.8 m/s 2 everywhere, even when the velocity is zero at the top of the path. Although the velocity is zero at the top, it is changing at the rate of 9.8 m/s 2 downward.
• The velocity at t = 5.0 s t = 5.0 s can be determined with Equation 3.15 : v = v 0 − g t = 24.5 m/s − 9.8 m/s 2 ( 5.0 s ) = −24.5 m/s . v = v 0 − g t = 24.5 m/s − 9.8 m/s 2 ( 5.0 s ) = −24.5 m/s .

Check Your Understanding 3.7

A chunk of ice breaks off a glacier and falls 30.0 m before it hits the water. Assuming it falls freely (there is no air resistance), how long does it take to hit the water? Which quantity increases faster, the speed of the ice chunk or its distance traveled?

Example 3.16

Rocket booster.

• From Equation 3.17 , v 2 = v 0 2 − 2 g ( y − y 0 ) v 2 = v 0 2 − 2 g ( y − y 0 ) . With v = 0 and y 0 = 0 v = 0 and y 0 = 0 , we can solve for y : y = v 0 2 2 g = ( 2.0 × 10 2 m / s ) 2 2 ( 9.8 m / s 2 ) = 2040.8 m . y = v 0 2 2 g = ( 2.0 × 10 2 m / s ) 2 2 ( 9.8 m / s 2 ) = 2040.8 m . This solution gives the maximum height of the booster in our coordinate system, which has its origin at the point of release, so the maximum height of the booster is roughly 7.0 km.
• An altitude of 6.0 km corresponds to y = 1.0 × 10 3 m y = 1.0 × 10 3 m in the coordinate system we are using. The other initial conditions are y 0 = 0 , and v 0 = 200.0 m/s y 0 = 0 , and v 0 = 200.0 m/s . We have, from Equation 3.17 , v 2 = ( 200.0 m / s ) 2 − 2 ( 9.8 m / s 2 ) ( 1.0 × 10 3 m ) ⇒ v = ± 142.8 m / s . v 2 = ( 200.0 m / s ) 2 − 2 ( 9.8 m / s 2 ) ( 1.0 × 10 3 m ) ⇒ v = ± 142.8 m / s .

Interactive

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Free Fall Practice Problems for High Schools: Complete Guide

In this long article, we are going to practice some problems about a freely falling object in the absence of air resistance. All these questions are suitable for high school or college students, or even the AP Physics 1 exam. 

Freely Falling Motion Problems

Problem (1): A tennis ball is thrown vertically upward with an initial speed of 17 m/s and caught at the same level above the ground.  (a) How high does the ball rise? (b) How long was the ball in the air? (c) How long does it take to reach its highest point?

Solution : Take up as the positive direction and the throwing point as the origin, so $y_0=0$. 

(a) The ball goes up so high that its vertical velocity becomes zero. For this part of ascending motion, we can use the free fall kinematic equation $v^2-v_0^2=-2g(y-y_0)$. Substituting the known values into it and solving for $y$, we get \begin{gather*} v^2-v_0^2=-2g(y-y_0) \\\\ 0-17^2 = -2(10)(y_{max}-0) \\\\ \Rightarrow \boxed{y_{max}=14.45\,\rm m} \end{gather*}  (b) In all free fall practice problems, the best way to find the total flight time that the object was in the air is to use the kinematic equation $y-y_0=-\frac 12 gt^2+v_0t$. Then, substitute the coordinates of where the object landed on the ground into it. 

As a rule of thumb, if the object returns to the same point of launch, its displacement vector is always zero, so $y-y_0=0$. Therefore, we have \begin{gather*} y-y_0=-\frac 12 gt^2+v_0t \\\\ 0=-\frac 12 (10)t^2+17t \\\\ \Rightarrow \boxed{5t^2-17t=0} \end{gather*} Solving this equation by factoring out the time and setting the remaining expression to zero, we get  \begin{gather*} 5t^2-17t=0 \\\\ t(5t-17)=0 \\\\ \Rightarrow t=0 \quad ,\quad t=3.4\,\rm s \end{gather*} The first result corresponds to the initial time, and the other time, $\boxed{t_{tot}=3.4\,\rm s}$, is the amount of time the ball is in the air until it reaches the ground. 

(c) At the highest point the vertical velocity is always zero, $v=0$. Using the equation $v=v_0-gt$, and solving for $t$, we have \begin{gather*} v=v_0-gt \\ 0=17-(10)t \\ \Rightarrow t_{top}=1.7\,\rm s \end{gather*} As you can see, the duration of the ball's going up, in the absence of air resistance, is always half the total flight time. Therefore, \[t_{top}=\frac 12 t_{tot}\]

Problem (2): A ball is dropped directly downward from a height of 45 meters with an initial speed of 6 m/s. How many seconds later does it strike the ground?

Solution : Taking up as the positive direction and the dropping point as the origin, we have $y_0=0$. Since the ball is moving downward, we choose a negative sign for its initial velocity, so $v_0=-6\,\rm m/s$.

The ball strikes the ground $45\,\rm m$ below the chosen origin, so its correct coordinate is $y=-45\,\rm m$. 

The only kinematic equation that relates all these variables to the time is $y-y_0=-\frac 12 gt^2+v_0t$. Substituting the numerical values into this equation, yields \begin{gather*} y-y_0 =-\frac 12 gt^2+v_0t \\\\ -45-0 =-\frac 12 (10)t^2+(-6)t \\\\ \Rightarrow \boxed{5t^2+6t-45=0} \end{gather*} In the last step, after rearranging, we arrived at a quadratic equation, like $at^2+bt+c=0$, that its solution is found using the below formula \[t=\frac{-b\pm\sqrt{b^2-4ac}}{2a} \] where $a,b,c$ are constants. In this case, we have \[a=5, b=5, c=-45\] Substituting these values into the above formula, we get \begin{gather*} t=\frac{-5\pm\sqrt{5^2-4(5)(-45)}}{2(5)} \\\\ t=2.46\,\rm s \quad , \quad t=-3.66 \end{gather*} Since we choose the positive time for free fall problems, the ball would reach the ground approximately $2.5\,\rm s$ after being dropped.

