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1.2 Representations of Motion

13 min readdecember 22, 2022

Peter Apps

Peter Apps

Daniella Garcia-Loos

Daniella Garcia-Loos

Peter Apps

Peter Apps

Daniella Garcia-Loos

Daniella Garcia-Loos

Attend a live cram event

Review all units live with expert teachers & students

Understanding Representations of Motion

In AP Physics 1, we study different representations of motion to understand and analyze the movement of objects. Some common types of representations include:

  1. Graphical representations: These include , , and . These graphs can be used to represent the motion of an object and to understand its characteristics, such as its speed and .

  2. Numerical representations: These include tables or lists of that describe the motion of an object. These data may include position, , and at different times.

  3. Analytical representations: These include mathematical equations that describe the motion of an object. These equations may involve variables such as position, , and , and can be used to make predictions about the motion of an object.

  4. Diagrammatic representations: These include sketches or diagrams that show the position, , and of an object at different times. These representations can help visualize and understand the motion of an object.

Overall, different representations of motion can be used to help understand the characteristics and behavior of moving objects, and can be useful tools for predicting and analyzing the motion of objects in different situations.

Center of Mass

The of the of a system is related to the net force exerted on the system, where a = F/m.

Key Vocabulary: - a point on an object or system that is the mean position of the matter.

⟶ A force may be applied to this point to cause a linear without angular occurring.

4.A.1 Essential Knowledge

The linear motion of a system can be described by the , , and of its .

4.A.2 Essential Knowledge

The is equal to the rate of change of with time, and is equal to the rate of change of position with time.

Graphical Representations of Motion

As we covered in 1.1 Position, and there are ways to represent all three quantities graphically. We also reviewed how to interpret these graphs,  but it is imperative to understand the relationships each graph has to one another. 

https://firebasestorage.googleapis.com/v0/b/fiveable-92889.appspot.com/o/images%2F-dauK8tAyoB0M.JPG?alt=media&token=0672f287-0935-42fc-bcc3-7c7b3862a481

Image Courtesy of geogebra

From the image above, we can use position to find for any given period of time by looking at the slope of the Position vs. Time Graph. Working from to position we can look at the area underneath the curve to find , but it is not possible to determine how far from the detector the object is located from a vs. Time Graph.  

When working from to we look at the slope of the vs. Time Graph. Similarly, when working from to we look at the area under the curve to find

Linearization

Many graphs in physics will not be perfectly straight lines, but we can turn the curve into a straight sloped line! We accomplish this by squaring the x-axis value. This is called Linearization. This is a concept that will be used throughout the course, so get comfortable with it now!

https://firebasestorage.googleapis.com/v0/b/fiveable-92889.appspot.com/o/images%2FCapture.PNG?alt=media&token=3afc8550-857b-4fda-9d7b-34aeb3109163

Image Courtesy of x-engineer.org

Linearization of a graph is a method of approximating the behavior of a nonlinear function with a straight line. This can be useful for making predictions or understanding the general trend of the function. Here are some key points to consider when linearizing a graph:

  • A linear function is defined by a straight line, which can be represented by the equation y = mx + b, where m is the slope and b is the y-intercept.

  • A nonlinear function is any function that is not a straight line. Nonlinear functions can take many different shapes, including curves, loops, and jumps.

  • To linearize a nonlinear function, we need to find a straight line that is a good approximation of the function. This can be done by finding two points on the nonlinear function and drawing a straight line through them.

  • The equation of the tangent line can be found using the point-slope form of a linear equation: y - y1 = m(x - x1), where (x1, y1) is a point on the function and m is the slope of the tangent line at that point.

  • The accuracy of the linear approximation depends on how closely the straight line follows the shape of the nonlinear function. The closer the line is to the function, the more accurate the approximation will be.

  • Linearization can be useful in physics because many physical systems can be modeled with nonlinear functions. By linearizing these functions, we can make predictions and understand the general trend of the system's behavior.

Example Problem

A ball is dropped from a height of 10 meters, and we want to find the time it takes to hit the ground. We know that the equation for the position of the ball as a function of time is:

y = 10 - 4.9t^2

However, this is a nonlinear equation and it is difficult to solve for t. Instead, we can use linearization to find an approximate solution.

