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Applying Free Fall Concepts to Problem 2

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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 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 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 (vf) 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..

Example Problem A

Luke drops a pile of roof shingles from the top of a roof located 8 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 meters. The displacement (d) of the shingles is -8. 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 vi value can be inferred to be 0 m/s since the shingles are dropped (released from rest). And the acceleration (a) of the shingles can be inferred to be -9 m/s 2 since the shingles are free-falling). (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.

Diagram: Given: Fi vi = 0 m/s d = -8 m

a = - 9 m/s 2

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, vi, 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 = vi • 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.

height). So d is the unknown information. The results of the first three steps are shown in the table below.

Diagram: Given: Find:

vi = 26 m/s vf = 0 m/s

a = -9 m/s 2

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 vi, vf, a, and d. An inspection of the four equations above reveals that the equation on the top right contains all four variables.

vf 2 = vi 2 + 2 • a • d

(0 m/s) 2 = (26 m/s) 2 + 2 •(-9/s 2 ) •d

0 m 2 /s 2 = 686 m 2 /s 2 + (-19 m/s 2 ) •d

(-19 m/s 2 ) • d = 0 m 2 /s 2 -686 m 2 /s 2

(-19 m/s 2 ) • d = -686 m 2 /s 2

d = (-686 m 2 /s 2 )/ (-19 m/s 2 )

d = 35. 0 m

The solution above reveals that the vase will travel upwards for a displacement of 35. 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!

Kinematic equations provide a useful means of determining the value of an unknown motion parameter if three motion parameters are known. In the case of a free-fall motion, the acceleration is often known. And in many cases, another motion parameter can be inferred through a solid knowledge of some basic kinematic principles.

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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 Motion: Explanation, Review, and Examples

• The Albert Team
• Last Updated On: February 16, 2023

Free fall and projectile motion describe objects that are moving through the air and acted on only by gravity. In this post, we will describe this type of motion using both graphs and kinematic equations. Since projectile motion involves two dimensions, these problems can be complex. We will explain many examples so you can see how to solve different types of projectile motion. 

What We Review

An object that is moving under only the influence of gravity is in free fall. In order for an object to be in free fall, wind and air resistance must be ignored. On Earth, all objects in free fall accelerate downward at the rate of gravity or 9.81\text{ m/s}^2 .

Applying Free Fall to Kinematic Equations

When analyzing free fall motion, we can apply the same kinematic equations as we did for motion on the ground. We can then use these equations to determine properties such as distance, time, and velocity. 

How to Find Distance Fallen for an Object in Free Fall

If an object is in free fall, we can use kinematic equations to find the distance it falls during a certain time. You will typically use the following kinematic equation to calculate the distance fallen:

In order to use this equation, you need to know the initial velocity of the object and the time of flight. Remember that the acceleration of a free falling object is always equal to the acceleration due to gravity, 9.81\text{ m/s}^2 . 

Many free fall physics problems will include scenarios where objects are dropped from rest. In this case, the initial velocity is zero and the first term of the kinematic equation above will cancel out. 

If the time is not known, another method for calculating the distance fallen is to use the following kinematic equation:

In this case, you must know the final velocity v_f of the object. Then, you can solve the equation for the distance d .

How to Find Time for an Object in Free Fall

The amount of time an object is in free fall will depend on its velocity and the distance it falls. Similar to distance, there are two equations you can use to find the time, depending on what you know. 

If you know the initial and final velocity of the object, then the simplest way to calculate time is using the kinematic equation:

This equation can be solved for time. Then, you’ll only need to substitute the values for the velocities and the acceleration due to gravity.

Another method to find time if you do not know the object’s final velocity is to use the equation:

Note that in this equation there are two terms that include the time t . Unless the initial velocity is zero, this can make it more challenging to solve this equation for time. If using this equation, you may need to use the quadratic formula to solve for time.

How to Find Final Velocity for an Object in Free Fall

The final velocity of an object in free fall depends on the amount of time it falls. Due to the acceleration of gravity, the velocity will increase every second by 9.81\text{ m/s} . The final velocity can be calculated using the equation:

If you do not know the amount of time the object is falling, another method for calculating the final velocity is using the kinematic equation: 

This equation requires that you instead know the distance that the object falls. If you are using this equation to find the final velocity, remember that the final velocity is squared in this equation. That means you will need to take a square root as your final step to solve for the final velocity. 