Problem (3): A coin is tossed vertically upward and remains $\rm 0.6\, s$ in the air before it comes back down.  (a) How fast is the coin tossed? (b) How high does the coin rise? 

Solution : Let the tossing point be the origin, so $y_0=0$. The time duration the coin is in the air until it reaches the highest point is $t=0.6\,\rm s$. Recall that in all free fall problems, the object at the highest point has a velocity $v=0$. 

(a) Use the kinematic equation $v=v_0-gt$, substitute the above-known values, and solve for the unknown initial speed $v_0$ \begin{align*} v&=v_0-gt \\ 0&=v_0-(10)(0.6) \\ \Rightarrow v_0&=6\,\rm m/s \end{align*} Therefore, the coin was tossed vertically upward with an initial speed of $6\,\rm m/s$. 

(b) In this part, we need a kinematic equation that relates the distance traveled to the time taken, $y-y_0=-\frac 12 gt^2+v_0t$, or to the initial and final velocities, $v^2-v_0^2=-2g(y-y_0)$. We want to use the second equation as follows. \begin{align*} v^2-v_0^2&=-2g(y-y_0) \\\\ (0)^2-(60)^2 &= -2(10)(y-0) \\\\ y&=\frac{-6^2}{-20} \\\\ &=\boxed{1.8\,\rm m} \end{align*} You can also use the first equation and arrive at the same result. Check it out yourself. 

Problem (4): A small stone is shot straight up with an initial speed of $\rm 15\,m/s$.  (a) How long does the stone take to reach its maximum height? (b) How high does the stone go? (c) With what speed would the stone hit the ground?

Solution : The known data is $y_0=0$ and $v_0=15\,\rm m/s$. 

(a) When an object is thrown vertically upward in the air, it rises until it reaches a point where its vertical velocity becomes zero; otherwise, it continues on its way. So, in this part, at the maximum height, we have $v=0$. Substituting all these known numerical values into the equation $v=v_0-gt$, and solving for the unknown time $t$, we get \begin{align*} 0 &= 15-(10)t \\ \Rightarrow t&=\boxed{1.5\,\rm s} \end{align*} Thus, it takes $1.5\,\rm s$ for  the stone to reach the highest point. 

(b) For this part, we have the time taken to reach the maximum height $t=1.5\,\rm s$, the initial and final velocities. Thus, we can use either the equation $y-y_0=-\frac 12 gt^2+v_0t$ or $v^2-v_0^2=-2g(y-y_0)$. We chose the second equation. \begin{align*} 0-(15)^2 &= -2(10)(y_{max}-0) \\\\ y_{max} &=\frac{-15^2}{-20} \\\\ \Rightarrow y_{max}&= \boxed{11.25\,\rm m} \end{align*} Thus, the stone reaches the maximum height of $11.25\,\rm m$. 

(c) When the stone hits the ground at the same level as the throwing point, then according to the definition of displacement, it displaces nothing. This means that the displacement between initial and final points $\Delta y=y-y_0$ is zero. 

Using this fact, we can use the equation $v^2-v_0^2=-2g(y-y_0)$ to get the velocity at the moment of hitting the ground. \begin{align*} v^2-(15)^2 &=-2(10)(0) \\ \Rightarrow v^2 = 15^2 \end{align*} Taking the square root of both sides, yields two roots of $v=\pm 15\,\rm m/s$. Recall that velocity is a vector in physics that has both magnitude and direction. 

The stone is hitting the ground, so the correct sign is a negative for its velocity, i.e., $v=-15\,\rm m/s$. 

Problem (5): From the top of a high cliff, a stone is released. It is seen after $2.5\,\rm s$, the stone strikes the ground. How high is the cliff? 

Solution : The stone is released, so its initial speed is zero, $v_0=0$. The total flight time is also $t=2.5\,\rm s$. Taking the releasing point as the origin, we will have $y_0=0$. In the kinematic equations, there is vertical displacement in only two of them, i.e., $v^2-v_0^2=-2g(y-y_0)$ and $y-y_0=-\frac 12 gt^2+v_0t$. As you can see, to use the first equation, we must have the final velocity where the stone hits the ground, and for the second equation, the total flight time is needed. Hence, it is simpler to apply the second equation and solve for the unknown vertical distance $y$.  \begin{gather*} y-y_0=-\frac 12 gt^2+v_0t \\\\ y-0=-\frac 12 (10)(2.5)^2+(0)(2.5) \\\\ \Rightarrow \quad \boxed{y=-31.25\,\rm m} \end{gather*} The negative indicates that the stone strikes the ground $31.25\,\rm m$ below our chosen origin. 

Problem (6): A stone is released at rest from a height and falls freely for $4\,\rm s$.  (a) What is the stone's velocity $1.5\,\rm s$ after releasing? (b) How high is the height?

Solution : The stone is released at rest, so $v_0=0$. Take upward as the positive direction. The total flight time is $t_{tot}=4\,\rm s$. 

(a) With this known information, use the equation $v=v_0-gt$ to find the stone's velocity at each instant of time. \begin{gather*} v=v_0-gt \\ v=0-(10)(1.5) \\ \Rightarrow \boxed{v=-15\,\rm m/s} \end{gather*} The negative indicates the direction of the stone at that moment, which is facing down. 

(b) The time between releasing the stone and hitting the ground is given. With this known information, use the equation $y-y_0=-\frac 12 gt^2 +v_0t$ and solve for the total distance fallen by the stone. \begin{gather*} y-0=-\frac 12 (10)(4)^2+(0)(4) \\\\ \Rightarrow \quad \boxed {y=-80\,\rm m} \end{gather*} The minus sign reminds us that the stone strikes the ground $80\,\rm m$ below our chosen origin. 

Problem (7): A person throws a light stone straight up and catches it $2.6\,\rm s$ later. With what speed did he throw the stone, and to what height does the stone go up? 

Solution : As with any other free-fall problem, let upward be the positive direction and the throwing point be the origin of the coordinate system so that $y_0=0$. 