To linearize the equation, we need to find the equation of the tangent line at a specific point on the curve. Let's choose the point where t = 1 second. At this point, the position of the ball is:

y = 10 - 4.9(1^2) = 5.1

Now we need to find the equation of the tangent line at this point. We can do this by using the point-slope form of a linear equation: y - y1 = m(x - x1), where (x1, y1) is the point on the curve and m is the slope of the tangent line at that point.

Since the point on the curve is (1, 5.1), we can set y1 = 5.1 and x1 = 1. To find the slope of the tangent line, we can plot a second point on the curve that is close to (1, 5.1) and find the slope between the two points. For example, we can choose the point where t = 1.1 seconds:

y = 10 - 4.9(1.1^2) = 4.61

The slope between the two points is (4.61 - 5.1)/(1.1 - 1) = -0.49.

Now we can plug the values into the point-slope form of the linear equation to find the equation of the tangent line:

y - 5.1 = -0.49(x - 1)

y = -0.49x + 5.6

We can use this linear equation to approximate the position of the ball as it falls. For example, if we want to find the position of the ball after 0.5 seconds, we can substitute t = 0.5 into the linear equation to get:

y = -0.49*0.5 + 5.6 = 5.355

This means that the ball has fallen about 5.355 meters after 0.5 seconds.

This is just an approximate solution, but it is a much simpler calculation than solving the nonlinear equation for the position of the ball. Linearization can be useful for finding approximate solutions to problems involving nonlinear functions, especially when the nonlinear function is difficult to work with or when we only need an approximate solution.

⟶ Still feeling a little confused on Linearization? Don’t worry! Check out this video from AP Physics 1 Online for more practice!

Mathematical Representations of Motion

In Kinematics there are four major equations you must understand to begin calculations. They relate , , initial and final , and time together

https://firebasestorage.googleapis.com/v0/b/fiveable-92889.appspot.com/o/images%2F-EkG9qd5ZdaaO.png?alt=media&token=a5e79975-bc15-4fac-9057-d9f42ac5a3fb

Variable Interpretation: Δx is horizontal in meters, Vf is final in meters/second, Vo is initial in meters/second, t is time in seconds, and a is in m/s/s. 

⟶ In order to solve for a variable without having all four other quantities known, we look at the ‘Variable Missing’ column to pick the equation that best suits our question. 

EXAMPLE:

A super car races by at a speed of 68 m/s and slows down as a rate of 4 m/s/s. 

How much runway is needed to stop the plane?

  • We are given the variables a, Vo, and Vf but we are missing Δx and t

    • Vf^2 =Vo^2 + 2aΔx

  • STEP 1: Cross out Vf because the car will stop at a of 0 m/s

  • STEP 2: Plug in the known variables: 0 = (68 m/s^2) + 2 (-4 m/s/s^2) (Δx)

Your final answer should be Δx = 578 m

Free Fall

Key Vocabulary: - an object only under the influence of gravity

Equation: = force of gravity x time

Key Vocabulary: due to Gravity - 9.8 m/s/s (it is acceptable to round up to 10 m/s/s on the AP Physics 1 exam)

Equation

Formula

Variable Missing

Big Four #2

Vf = Vo + gt

Δy

Big Four #3

y = Vot + 1/2 gt2

Vf

Big Four #4

Vf2 = Vo2 + 2gy

t

Variable Interpretation: Δy is vertical in meters, Vf is final in meters/second, Vo is initial in meters/second, t is time in seconds, and g is due to gravity in m/s/s. 

⟶ In equations, we now replace Δx with Δy and a with g giving us a modified list of The Big Four as seen in the table above. 

Object Dropped (trip down)

  • Vo (initial ) = 0 m/s

  • g works in the direction of motion

Object Tossed (trip up)

  • Vf (final ) = 0 m/s

    • At maximum height of its trip, an object has a of 0 m/s

  • g works against the direction of motion

EXAMPLE #1:

A ball is dropped from the top of a building. It falls 2.8s. 

What is the of the ball?

  • We are given the variables g, t, and Vo, but we are missing Δy and Vf

    • Δy = Vot + ½ gt2

  • STEP 1: Cross out Vo because an object that is dropped has an initial of 0 m/s

  • STEP 2: Plug in the known variables: Δy = ½ (10 m/s/s) (2.8 s2)

Your final answer should be Δy = 39.2 m

What is the ball’s final ?