Examples of Free Fall

In this next section, we’ll apply the methods you just learned to solve some problems about free fall motion.

Example 1: How to Find the Distance for an Object Dropped from Rest

For example, an object is dropped from rest from the top of a tall building. It hits the ground 5\text{ s} after it is dropped. What is the height of the building? 

In this scenario, we know that the object’s initial velocity is zero because it was dropped from rest. We also know that the acceleration is 9.81\text{ m/s}^2 . This problem is asking us to find the distance the object falls. This will be equal to the height of the building.

Based on this information, we can use the following kinematic equation to find the distance:

Substituting the given values produces:

Therefore, the height of the building is about 123\text{ m} .

Example 2: How to Find the Final Velocity for an Object with Initial Velocity

In another example, an object in free fall has an initial, downward velocity of 2\text{ m/s} and falls a distance of 45\text{ m} . What is the object’s final velocity? 

In this scenario, we are given the object’s initial velocity, v_i and the distance d . We also know that the acceleration is 9.81\text{ m/s}^2 . Based on this information, we can use the following kinematic equation to find the final velocity:

Since the initial velocity is in the same direction as the acceleration (downward) we can use the same sign for both values.

Our last step is to eliminate the square by taking the square root:

Therefore, the final velocity of the object is about 30\text{ m/s} .

Motion Graphs for Objects in Free Fall

In addition to using physics equations, we can also represent free fall motion with motion graphs. Position-time graphs, velocity-time graphs, and acceleration-time graphs can tell us a lot about the object’s motion over time. Want a more in-depth review of motion graphs? Check out this blog post !

Position-Time Graph for an Object in Free Fall

In terms of position, many objects in free fall start at a high position, or height off the ground, and move downward. Objects in free fall accelerate due to gravity. Therefore, the position-time graph for free fall motion must be curved. This means that objects in free fall start with a slow velocity and gradually speed up which is represented by the steep downward curve of the graph. 

Velocity-Time Graph for an Object in Free Fall

As an object falls, its velocity increases due to the acceleration of gravity. This means that the velocity starts slow and steadily increases in the downward direction. The graph below shows the velocity-time for an object in free fall:

Note that the slope of this graph is constant and represents the acceleration due to gravity, or -9.81\text{ m/s}^2 .

Acceleration-Time Graph for an Object in Free Fall

Free fall acceleration is constant. Throughout the entire time that an object is falling, it is accelerating at a rate equal to the acceleration due to gravity, -9.81\text{ m/s}^2 . As shown in the graph below, the acceleration-time graph is a constant negative line. 

Projectile Motion

A projectile is an object that is launched or thrown into the air and then only influenced by gravity. Projectile motion has many similarities to free fall motion, however, projectiles may also travel a horizontal distance in addition to falling vertically down. 

Examples of Projectile Motion

The exact trajectory, or path, a projectile will take depends on how it is launched. However, all projectiles follow a curved trajectory such as in the image shown below:

If you play or watch sports, you likely have already observed projectile motion. Projectile motion describes the arc of a basketball in a free throw, a fly ball in baseball, or a volleyball bumped over the net. 

Horizontal Component of Velocity

To analyze projectile motion, we must separate the motion into horizontal and vertical components. The horizontal component of a projectile’s velocity is independent of the vertical component of velocity. Since gravity acts vertically, there are no horizontal forces acting on projectiles. This means that the horizontal component of a projectile’s velocity remains constant throughout the entire flight. 

Example: Finding the Horizontal Component

For example, a projectile is launched from the ground with an initial speed of 8\text{ m/s} at a 60^{\circ} angle. What is the horizontal component of the projectile’s velocity?

We will need to use trig identities to determine the components of the velocity. We can visualize the components as a triangle where the hypotenuse is the initial velocity and the sides represent the horizontal, v_{ix} , and vertical, v_{iy} , components of the velocity.