The stone is caught at the same level of throwing, so its vertical displacement is zero, $\Delta y=y-y_0=0$. The total flight time is also known. Hence, applying the vertical displacement kinematic equation, $y-y_0=-\frac 12 gt^2+v_0t$, and solving for the initial speed $v_0$ gives us \begin{gather*} y-y_0=-\frac 12 gt^2+v_0t \\\\ 0=-\frac 12 (10)(2.6)^2+v_0 (2.6) \end{gather*} Factoring out $2.6$ from the last expression, would get \begin{gather*} (2.6)(-5(2.6)+v_0)=0 \\\\ \Rightarrow \quad \boxed{v_0=13\,\rm m/s} \end{gather*} As a side note, if an object is thrown vertically upward and caught at the same level, by knowing the time interval between these two moments, we can use the following formula to find its initial speed \[v_0=\frac 12 g t\] The stone goes up until it reaches a point where its vertical velocity is zero, i.e., $v=0$. Now that we know the stone's initial speed, applying the equation $v^2-v_0^2=-2g(y-y_0)$ gives us \begin{gather*} 0-(13)^2=-2(10)(y-0) \\\\ \Rightarrow \quad \boxed{y=8.45\,\rm m} \end{gather*} 

Problem (8): A baseball is thrown straight up with a speed of $25\,\rm m/s$.  (a) With what speed is it moving when it is at a height of $10\,\rm m$? (b) How much time does it take to reach that point?

Solution : As usual, take up the positive $y$-direction and set $y_0=0$. The initial speed is also $v_0=25\,\rm m/s$. 

(a) The only time-independent kinematic equation that relates these known values to each other is $v^2-v_0^2=-2g(y-y_0)$. Substituting the known numerical values into this, we get \begin{gather*} v^2-(25)^2=-2(9.8)(10-0) \\\\ v^2=429 \\\\ \Rightarrow \quad v=\pm 20.7\,\rm m/s \end{gather*} As you can see, we obtained a speed with two different signs. Here, the speed with a positive sign indicates that the baseball at that desired height is being moved upward, while the negative sign hints to us that the baseball is at that height when it is moving down. 

Thus, we arrive at the fact that the baseball reaches that height twice, once when it is going up and once when it is going down.

(b) In the previous part, we found out that we were at that height twice. Thus, there are two corresponding times for this situation. Applying the equation $y-y_0=-\frac 12 gt^2+v_0t$ and solving for the required time $t$, we get \begin{gather*} 10-0=-\frac 12 (10)t^2+25t \\\\ \Rightarrow \quad 4.9t^2-25t+10=0 \end{gather*} The quadratic equation above has two solutions as below \begin{gather*} t_1=0.44\,\rm s \quad , \quad t_2=4.66\,\rm s \end{gather*} the first time, $t_1$ is when the ball is going up, and the second one corresponds to when it is moving down. 

Note : The quadratic equation of $at^2+bt+c=0$ has the following solution formula \[t=\frac{-b\pm\sqrt{b^2-4ac}}{2a}\]

Problem (9): A helicopter is ascending vertically upward with a constant speed of $6.5\,\rm m/s$ and carrying a package of $2\,\rm kg$. When it reaches a height of $150\,\rm m$ above the surface, it drops the package. How much time does it take for the package to hit the surface?

Solution : Take up the positive direction, as always. At the moment that the package is released, it has the speed of its carrier. 

There is a subtle point in this case. Since the object is moving down, in the opposite direction of our positive direction, a negative must accompany it. Therefore, the correct input for the initial speed is $v_0=-6.5\,\rm m/s$.

Applying the equation $y-y_0=-\frac 12 gt^2+v_0t$, we will have \begin{gather*} y-y_0=-\frac 12 gt^2+v_0t \\\\ -150-0=-\frac 12 (10)t^2+(-6.5)t \\\\ \Rightarrow 5t^2+6.5t-150=0 \end{gather*} Using a graphing calculator or the formula in the previous question, we find that it takes about $4.9\,\rm s$ for the package to reach the ground. 

Note that in the above, we inserted a negative for the vertical height as $y=-150\,\rm m$, since the package hits the ground $150\,\rm m$ below our chosen origin. 

Problem (10): A stone is thrown vertically upward from a building $15\,{\rm m}$ high with an initial velocity of $10\,{\rm m/s}$. What is the stone's velocity just before hitting the ground?

Solution : In all freely falling practice problems, the important note is choosing the origin. Usually, the throwing (releasing or dropping) point is the best choice. In this case, the hitting point is below the origin, so its vertical displacement ($y$) is negative.

Applying the time-independent free fall kinematic equation, we have \begin{align*}v_f^{2}-v_i^{2}&=2\,(-g)\Delta y\\v_f^{2}-(10)^{2}&=2\,(-10)(-15) \\ \Rightarrow v_f&=\pm 20\,{\rm m/s}\end{align*} Since velocity is a vector quantity and just before striking the ground its direction is vertically downward, the negative value must be chosen, i.e., $v_f=-20\,{\rm m/s}$. 

Problem (11): A bullet is fired vertically downward with an initial speed of $15\,{\rm m/s}$ from the top of a tower of $20\,{\rm m}$ high. What is its velocity at the instant of striking the ground? 

Solution : Let the origin be the firing point. Using the below kinematic equation, we have \begin{align*}v_f^{2}-v_i^{2}&=-2g(y-y_0) \\\\ v_f^{2}-0&=-2(10)(-20)\\\Rightarrow v_f&=\pm 20\,{\rm m/s}\end{align*} Since the direction of velocity at the moment of striking the ground is downward, we must choose the negative correct sign, so we have $v_f=-20\,\rm m/s$. 

Problem (12): There is a well with a depth of $34\,{\rm m}$. A person drops a stone vertically into it with an initial velocity of $7\,{\rm m/s}$. What is the time interval between dropping the stone and hearing its impact sound? (Assume $g=10\,{\rm m/s^2}$ and the speed of the sound in the air is $340\,{\rm m/s}$).

Solution : This motion has two parts. One is descending into the well, which is a constant acceleration motion, and the other is ascending to the sound of impact, which is a uniform motion with constant speed. 