  • We are given the variables g, t, and Vo, but we are missing Δy and Vf

    • Vf = Vo + gt

  • STEP 1: Cross out Vo because an object that is dropped has an initial of 0 m/s

  • STEP 2: Plug in the known variables: Vf = (10 m/s/s) (2.8 s)

Your final answer should be Vf = 28 m/s

EXAMPLE #2:

A soccer ball is thrown straight up with an initial of 20 m/s.

What height did the soccer ball reach?

  • We are given the variables g, Vo, and Vf, but we are missing Δy and t

    • Vf2 = Vo2 + 2gΔy

  • STEP 1: Cross out Vf because an object thrown up has a final of 0 m/s

  • STEP 2: Plug in known variables: 0 = (20 m/s2) + 2(10 m/s/s) (Δy)

Your final answer should be Δy = 20 m

Projectile Motion

Projectile motion is the motion of an object that is thrown, launched, or projected into the air and is then subject only to the force of gravity. Here are some key points to consider when solving projectile motion problems:

  • In projectile motion, the object moves in two dimensions: horizontally and vertically.

    • The horizontal motion of a projectile is uniform, which means it moves at a constant speed in a straight line. This is because there are no forces acting on the projectile in the horizontal direction.

    • The vertical motion of a projectile is affected by the force of gravity, which pulls the projectile downward. The force of gravity causes the projectile to accelerate downward at a rate of 9.8 m/s^2.

  • The path of a projectile is called its , and it is a parabolic curve.

  • The initial of a projectile is a vector that specifies both the magnitude and direction of the projectile's motion. It can be broken down into its horizontal and vertical components.

  • The range of a projectile is the horizontal distance it travels from the starting point to where it lands.

  • The maximum height of a projectile is the highest point it reaches during its motion.

  • The time of flight of a projectile is the total time it takes to complete its motion from the starting point to the landing point.

  • The angle at which a projectile is launched has an effect on its range, maximum height, and time of flight.

  • Air resistance is ignored

When dealing with horizontal projectile launches, launches that have an entirely horizontal initial (ex. a ball rolling off a table), we break it up into two sets of equations: vertical and horizontal. 

Formula

Type

Variable Missing

y = Voyt + 1/2 gt2

Vertical

Vfy

Vfy = Voy + gt

Vertical

Δy

Vfy2 = Voy2 + 2gy

Vertical

t

x = Vxt

Horizontal

N/A

Variable Interpretation: Δy is vertical in meters, Δx is horizontal in meters, Vfy is vertical final in meters/second, Voy is vertical initial in meters/second, Vx is horizontal in m/s, t is time in seconds, and g is due to gravity in m/s/s. 

As shown in the chart above, there are three equations we use for the vertical component of launches and one for the horizontal. You cannot put an x-component into a formula without a y-component!

EXAMPLE:

A tennis ball is rolling on a ledge with a of 5 m/s. If the tennis ball rolls off the table, which is 1.5m high:

(a) How long would it take to hit the ground?

  • We are given variables Δy, Voy and Vx, but we are missing t, Δx, and Vfy

    • Δy = Voyt + ½ gt2

  • STEP 1: Cross out Voy because there is no initial vertical

  • STEP 2: Plug in known variables: 1.5m = ½ (10 m/s/s) t2

Your final answer should be 0.55 seconds

(b) What will be the distance it travels before reaching the ground?

  • We are given variables Δy, Voy and Vx, but we are missing Δx and Vfy

    • Δx  = Vxt

  • STEP 1: Plug in known variables: Δx  = (5 m/s) (0.55 s)

Your final answer should be 2.75 meters

(c) What is the magnitude of the right before the tennis ball reaches the ground?

  • We are given variables Δy, Voy and Vx, but we are missing Vfy 

    • Vfy = Voy + gt OR Vfy2 = Voy2 + 2gΔy

  • STEP 1: Plug in known variables: Vfy = (10 m/s/s) (0.55 s) OR Vfy2 = 2 (10 m/s/s) (1.5 m)

Your final answer should be 5.5 m/s

⟶ Still feeling a little confused on Projectile Launches? Don’t worry! Check out this live stream from Fiveable for more practice!

Angled Motion

Key Vocabulary: - launches at an angle that includes both a horizontal and vertical component of initial

Key Vocabulary: - the horizontal and vertical parts of a vector

require you to find the Vox and Voy, or , based on the initial Vo and the angle Ө. Now you can solve for the following: Vo, Vox, Voy, Ө, t, X, Ymax, and Vf. 