Cosine is defined as the adjacent side of the triangle divided by the hypotenuse. Since the horizontal component is adjacent to the angle, we can use cosine to find the horizontal component of velocity:

Therefore, the horizontal component of the initial velocity is 4\text{ m/s} .

Need to review your trig identities? Try out this resource from Khan Academy .

Vertical Component of Velocity

The vertical component of a projectile’s velocity will be influenced by gravity, which acts vertically on the object causing it to accelerate downward. Therefore, the vertical component of velocity will change throughout the projectile’s flight. We can calculate the vertical component of velocity at a particular time in a method similar to calculating the horizontal component. 

Example: Finding the Vertical Component

In the same example, a projectile is launched from the ground with an initial speed of 8\text{ m/s} at a 60^{\circ} angle. What is the vertical component of the projectile’s velocity?

As we visualize the velocity components, we are solving this time for the opposite side of the triangle. Sine is defined as the opposite side of the triangle divided by the hypotenuse. Therefore, the initial vertical velocity is:

Solving Projectile Motion Questions

Let’s apply what we’ve learned to some examples of projectile motion!

Example 1: Finding the Range of a Projectile

In this example, a projectile is fired horizontally with a speed of 5\text{ m/s} from a cliff with a height of 60\text{ m} . How far from the base of the cliff will the projectile land? 

In this scenario, we are given the initial horizontal velocity v_{ix}=5\text{ m/s} and the vertical change in position d_y=-60\text{ m} . Since the projectile is launched horizontally, the initial vertical velocity, v_{iy} , is zero. We also always know in projectile motion that the vertical acceleration is a_y=-9.81\text{ m/s}^2 and the horizontal acceleration, a_x , is zero.

This problem is asking us to find the horizontal displacement, or d_x . This is also referred to as the range . We can use the following kinematic equation to find the projectile’s final horizontal position:

Since the horizontal acceleration of a projectile is zero, this equation can be simplified to:

Before we can solve this equation, we must first determine the time of the projectile’s flight. We can actually use this same equation in the vertical direction to solve for time:

Since the initial vertical velocity is zero, this equation can be simplified to:

Solving for t :

Substituting the given values:

Now we can use this time to calculate the horizontal displacement of the projectile:

Therefore, the projectile will land about 17.5\text{ m} from the base of the cliff. 

Example 2: Finding the Maximum Height of a Projectile

As another example, a projectile is launched from the ground with an initial velocity of 25\text{ m/s} at an angle of 50^{\circ} . What is the projectile’s maximum height?

As a projectile travels upward, its vertical velocity becomes slower and slower due to the negative acceleration of gravity. At the maximum height of the trajectory, the projectile’s vertical velocity will momentarily be zero as the projectile stops and turns to move downward. Therefore, in this scenario, our final vertical velocity, v_{fy} , is zero.

We can use the following kinematic equation to solve for the maximum height, d_y :

Solving for d_y :

Before we can use this equation to calculate the height, we will need to use the sine trig identity to find the vertical component of the initial velocity:

Since the initial velocity is in the opposite direction as the acceleration, it’s really important to remember the sign here. If we define moving up as positive, then the initial velocity is positive and the acceleration is negative. Substituting this initial vertical velocity and the given values into the equation above gives:

Therefore, the projectile will reach a maximum height of about 18.7\text{ m} .

For more examples and an explanation of solving these types of projectile motion problems, check out this youtube video from Professor Dave . 

Understanding free fall and projectile motion allows you to solve some of the most complex problems you will encounter in introductory physics. All projectiles are acted on only by gravity, and the vertical and horizontal components of motion are independent of each other. This allows us to apply our kinematic equations to solve for a projectile’s time of flight, velocity, and displacement in each direction.  

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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
• MP.1     Make sense of problems and persevere in solving them.
• MP.2     Reason abstractly and quantitatively
• A-CED.4   Rearrange formulas to highlight a quantity of interest, using the same reasoning as in solving equations.
• F-IF.4      For a function that models a relationship between two quantities, interpret key features of graphs and tables in terms of the quantities, and sketch graphs showing key features given a verbal description of the relationship.
• 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.

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Question: 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

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