For the first part, use the kinematic equation $y-y_0=-\frac 12 gt^{2}+v_0 t$ to find the falling time as \begin{gather*}y-y_0=-\frac 12 gt^{2}+v_0 t \\\\ -34-0=-\frac 12 (10)t^{2}+(-7)t \\\\ \Rightarrow 5t^{2}+7t-34=0 \end{gather*} Since initial velocity is a vector and its direction is initially downward, a negative is included in front of it. The above quadratic equation has two solutions: $t_1=2\,{\rm s}$ and $t_2=-3.4\,{\rm s}$. Obviously, a negative value for time is not accepted because time in all kinematic questions must be a positive quantity.

The second part is a uniform motion as the speed of sound is constant, so we can calculate its rising time using the definition of average velocity: \begin{align*} t&=\frac{\Delta y}{v} \\\\ &=\frac{34}{340}=0.1\,{\rm s}\end{align*} Thus, the total time is obtained as $t_T=2+0.1=2.1\,{\rm s}$. 

Problem (13): A stone is dropped vertically downward from the top of a building with a height of $h$. What is its speed at the height of $\frac h2$?

Solution : Let the dropping point be the origin. Thus, in the kinematic equations, the vertical displacement must be negative, i.e., $y-y_0=-\frac h2$. Use the below equation to find the speed at the desired level \begin{align*}v_2^2-v_1^2&=-2g(y-y_0) \\\\ v_2^2 -0&=-2g(-\frac h2) \\\\ \Rightarrow v_2&=\sqrt{gh}\end{align*}

Problem (14): You throw a baseball straight up in the air and catch it 3.4 seconds later at the same place at which you threw it. How high did it go? (DIFFICULT)

Solution : First of all, choose a coordinate system with the upward direction as positive. In the absence of air resistance, when you throw an object upward, the time it takes to rise is equal to the time it takes to fall. In other words, the time of reaching the highest point is half the total time the object is in the air \[t_{rise}=t_{fall}=\frac 12 t_{tot}\] Other than the total time of flight, no other relevant information is given. Substitute the time taken to reach the highest point into $v=v_0-gt$ and solve for the initial velocity $v_0$ as below \begin{gather*} v_{top}=v_0-gt_{rise} \\\\ 0=v_0-(9.8)(1.7) \\\\ \Rightarrow v_0=16.6\,\rm m/s\end{gather*} Note that at the highest point $v_{top}=0$. 

Now, apply the equation $v^2-v_0^2=-2g\Delta y$ between the initial point and the highest point so that $\Delta y=h$ and solve for $h$. \begin{gather*} v^2_{top}-v_0^2=-2gh \\\\ (0)^2-(16.6)^2=-2(9.8)h \\\\ \Rightarrow \boxed{h=14.0\,\rm m} \end{gather*} Therefore, the ball rises as high as $14.0\,\rm m$.

Problem (15): A bullet is fired vertically upward from a height of $90\,{\rm m}$ and after $10\,{\rm s}$ reaches the ground. After $2\,{\rm s}$ from the throwing point, the bullet is how far away from the surface? ($g=9.8\,{\rm m/s^2}$)

Solution : In a free-fall problem, the vertical position of an object in an instant of time is given by the kinematic equation $y=-\frac 12 gt^{2}+v_0 t+y_0$. Let the firing point be the origin, so $y_0=0$. To complete the equation above, you must have an initial speed. 

To find the initial speed, apply the kinematic formula $y=-\frac 12 gt^{2}+v_0 t+y_0$ between the origin and striking point. \begin{gather*}y-y_0=-\frac 12 gt^{2}+v_0 t \\\\ -90-0=-\frac 12 (9.8)(10)^{2}+v_0(10) \\\\ \Rightarrow \boxed{v_0=40\,\rm m/s}\end{gather*} Notice that the striking point is $-90\,{\rm m}$ below the origin that's why the negative is entered for the vertical displacement $y$.  

Having the initial velocity, substitute it into the above equation again but with time $t=2\,{\rm s}$ to find the position of the bullet after $2\,{\rm s}$ with respect to the firing point \begin{align*}y-y_0&=-\frac 12 gt^{2}+v_0 t\\\\ &=-\frac 12\,(9.8)(2)^{2}+(40)(2) \\\\ \Rightarrow y&=\boxed{60.4\,\rm m} \end{align*}

Problem (16): A ball is thrown vertically upward with an initial velocity of $18\,\rm m/s$. How many seconds after throwing, the ball's speed is $9\,\rm m/s$ downward?

Solution : In all kinematic equations, $x,y,v,v_0$, and $a$ are vectors, so their signs matter. A speed of $9\,{\rm m/s}$ downward means a velocity of $-9\,{\rm m/s}$. Downward or upward indicates the direction of velocity. 

Take up as the positive $y$ direction. Now use the equation $v=v_0-gt$ to find the velocity at any later time.\begin{gather*}v=v_0-gt\\-9=+18-(10)t\\ \Rightarrow \quad \boxed{t=2.7\,\rm s}\end{gather*}

Problem (17): An object is thrown vertically upward in the air from a $100\,{\rm m}$ height with an initial velocity of $v_0$. After $5\,{\rm s}$, it reaches the ground. Determine the magnitude and direction of the initial velocity.

Solution : Let the origin be the throwing point (so $y_0=0$) and the upward direction positive. Substitute the given total time into the vertical displacement kinematic equation $y-y_0=-\frac 12 gt^2+v_0t$ with $y-y_0=-100\,{\rm m}$ (since the impact point is below the origin). \begin{gather*} y-y_0 =-\frac 12 gt^2+v_0t \\\\ -100=-\frac 12\,(10)(5)^2+v_0 (5) \\\\ \Rightarrow \boxed{v_0=-5\,\rm m/s}\end{gather*} the minus sign indicates the initial velocity is downward with the magnitude (speed) of $5\,\rm m/s$.

Problem (18): From a height of $15\,{\rm m}$, a ball is kicked vertically up into the air with an initial speed of $v_0$. It reaches the highest point of its path with an elevation of $20\,{\rm m}$ from the surface. Find the initial velocity $v_0$. 