  • If an object is shooting upward, Voy is (+)

  • If an object is shooting downward, Voy is (-)

  • Vertical is 0 m/s at the top

  • Flight is symmetric if the projectile starts and ends at the same height

Angled Launch Formulas

cos(Vo) = Vox/Ө

sin(Vo) = Voy/Ө

Vox = Vocos(Ө)

Voy = Vosin(Ө)

t = 2(Voy)/g(This only applies if the starting and ending heights are the same)

Variable Interpretation: Voy is vertical initial in meters/second, Vox is initial horizontal in m/s, t is time in seconds, t is time in seconds, and g is due to gravity in m/s/s. 

EXAMPLE:

  • A cannonball is shot at a 30 degree angle above the horizontal at 20 m/s.

(a) How much time will the cannonball travel for in the air?

  • We are given variables Ө and Vo, but we are missing t, Vox, and Voy

    • t = 2(Voy)/g AND Voy = Vosin(Ө)

  • STEP 1: Solve for Voy: Voy = 20sin(30), Voy = 10 m/s

  • STEP 2: Plug in known variables: t = 2(10 m/s)/(10 m/s/s)

Your final answer should be 2 seconds

(b) How far will the cannonball travel?

  • We are given variables Ө and Vo, but we are missing Vox

    • Δx  = Vxt

  • STEP 1: Solve for Vx: Vox = 20cos(30), Vox = 17.3 m/s

  • STEP 2: Plug in known variables: Δx  = (17.3 m/s)(2 s)

Your final answer should be 34.6 meters

(c) What is the maximum height the cannonball will reach?

  • We are given variables Ө, Vfy, and Vo, but we are missing Vox

    • Vfy2 = Voy2 + 2gΔy 

  • STEP 1: Plug in known variables: 0 = (10 m/s)2 + 2(10 m/s/s)Δy

Your final answer should be 5 meters

⟶ Still feeling a little confused on ? Don’t worry! Check out this video from Khan Academy for more practice!  Want more practice - Check out the Fiveable Live streams on this topic:

Key Terms to Review (15)

Acceleration

: Acceleration refers to the rate at which an object's velocity changes over time. It can be positive (speeding up), negative (slowing down), or zero (constant speed).

Acceleration-Time Graphs

: Acceleration-time graphs show the relationship between an object's acceleration and time. They display how an object's acceleration changes over a specific period.

Angled Launches

: Angled launches refer to the motion of an object that is projected into the air at an angle, rather than straight up or straight down. It involves both horizontal and vertical components of motion.

Center of mass

: The center of mass is the point in an object or system where its mass can be considered to be concentrated. It is the average position of all the parts of the object, weighted according to their masses.

Cosine Function (cos)

: The cosine function (cos) is a mathematical function that relates an angle within a right triangle to the ratio between the length of the adjacent side and hypotenuse. It is commonly used in physics to determine component vectors.

Displacement

: Displacement refers to the change in position of an object from its initial point to its final point, taking into account both distance and direction.

Free Fall

: Free fall occurs when an object falls under the sole influence of gravity, without any other forces acting upon it. During free fall, all objects experience acceleration due to gravity at approximately 9.8 m/s².

Initial Velocity Vector Components

: The initial velocity of an object can be broken down into two components - one in the horizontal direction (x-axis) and one in the vertical direction (y-axis).

Numerical Data

: Numerical data refers to information that is expressed in numbers. It can be collected through measurements, observations, or calculations.

Position-Time Graphs

: Position-time graphs show the relationship between an object's position and time. They display how an object's position changes over a specific period.

Sine Function (sin)

: The sine function (sin) is another mathematical function that relates an angle within a right triangle to the ratio between the length of opposite side and hypotenuse. It is commonly used in physics for analyzing periodic motion.

Trajectory

: The path followed by an object in motion under the influence of gravity. It is determined by the initial velocity and launch angle.

Vector Components

: Vector components refer to the individual parts of a vector that represent its magnitude and direction along specific axes. They are typically represented as horizontal and vertical components.

Velocity

: Velocity refers to the rate at which an object changes its position in a specific direction. It includes both speed and direction.

Velocity-Time Graphs

: Velocity-time graphs show the relationship between an object's velocity and time. They display how an object's velocity changes over a specific period.