Solution : The highest point is $5\,{\rm m}$ above the kicking point. Apply the time-independent kinematic equation below to find the initial velocity \begin{gather*}v^2-v_0^2=-2g(y-y_0) \\\\ 0-v_0^2=-2(10)(5) \\\\ \Rightarrow v_0=\pm 10\,{\rm m/s}\end{gather*} Because the ball kicked upward so we must choose the plus sign, i.e. $v_0=+10\,{\rm m/s}$. In the above, we used the fact that in all freely falling problems, at the highest point (apex) the velocity is zero ($v=0$).

Recall that projectiles are a particular type of freely falling motion with a launch angle of $\theta=90$ with its own formulas.

Problem (19): From the bottom of a $25\,{\rm m}$-depth well, a stone is thrown vertically upward with an initial speed of $30\,{\rm m/s}$.  (a) How high does the stone rise out of the well?  (b) Before the stone returns into the well, how many seconds are outside the well?

Solution :  (a) Let the bottom of the well be the origin, so $y_0=0$. First, we find how much distance the ball rises. Recall that the highest point is where $v=0$, so we have \begin{gather*}v^2-v_0^2=-2g(y-y_0) \\\\ 0-(30)^2=-2(10)(y-0) \\\\ \Rightarrow \boxed{y=45\,\rm m}\end{gather*} Of this height, $25\,{\rm m}$ is for the well's depth, so the stone is $20\,{\rm m}$ outside of the well. 

(b) We want to examine the duration between exiting and reentering the stone into the well. During this time interval, the ball returns to its initial position, so its displacement vector is zero, i.e., $\Delta y=y-y_0=0$. If we want to use the equation, $y-y_0=-\frac 12 gt^2+v_0t$, the speed of the stone is required exactly when it leaves the well. 

The speed at the bottom of the well is known. Thus, apply the equation $v^2-v_0^2=-2g\Delta y$ to find the speed just before it leaves the well. \begin{gather*}v^2-v_0^2=-2g\Delta y \\\\ v^2-(30)^{2}=-2(10)(25) \\\\ \Rightarrow v=+20\,{\rm m/s}\end{gather*} This speed can be used as the initial speed for the part where the stone is outside the well. Hence, the total time the stone is out of the well is obtained as below \begin{gather*} \Delta y=-\frac 12 gt^{2}+v_0 t\\ 0=-\frac 12 (10)t^{2}+(20)(2) \end{gather*} Solving for $t$, one can obtain the required time as $t=4\,{\rm s}$.

Problem (20): From the top of a $20-{\rm m}$-high tower, a small ball is thrown vertically upward. If $4\,{\rm s}$ after throwing, it hit the ground. How many seconds before striking the surface does the ball again meet the original throwing point? (Air resistance is neglected and $g=10\,{\rm m/s^2}$). 

Solution : Let the origin be the throwing point. The ball strikes the ground $20\,\rm m$ below our chosen origin, so its total displacement between initial position $y_i=0$ and final position is $\Delta y=y_f-y_i=-20\,\rm m$. The total time in which the ball is in the air is also $4\,{\rm s}$. With these known values, one can find the initial velocity as \begin{gather*}\Delta y=-\frac 12 gt^{2}+v_0t \\\\ -25=-\frac 12 (10)(4)^{2}+v_0(4) \\\\ \Rightarrow \boxed{v_0=15\,\rm m/s} \end{gather*} When the ball returns to its initial position, its total displacement is zero, i.e., $\Delta y=0$ so we can use the following kinematic equation to find the total time it takes the ball to return to the starting point \begin{gather*}\Delta y=-\frac 12 gt^{2}+v_0t \\\\ 0=-\frac 12 (10)t^{2}+(15)t \end{gather*} Rearranging and solving for $t$, we get $t=3\,{\rm s}$.

Problem (21): A rock is thrown vertically upward in the air. It reaches the height of $40\,{\rm m}$ from the surface at times $t_1=2\,{\rm s}$ and $t_2$. Find $t_2$ and determine the greatest height reached by the rock (neglect air resistance and assume $g=10\,{\rm m/s^2}$).

Solution : Let the throwing point (surface of the ground) be the origin. Between our chosen origin and the point with known values $h=4\,{\rm m}$, $t=2\,{\rm s}$ one can write down the kinematic equation $\Delta y=-\frac 12 gt^{2}+v_0\,t$ to find the initial velocity as \begin{gather*} \Delta y=-\frac 12 gt^{2}+v_0t \\\\ 40=-\frac 12 (10)(2)^2 +v_0(2) \\\\ \Rightarrow v_0=30\,{\rm m/s}\end{gather*} Now we are going to find the times when the rock reaches the height $40\,{\rm m}$ (Recall that when an object is thrown upward, it passes through every point twice). Applying the same equation above, we get \begin{gather*} \Delta y=-\frac 12 gt^{2}+v_0t \\\\ 40=-\frac 12\,(10)t^2+30t \\\\ \Rightarrow \boxed{5t^2-30t+40=0} \end{gather*} Rearranging and solving for $t$ using quadratic equation formula, two times are obtained i.e. $t_1=2\,{\rm s}$ and $t_2=4\,{\rm s}$. Thus, again after $4\,\rm s$ the ball again is at the same height of $40\,\rm m$ from the surface. 

The greatest height is where the vertical velocity becomes zero so we have \begin{gather*}v^2-v_0^2 =-2g\Delta y \\\\ 0-(30)^2=-2(10)\Delta y\\\\ \Rightarrow \boxed{\Delta y=45\,\rm m}\end{gather*} Thus, the highest point where the rock can reach is located at $H=45\,{\rm m}$ above the ground.

Problem (22): A ball is launched with an initial velocity of $30\,{\rm m/s}$ straight upward. How long will it take the ball to reach $20\,{\rm m}$ below the highest point for the first time? (neglect air resistance and assume $g=10\,{\rm m/s^2}$).