1.2 Representations of Motion

13 min readdecember 22, 2022

Peter Apps

Peter Apps

Daniella Garcia-Loos

Daniella Garcia-Loos

Peter Apps

Peter Apps

Daniella Garcia-Loos

Daniella Garcia-Loos

Attend a live cram event

Review all units live with expert teachers & students

Understanding Representations of Motion

In AP Physics 1, we study different representations of motion to understand and analyze the movement of objects. Some common types of representations include:

  1. Graphical representations: These include , , and . These graphs can be used to represent the motion of an object and to understand its characteristics, such as its speed and .

  2. Numerical representations: These include tables or lists of that describe the motion of an object. These data may include position, , and at different times.

  3. Analytical representations: These include mathematical equations that describe the motion of an object. These equations may involve variables such as position, , and , and can be used to make predictions about the motion of an object.

  4. Diagrammatic representations: These include sketches or diagrams that show the position, , and of an object at different times. These representations can help visualize and understand the motion of an object.

Overall, different representations of motion can be used to help understand the characteristics and behavior of moving objects, and can be useful tools for predicting and analyzing the motion of objects in different situations.

Center of Mass

The of the of a system is related to the net force exerted on the system, where a = F/m.

Key Vocabulary: - a point on an object or system that is the mean position of the matter.

⟶ A force may be applied to this point to cause a linear without angular occurring.

4.A.1 Essential Knowledge

The linear motion of a system can be described by the , , and of its .

4.A.2 Essential Knowledge

The is equal to the rate of change of with time, and is equal to the rate of change of position with time.

Graphical Representations of Motion

As we covered in 1.1 Position, and there are ways to represent all three quantities graphically. We also reviewed how to interpret these graphs,  but it is imperative to understand the relationships each graph has to one another. 

https://firebasestorage.googleapis.com/v0/b/fiveable-92889.appspot.com/o/images%2F-dauK8tAyoB0M.JPG?alt=media&token=0672f287-0935-42fc-bcc3-7c7b3862a481

Image Courtesy of geogebra

From the image above, we can use position to find for any given period of time by looking at the slope of the Position vs. Time Graph. Working from to position we can look at the area underneath the curve to find , but it is not possible to determine how far from the detector the object is located from a vs. Time Graph.  

When working from to we look at the slope of the vs. Time Graph. Similarly, when working from to we look at the area under the curve to find

Linearization

Many graphs in physics will not be perfectly straight lines, but we can turn the curve into a straight sloped line! We accomplish this by squaring the x-axis value. This is called Linearization. This is a concept that will be used throughout the course, so get comfortable with it now!

https://firebasestorage.googleapis.com/v0/b/fiveable-92889.appspot.com/o/images%2FCapture.PNG?alt=media&token=3afc8550-857b-4fda-9d7b-34aeb3109163

Image Courtesy of x-engineer.org

Linearization of a graph is a method of approximating the behavior of a nonlinear function with a straight line. This can be useful for making predictions or understanding the general trend of the function. Here are some key points to consider when linearizing a graph:

  • A linear function is defined by a straight line, which can be represented by the equation y = mx + b, where m is the slope and b is the y-intercept.

  • A nonlinear function is any function that is not a straight line. Nonlinear functions can take many different shapes, including curves, loops, and jumps.

  • To linearize a nonlinear function, we need to find a straight line that is a good approximation of the function. This can be done by finding two points on the nonlinear function and drawing a straight line through them.

  • The equation of the tangent line can be found using the point-slope form of a linear equation: y - y1 = m(x - x1), where (x1, y1) is a point on the function and m is the slope of the tangent line at that point.

  • The accuracy of the linear approximation depends on how closely the straight line follows the shape of the nonlinear function. The closer the line is to the function, the more accurate the approximation will be.

  • Linearization can be useful in physics because many physical systems can be modeled with nonlinear functions. By linearizing these functions, we can make predictions and understand the general trend of the system's behavior.

Example Problem

A ball is dropped from a height of 10 meters, and we want to find the time it takes to hit the ground. We know that the equation for the position of the ball as a function of time is:

y = 10 - 4.9t^2

However, this is a nonlinear equation and it is difficult to solve for t. Instead, we can use linearization to find an approximate solution.

To linearize the equation, we need to find the equation of the tangent line at a specific point on the curve. Let's choose the point where t = 1 second. At this point, the position of the ball is:

y = 10 - 4.9(1^2) = 5.1

Now we need to find the equation of the tangent line at this point. We can do this by using the point-slope form of a linear equation: y - y1 = m(x - x1), where (x1, y1) is the point on the curve and m is the slope of the tangent line at that point.

Since the point on the curve is (1, 5.1), we can set y1 = 5.1 and x1 = 1. To find the slope of the tangent line, we can plot a second point on the curve that is close to (1, 5.1) and find the slope between the two points. For example, we can choose the point where t = 1.1 seconds:

y = 10 - 4.9(1.1^2) = 4.61

The slope between the two points is (4.61 - 5.1)/(1.1 - 1) = -0.49.

Now we can plug the values into the point-slope form of the linear equation to find the equation of the tangent line:

y - 5.1 = -0.49(x - 1)

y = -0.49x + 5.6

We can use this linear equation to approximate the position of the ball as it falls. For example, if we want to find the position of the ball after 0.5 seconds, we can substitute t = 0.5 into the linear equation to get:

y = -0.49*0.5 + 5.6 = 5.355

This means that the ball has fallen about 5.355 meters after 0.5 seconds.

This is just an approximate solution, but it is a much simpler calculation than solving the nonlinear equation for the position of the ball. Linearization can be useful for finding approximate solutions to problems involving nonlinear functions, especially when the nonlinear function is difficult to work with or when we only need an approximate solution.

⟶ Still feeling a little confused on Linearization? Don’t worry! Check out this video from AP Physics 1 Online for more practice!

Mathematical Representations of Motion

In Kinematics there are four major equations you must understand to begin calculations. They relate , , initial and final , and time together

https://firebasestorage.googleapis.com/v0/b/fiveable-92889.appspot.com/o/images%2F-EkG9qd5ZdaaO.png?alt=media&token=a5e79975-bc15-4fac-9057-d9f42ac5a3fb

Variable Interpretation: Δx is horizontal in meters, Vf is final in meters/second, Vo is initial in meters/second, t is time in seconds, and a is in m/s/s. 

⟶ In order to solve for a variable without having all four other quantities known, we look at the ‘Variable Missing’ column to pick the equation that best suits our question. 

EXAMPLE:

A super car races by at a speed of 68 m/s and slows down as a rate of 4 m/s/s. 

How much runway is needed to stop the plane?

  • We are given the variables a, Vo, and Vf but we are missing Δx and t

    • Vf^2 =Vo^2 + 2aΔx

  • STEP 1: Cross out Vf because the car will stop at a of 0 m/s

  • STEP 2: Plug in the known variables: 0 = (68 m/s^2) + 2 (-4 m/s/s^2) (Δx)

Your final answer should be Δx = 578 m

Free Fall

Key Vocabulary: - an object only under the influence of gravity

Equation: = force of gravity x time

Key Vocabulary: due to Gravity - 9.8 m/s/s (it is acceptable to round up to 10 m/s/s on the AP Physics 1 exam)

Equation

Formula

Variable Missing

Big Four #2

Vf = Vo + gt

Δy

Big Four #3

y = Vot + 1/2 gt2

Vf

Big Four #4

Vf2 = Vo2 + 2gy

t

Variable Interpretation: Δy is vertical in meters, Vf is final in meters/second, Vo is initial in meters/second, t is time in seconds, and g is due to gravity in m/s/s. 

⟶ In equations, we now replace Δx with Δy and a with g giving us a modified list of The Big Four as seen in the table above. 

Object Dropped (trip down)

  • Vo (initial ) = 0 m/s

  • g works in the direction of motion

Object Tossed (trip up)

  • Vf (final ) = 0 m/s

    • At maximum height of its trip, an object has a of 0 m/s

  • g works against the direction of motion

EXAMPLE #1:

A ball is dropped from the top of a building. It falls 2.8s. 

What is the of the ball?

  • We are given the variables g, t, and Vo, but we are missing Δy and Vf

    • Δy = Vot + ½ gt2

  • STEP 1: Cross out Vo because an object that is dropped has an initial of 0 m/s

  • STEP 2: Plug in the known variables: Δy = ½ (10 m/s/s) (2.8 s2)

Your final answer should be Δy = 39.2 m

What is the ball’s final ?