Solution : Between the origin (surface level) and the highest point ($v=0$) apply the time-independent kinematic equation below to find the greatest height $H$ where the ball reaches.\begin{gather*}v^2-v_0^2=-2g\Delta y \\\\ 0-(30)^2=-2(10)H \\\\ \Rightarrow \boxed{H=45\,\rm m}\end{gather*} This is the maximum height that the ball can reach. The $20\,{\rm m}$ below this maximum height $H$ has a height of $h=45-20=25\,{\rm m}$. Now use the vertical displacement kinematic equation between the throwing point and the desired position to find the required time taken. \begin{gather*} \Delta y=-\frac 12 gt^2+v_0t \\\\ 25=-\frac 12(10)t^2+30(t)\end{gather*} Solving for $t$ (using quadratic formula), we get $t_1=1\,{\rm s}$ and $t_2=5\,{\rm s}$ one for up way and the second for down way.  

Problem (23): A stone is launched directly upward from the surface level with an initial velocity of $20\,{\rm m/s}$. How many seconds after launch is the stone's velocity $5\,{\rm m/s}$ downward?

Solution : Let the origin be at the surface level and take the positive direction up. Therefore, we have initial velocity $v_0=+20\,{\rm m/s}$ and final velocity $v=-5\,{\rm m/s}$. Use the velocity kinematic equation $v=v_0-gt$ to find the desired time as below \begin{gather*}v=v_0-gt \\-5=+20-10\times t \\ \Rightarrow \boxed{t=2.5\,\rm s}\end{gather*}

Problem (24): From a $25-{\rm m}$ building, a ball is thrown vertically upward at an initial velocity of $20\,{\rm m/s}$. How long will it take the ball to hit the ground?

Solution : Origin is considered to be at the throwing point, so $y_0=0$. Apply the position kinematic equation below to find the desired time \begin{gather*} y-y_0=-\frac 12 gt^2+v_0 t \\\\ -25=-\frac 12(10)t^2+20t \\\\ \Rightarrow 5t^2-20t-25=0 \end{gather*} Rearranging and converting it into the standard form of a quadratic equation $at^2+bt+c=0$, its solutions are obtained as \begin{align*}t_{1,2}&=\frac{-b\pm \sqrt{b^2-4\,ac}}{2a} \\\\ &=\frac{-(-4)\pm\sqrt{(-4)^2-4(1)(-5)}}{2(1)}\\\\ &=-1 \, \text{and} \, 5 \end{align*} Therefore, the time needed for the ball to hit the ground is $5\,{\rm s}$.

Problem (25): From the top of a building with a height of $60\,{\rm m}$, a rock is thrown directly upward at an initial velocity of $20\,{\rm m/s}$. What is the rock's velocity at the instant of hitting the ground? 

Solution : Apply the time-independent kinematic equation as \begin{gather*} v^2-v_0^2 =-2g(y-y_0) \\\\ v^2-(20)^2 =-2(10)(-60) \\\\ v^2 =1600\\\\ \Rightarrow \quad \boxed{v=\pm 40\,\rm m/s} \end{gather*} Therefore, the rock's velocity when it hit the ground is $v=-40\,{\rm m/s}$.

In this article, you learned how to solve free fall problems step-by-step using simple kinematic equations. 

Author : Dr. Ali Nemati Published : 8/11/2022  

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Look at the given example below and try to understand what I tried to explain above.

Kinematics Exams and Solutions

Mechanical Engineering

Kinematic equations and free fall.

As mentioned in Lesson 5, a free-falling object is an object that is falling under the sole influence of gravity. That is to say that any object that is moving and being acted upon only be the force of gravity is said to be “in a state of  free fall .” Such an object will experience a downward acceleration of 9.8 m/s/s. Whether the object is falling downward or rising upward towards its peak, if it is under the sole influence of gravity, then its acceleration value is 9.8 m/s/s.

Like any moving object, the motion of an object in free fall can be described by four kinematic equations. The kinematic equations that describe any object’s motion are:

The symbols in the above equation have a specific meaning: the symbol  d  stands for the  displacement ; the symbol  t  stands for the  time ; the symbol  a  stands for the  acceleration  of the object; the symbol  v i  stands for the  initial velocity  value; and the symbol  v f  stands for the  final velocity .  

Applying Free Fall Concepts to Problem-Solving

There are a few conceptual characteristics of free fall motion that will be of value when using the equations to analyze free fall motion. These concepts are described as follows:

·         An object in free fall experiences an acceleration of -9.8 m/s/s. (The – sign indicates a downward acceleration.) Whether explicitly stated or not, the value of the acceleration in the kinematic equations is -9.8 m/s/s for any freely falling object.

·         If an object is merely dropped (as opposed to being thrown) from an elevated height, then the initial velocity of the object is 0 m/s.

·         If an object is projected upwards in a perfectly vertical direction, then it will slow down as it rises upward. The instant at which it reaches the peak of its trajectory, its velocity is 0 m/s. This value can be used as one of the motion parameters in the kinematic equations; for example, the final velocity ( v f ) after traveling to the peak would be assigned a value of 0 m/s.

·         If an object is projected upwards in a perfectly vertical direction, then the velocity at which it is projected is equal in magnitude and opposite in sign to the velocity that it has when it returns to the same height. That is, a ball projected vertically with an upward velocity of +30 m/s will have a downward velocity of -30 m/s when it returns to the same height.

These four principles and the four kinematic equations can be combined to solve problems involving the motion of free falling objects. The two examples below illustrate application of free fall principles to kinematic problem-solving. In each example, the problem solving strategy that was introduced earlier in this lesson will be utilized.  

Example Problem A

Luke Autbeloe drops a pile of roof shingles from the top of a roof located 8.52 meters above the ground. Determine the time required for the shingles to reach the ground.