  • We are given the variables g, t, and Vo, but we are missing Δy and Vf

    • Vf = Vo + gt

  • STEP 1: Cross out Vo because an object that is dropped has an initial of 0 m/s

  • STEP 2: Plug in the known variables: Vf = (10 m/s/s) (2.8 s)

Your final answer should be Vf = 28 m/s

EXAMPLE #2:

A soccer ball is thrown straight up with an initial of 20 m/s.

What height did the soccer ball reach?

  • We are given the variables g, Vo, and Vf, but we are missing Δy and t

    • Vf2 = Vo2 + 2gΔy

  • STEP 1: Cross out Vf because an object thrown up has a final of 0 m/s

  • STEP 2: Plug in known variables: 0 = (20 m/s2) + 2(10 m/s/s) (Δy)

Your final answer should be Δy = 20 m

Projectile Motion

Projectile motion is the motion of an object that is thrown, launched, or projected into the air and is then subject only to the force of gravity. Here are some key points to consider when solving projectile motion problems:

  • In projectile motion, the object moves in two dimensions: horizontally and vertically.

    • The horizontal motion of a projectile is uniform, which means it moves at a constant speed in a straight line. This is because there are no forces acting on the projectile in the horizontal direction.

    • The vertical motion of a projectile is affected by the force of gravity, which pulls the projectile downward. The force of gravity causes the projectile to accelerate downward at a rate of 9.8 m/s^2.

  • The path of a projectile is called its , and it is a parabolic curve.

  • The initial of a projectile is a vector that specifies both the magnitude and direction of the projectile's motion. It can be broken down into its horizontal and vertical components.

  • The range of a projectile is the horizontal distance it travels from the starting point to where it lands.

  • The maximum height of a projectile is the highest point it reaches during its motion.

  • The time of flight of a projectile is the total time it takes to complete its motion from the starting point to the landing point.

  • The angle at which a projectile is launched has an effect on its range, maximum height, and time of flight.

  • Air resistance is ignored

When dealing with horizontal projectile launches, launches that have an entirely horizontal initial (ex. a ball rolling off a table), we break it up into two sets of equations: vertical and horizontal. 

Formula

Type

Variable Missing

y = Voyt + 1/2 gt2

Vertical

Vfy

Vfy = Voy + gt

Vertical

Δy

Vfy2 = Voy2 + 2gy

Vertical

t

x = Vxt

Horizontal

N/A

Variable Interpretation: Δy is vertical in meters, Δx is horizontal in meters, Vfy is vertical final in meters/second, Voy is vertical initial in meters/second, Vx is horizontal in m/s, t is time in seconds, and g is due to gravity in m/s/s. 

As shown in the chart above, there are three equations we use for the vertical component of launches and one for the horizontal. You cannot put an x-component into a formula without a y-component!

EXAMPLE:

A tennis ball is rolling on a ledge with a of 5 m/s. If the tennis ball rolls off the table, which is 1.5m high:

(a) How long would it take to hit the ground?

  • We are given variables Δy, Voy and Vx, but we are missing t, Δx, and Vfy

    • Δy = Voyt + ½ gt2

  • STEP 1: Cross out Voy because there is no initial vertical

  • STEP 2: Plug in known variables: 1.5m = ½ (10 m/s/s) t2

Your final answer should be 0.55 seconds

(b) What will be the distance it travels before reaching the ground?

  • We are given variables Δy, Voy and Vx, but we are missing Δx and Vfy

    • Δx  = Vxt

  • STEP 1: Plug in known variables: Δx  = (5 m/s) (0.55 s)

Your final answer should be 2.75 meters

(c) What is the magnitude of the right before the tennis ball reaches the ground?

  • We are given variables Δy, Voy and Vx, but we are missing Vfy 

    • Vfy = Voy + gt OR Vfy2 = Voy2 + 2gΔy

  • STEP 1: Plug in known variables: Vfy = (10 m/s/s) (0.55 s) OR Vfy2 = 2 (10 m/s/s) (1.5 m)

Your final answer should be 5.5 m/s

⟶ Still feeling a little confused on Projectile Launches? Don’t worry! Check out this live stream from Fiveable for more practice!

Angled Motion

Key Vocabulary: - launches at an angle that includes both a horizontal and vertical component of initial

Key Vocabulary: - the horizontal and vertical parts of a vector

require you to find the Vox and Voy, or , based on the initial Vo and the angle Ө. Now you can solve for the following: Vo, Vox, Voy, Ө, t, X, Ymax, and Vf. 