The solution to this problem begins by the construction of an informative diagram of the physical situation. This is shown below. The second step involves the identification and listing of known information in variable form. You might note that in the statement of the problem, there is only one piece of numerical information explicitly stated: 8.52 meters. The displacement (d) of the shingles is -8.52 m. (The – sign indicates that the displacement is downward). The remaining information must be extracted from the problem statement based upon your understanding of the above principles. For example, the v i  value can be inferred to be 0 m/s since the shingles are dropped (released from rest; see note above). And the acceleration (a) of the shingles can be inferred to be -9.8 m/s 2  since the shingles are free-falling (see note above). (Always pay careful attention to the + and – signs for the given quantities.) The next step of the solution involves the listing of the unknown (or desired) information in variable form. In this case, the problem requests information about the time of fall. So t is the unknown quantity. The results of the first three steps are shown in the table below.

The next step involves identifying a kinematic equation that allows you to determine the unknown quantity. There are four kinematic equations to choose from. In general, you will always choose the equation that contains the three known and the one unknown variable. In this specific case, the three known variables and the one unknown variable are d, v i , a, and t. Thus, you will look for an equation that has these four variables listed in it. An inspection of the four equations above reveals that the equation on the top left contains all four variables.

d = v i   • t + ½ • a • t 2

Once the equation is identified and written down, the next step involves substituting known values into the equation and using proper algebraic steps to solve for the unknown information. This step is shown below.

-8.52 m = (0 m/s) • (t) + ½ • (-9.8 m/s 2 ) • (t) 2

-8.52 m = (0 m) *(t) + (-4.9 m/s 2 ) • (t) 2

-8.52 m = (-4.9 m/s 2 ) • (t) 2

(-8.52 m)/(-4.9 m/s 2 ) = t 2

1.739 s 2  = t 2

The solution above reveals that the shingles will fall for a time of 1.32 seconds before hitting the ground. (Note that this value is rounded to the third digit.)

The last step of the problem-solving strategy involves checking the answer to assure that it is both reasonable and accurate. The value seems reasonable enough. The shingles are falling a distance of approximately 10 yards (1 meter is pretty close to 1 yard); it seems that an answer between 1 and 2 seconds would be highly reasonable. The calculated time easily falls within this range of reasonability. Checking for accuracy involves substituting the calculated value back into the equation for time and insuring that the left side of the equation is equal to the right side of the equation. Indeed it is!  

Example Problem B

Rex Things throws his mother’s crystal vase vertically upwards with an initial velocity of 26.2 m/s. Determine the height to which the vase will rise above its initial height.

Once more, the solution to this problem begins by the construction of an informative diagram of the physical situation. This is shown below. The second step involves the identification and listing of known information in variable form. You might note that in the statement of the problem, there is only one piece of numerical information explicitly stated: 26.2 m/s. The initial velocity (v i ) of the vase is +26.2 m/s. (The + sign indicates that the initial velocity is an upwards velocity). The remaining information must be extracted from the problem statement based upon your understanding of the above principles. Note that the v f value can be inferred to be 0 m/s since the final state of the vase is the peak of its trajectory (see note above). The acceleration (a) of the vase is -9.8 m/s 2  (see note above). The next step involves the listing of the unknown (or desired) information in variable form. In this case, the problem requests information about the displacement of the vase (the height to which it rises above its starting height). So d is the unknown information. The results of the first three steps are shown in the table below.

The next step involves identifying a kinematic equation that would allow you to determine the unknown quantity. There are four kinematic equations to choose from. Again, you will always search for an equation that contains the three known variables and the one unknown variable. In this specific case, the three known variables and the one unknown variable are v i , v f , a, and d. An inspection of the four equations above reveals that the equation on the top right contains all four variables.

v f 2   = v i 2   + 2 • a • d

(0 m/s) 2  = (26.2 m/s) 2  + 2 •(-9.8m/s 2 ) •d

0 m 2 /s 2  = 686.44 m 2 /s 2  + (-19.6 m/s 2 ) •d

(-19.6 m/s 2 ) • d = 0 m 2 /s 2  -686.44 m 2 /s 2

(-19.6 m/s 2 ) • d = -686.44 m 2 /s 2

d = (-686.44 m 2 /s 2 )/ (-19.6 m/s 2 )

The solution above reveals that the vase will travel upwards for a displacement of 35.0 meters before reaching its peak. (Note that this value is rounded to the third digit.)

The last step of the problem-solving strategy involves checking the answer to assure that it is both reasonable and accurate. The value seems reasonable enough. The vase is thrown with a speed of approximately 50 mi/hr (merely approximate 1 m/s to be equivalent to 2 mi/hr). Such a throw will never make it further than one football field in height (approximately 100 m), yet will surely make it past the 10-yard line (approximately 10 meters). The calculated answer certainly falls within this range of reasonability. Checking for accuracy involves substituting the calculated value back into the equation for displacement and insuring that the left side of the equation is equal to the right side of the equation. Indeed it is!

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Free Fall - Complete Toolkit

• To know the meaning of free fall and the conditions under which it occurs.
• To know the value for the acceleration of gravity (g) and to relate this value to velocity-time information.
• o relate the motion of a free-falling object to the graphical descriptions of its motion.
• To use equations to calculate how fast and how far a free-falling object will move in a given amount of time.
• To understand that the acceleration of gravity is independent of the mass of the free-falling object.

Readings from The Physics Classroom Tutorial

• The Physics Classroom Tutorial, 1-D Kinematics Chapter, Lesson 5 http://www.physicsclassroom.com/class/1DKin/Lesson-5/Introduction

Interactive Simulations

Video and Animations

Labs and Investigations

• The Physics Classroom, The Laboratory, Free Fall Students use a motion detector to determine the acceleration of a falling object.  
• The Physics Classroom, The Laboratory, Dune Buggy Challenge This challenging lab project requires that students combine information about free falling motion and linear motion in order to calculate when a marble must be released in order for it to land is a toy car that is moving along the table below. Link:  http://www.physicsclassroom.com/lab#1dk

Minds On Physics Internet Modules:

• Newton's Laws, Ass’t NL10 -  Free Fall Acceleration

Concept Building Exercises:

• The Curriculum Corner, 1-D Kinematics, Free Fall

Problem-Solving Exercises:

• The Calculator Pad, 1-D Kinematics, Problems #25 - #28

Science Reasoning Activities:

• Science Reasoning Center, 1-D Kinematics, Kinematics

Real Life Connections:

Common Misconception:

• Acceleration of Gravity and Mass Many students believe that a more massive object will free fall faster than a less massive object. While this dependence upon mass is relevant to situations in which objects fall under the influence of air resistance, it does not apply to free-falling objects. All objects, regardless of mass, will free fall at a rate of 9.8 m/s/s.   