  • If an object is shooting upward, Voy is (+)

  • If an object is shooting downward, Voy is (-)

  • Vertical is 0 m/s at the top

  • Flight is symmetric if the projectile starts and ends at the same height

Angled Launch Formulas

cos(Vo) = Vox/Ө

sin(Vo) = Voy/Ө

Vox = Vocos(Ө)

Voy = Vosin(Ө)

t = 2(Voy)/g(This only applies if the starting and ending heights are the same)

Variable Interpretation: Voy is vertical initial in meters/second, Vox is initial horizontal in m/s, t is time in seconds, t is time in seconds, and g is due to gravity in m/s/s. 

EXAMPLE:

  • A cannonball is shot at a 30 degree angle above the horizontal at 20 m/s.

(a) How much time will the cannonball travel for in the air?

  • We are given variables Ө and Vo, but we are missing t, Vox, and Voy

    • t = 2(Voy)/g AND Voy = Vosin(Ө)

  • STEP 1: Solve for Voy: Voy = 20sin(30), Voy = 10 m/s

  • STEP 2: Plug in known variables: t = 2(10 m/s)/(10 m/s/s)

Your final answer should be 2 seconds

(b) How far will the cannonball travel?

  • We are given variables Ө and Vo, but we are missing Vox

    • Δx  = Vxt

  • STEP 1: Solve for Vx: Vox = 20cos(30), Vox = 17.3 m/s

  • STEP 2: Plug in known variables: Δx  = (17.3 m/s)(2 s)

Your final answer should be 34.6 meters

(c) What is the maximum height the cannonball will reach?

  • We are given variables Ө, Vfy, and Vo, but we are missing Vox

    • Vfy2 = Voy2 + 2gΔy 

  • STEP 1: Plug in known variables: 0 = (10 m/s)2 + 2(10 m/s/s)Δy

Your final answer should be 5 meters

⟶ Still feeling a little confused on ? Don’t worry! Check out this video from Khan Academy for more practice!  Want more practice - Check out the Fiveable Live streams on this topic:

Key Terms to Review (15)

Acceleration

: Acceleration refers to the rate at which an object's velocity changes over time. It can be positive (speeding up), negative (slowing down), or zero (constant speed).

Acceleration-Time Graphs

: Acceleration-time graphs show the relationship between an object's acceleration and time. They display how an object's acceleration changes over a specific period.

Angled Launches

: Angled launches refer to the motion of an object that is projected into the air at an angle, rather than straight up or straight down. It involves both horizontal and vertical components of motion.

Center of mass

: The center of mass is the point in an object or system where its mass can be considered to be concentrated. It is the average position of all the parts of the object, weighted according to their masses.

Cosine Function (cos)

: The cosine function (cos) is a mathematical function that relates an angle within a right triangle to the ratio between the length of the adjacent side and hypotenuse. It is commonly used in physics to determine component vectors.

Displacement

: Displacement refers to the change in position of an object from its initial point to its final point, taking into account both distance and direction.

Free Fall

: Free fall occurs when an object falls under the sole influence of gravity, without any other forces acting upon it. During free fall, all objects experience acceleration due to gravity at approximately 9.8 m/s².

Initial Velocity Vector Components

: The initial velocity of an object can be broken down into two components - one in the horizontal direction (x-axis) and one in the vertical direction (y-axis).

Numerical Data

: Numerical data refers to information that is expressed in numbers. It can be collected through measurements, observations, or calculations.

Position-Time Graphs

: Position-time graphs show the relationship between an object's position and time. They display how an object's position changes over a specific period.

Sine Function (sin)

: The sine function (sin) is another mathematical function that relates an angle within a right triangle to the ratio between the length of opposite side and hypotenuse. It is commonly used in physics for analyzing periodic motion.

Trajectory

: The path followed by an object in motion under the influence of gravity. It is determined by the initial velocity and launch angle.

Vector Components

: Vector components refer to the individual parts of a vector that represent its magnitude and direction along specific axes. They are typically represented as horizontal and vertical components.

Velocity

: Velocity refers to the rate at which an object changes its position in a specific direction. It includes both speed and direction.

Velocity-Time Graphs

: Velocity-time graphs show the relationship between an object's velocity and time. They display how an object's velocity changes over a specific period.


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© 2024 Fiveable Inc. All rights reserved.

AP® and SAT® are trademarks registered by the College Board, which is not affiliated with, and does not endorse this website.