Elsewhere on the Web:

• HS-PS2-1 :     Analyze data to support the claim that Newton’s second law of motion describes the mathematical relationship among the new force on a macroscopic object, its mass, and its acceleration.  (Clarification statement:  Examples of data could include tables or graphs of position or velocity as a function of time for objects subject to a net unbalanced force, such as a falling object.)
• MS-PS2.B.iii      Forces that act at a distance (electric, magnetic, and gravitational) can be explained by fields that extend through space and can be mapped by their effect on a test object (a charged object or a ball, respectively).    Limited alignment
• HS-PS2.A.i        Newton’s second law accurately predicts changes in the motion of macroscopic objects.  
• Plan and conduct an investigation individually and collaboratively to produce data to serve as the basis for evidence … and consider limitations on the precision of the data  
• Create and/or revise a computational model or simulation of a phenomenon, designed device, process, or system.
• Use mathematical and/or algorithmic representations of phenomena or design solutions to describe and/or support claims and explanations.
• Use a computational representation of phenomena to describe and/or support claims and/or explanations
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• MP.2     Reason abstractly and quantitatively
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• F-IF.8.b      Use the properties of exponents to interpret expressions for exponential functions.
• N-VM.1      Recognize vector quantities as having both magnitude and direction. Represent vector quantities by directed line segments, and use appropriate symbols for vectors and their magnitudes.
• N-VM.3     Solve problems involving velocity and other quantities that can be represented by vectors.
• RST.11-12.2     Determine the central ideas or conclusions of a text; summarize complex concepts, processes, or information presented in a text by paraphrasing them in simpler but still accurate terms.
• RST.11-12.3     Follow precisely a complex multistep procedure when carrying out experiments, taking measurements, or performing technical tasks; analyze the specific results based on explanations in the text.
• RST.11-12.4     Determine the meaning of symbols, key terms, and other domain-specific words and phrases as they are used in a specific scientific or technical context relevant to grades 11—12 texts and topics.
• RST.11-12.9     Synthesize information from a range of sources (e.g., texts, experiments, simulations) into a coherent understanding of a process, phenomenon, or concept, resolving conflicting information when possible.
• RST.11-12.10   -- By the end of grade 12, read and comprehend science/technical texts in the grades 11-CCR text complexity band independently and proficiently.
• High School H.3.1.5     When the change in an object’s instantaneous velocity is the same in each successive unit time interval, the object has constant acceleration. For straight-line motion, constant acceleration can be represented by and calculated from a linear instantaneous velocity vs. time graph, a motion diagram, and the mathematical expression [a = (vf - vi)/(tf - ti)]. The sign (+ or -) of the constant acceleration indicates the direction of the change-of-velocity vector. A negative sign does not necessarily mean that the object is traveling in the negative direction or that it is slowing down. (Boundary: The term “deceleration” should be avoided because students tend to associate a negative sign of acceleration only with slowing down.)
• High School H.3.1.7       When the acceleration is constant, the magnitude of the average velocity during a time interval is one-half of the sum of the initial and final instantaneous velocities  [v = (vf + vi)/2].
• High School H.3.4.4      When people are in free fall, they feel “weightless” because people do not feel the extremely small gravitational force on each atom in their bodies. When standing, people feel the (normal) force of the ground pushing upwards on their feet, which produces the sensation of weight. 
• High School H.3.4.3      When an object’s distance from Earth’s surface is small compared to Earth’s radius, then a simplifying assumption is that the gravitational force on an object depends only on the mass of the object. In this case, objects fall with approximately the same acceleration: 9.8 m/sec/sec.  
• Explain what is “constant” when an object is moving with a constant acceleration, the two ways in which an object that has a positive constant acceleration can be moving (speeding up or slowing down) and the two ways in which an object that has a negative constant acceleration can be moving (speeding up or slowing down). Justify the explanations by constructing sketches of motion diagrams and using the shape of instantaneous velocity versus time graphs.
• Translate between different representations of the motion of objects: verbal and/or written descriptions, motion diagrams, data tables, graphical representations (position vs. time graphs and instantaneous velocity vs. time graphs) and mathematical representations.
• Evaluate the evidence for claims about the velocity or acceleration of objects in different experimental problems, using the criteria: (a) appropriate match of the evidence to the question or prediction; (b) adequate precision and accuracy; (c) correctness or data analysis and representation procedures (e.g., error bars on graphs for best estimate of slopes; and (d) the investigation was replicated (by other groups or classes).   Partial alignment
• Explain why all objects near Earth’s surface fall with approximately the same acceleration, despite having different masses and weights. Justify by using the universal law of gravitation and Newton’s Second Law.

• Physics Formulas

Free Fall Formula

Freefall as the term says, is a body falling freely because of the gravitational pull of our earth.

Imagine a body with velocity (v) is falling freely from a height (h) for time (t) seconds because of gravity (g).

Free Fall Formulas  are articulated as follows:

h = (1/2) gt 2

Freefall Related Solved Examples

Underneath are given questions on free fall which may be useful for you.

Problem 1:  Calculate  the  body  height  if  it  has  a  mass  of  2  kg  and  after  7 seconds  it  reaches  the  ground?

Given: Height h =? Time t = 7s We all are acquainted with the fact that free fall is independent of mass.

Hence, it is given as

h = 0.5 × 9.8 × (7) 2

h = 240.1 m

Problem  2: The cotton falls after 3 s and iron falls after 5 s. Which is moving with higher velocity? Answer:

The Velocity in free fall is autonomous of mass.

V (Velocity of iron) = gt = 9.8 m/s 2  × 5s = 49 m/s

V (Velocity of cotton) = gt = 9.8 m/s 2  × 3s = 29.4 m/s.

The Velocity of iron is more than cotton.

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