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High School Physics : Linear Motion

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Example Question #35 : Understanding Motion In Two Dimensions

Laurence throws a 0.5kg rock off the edge of a 12m tall building at an angle of $20^{\circ}$ from the horizontal with an initial speed of $31\frac{m}{s}$ .

$g=-9.8\frac{m}{s^2}$ .

How high will the rock go?

Possible Answers:

2.56m

12m

17.73m

11.46m

5.73m

Correct answer:

17.73m

Explanation:

Even though the problem gives us an initial velocity, we need to break it down into horizontal and vertical components.

Use the sine function with the initial velocity and angle to find the vertical velocity.

$v_{i}*\sin(\theta)=v_{iy}$

$31\frac{m}{s}*\sin(20^{\circ})=v_{iy}$

$31\frac{m}{s}*(0.34)=v_{iy}$

$10.6\frac{m}{s}=v_{iy}$

Remember that the vertical velocity at the highest point of a parabola is zero. We know the initial and final velocities, and the acceleration of gravity. Using these values in the appropriate motion equation, we can find the height that the rock travels.

$v_f^2=v_i^2+2a\Delta y$

$(0\frac{m}{s})^2=(10.6\frac{m}{s})^2+2(-9.8\frac{m}{s^2})\Delta y$

$(0\frac{m}{s})=(112.36\frac{m}{s})+(-19.6\frac{m}{s^2})\Delta y$

$-(112.36\frac{m}{s})=(-19.6\frac{m}{s^2})\Delta y$

$\frac{-112.36\frac{m}{s}}{-19.6\frac{m}{s^2}}=\Delta y$

$5.73m=\Delta y$

Remember, $\Delta y$ only tells us the CHANGE in the vertical direction. Since the rock started at the top of a 12m building, it rose an extra 5.73m after it was thrown. Its highest point it is 17.73m above the ground.

12m+5.73m=17.73m

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Example Question #36 : Understanding Motion In Two Dimensions

Laurence throws a 0.5kg rock off the edge of a 12m tall building at an angle of $20^{\circ}$ from the horizontal with an initial speed of $31\frac{m}{s}$ .

$g=-9.8\frac{m}{s^2}$ .

How long is the rock in the air?

Possible Answers:

1.12s

9.81s

4.71s

20.8s

2.98s

Correct answer:

2.98s

Explanation:

The given velocity won't help us much here. We need to break it down into horizontal and vertical components.

Use the sine function with the initial velocity and angle to find the vertical velocity.

$v_{i}*\sin(\theta)=v_{iy}$

$31\frac{m}{s}*\sin(20^{\circ})=v_{iy}$

$31\frac{m}{s}* (0.34)=v_{iy}$

$10.6\frac{m}{s}=v_{iy}$

Now we need to work on finding the time in the air. To do this, we need to break the rock's path into two parts. The first part is the time that the rock is rising to its maximum height, and the second part is the time that it is falling from the highest point to the ground.

For the first part, we can assume that the final vertical velocity is zero, since this will be the top of the parabola. Using the initial velocity, final velocity, and gravity, we can solve for the time to travel this portion of the path.

v_f=v_i+at_1

$0\frac{m}{s}=10.6\frac{m}{s}+(-9.8\frac{m}{s^2})t_1$

$-10.6\frac{m}{s}=-9.8\frac{m}{s^2}\cdot t_1$

$\frac{-10.6\frac{m}{s}}{-9.8\frac{m}{s^2}}=t_1$

1.08s=t_1 .

This is the time that the rock is traveling upward. Now we need to focus on the time that the rock travels downward. We don't know the final velocity at the end of the parabola, so we can't use the equation from the first part. If we can find the total height of the parabola, we can use a different equation to solve for time.

$\Delta y =v_it+\frac{1}{2}at^2$

Remember that the vertical velocity at the highest point of a parabola is zero. Now that we know the initial vertical velocity, we can plug this into an equation to solve for the distance that the rock travels from its maximum height to the original height.

$v_f^2=v_i^2+2a\Delta y$

$(0\frac{m}{s})^2=(10.6\frac{m}{s})^2+2(-9.8\frac{m}{s^2})\Delta y$

$(0\frac{m}{s})=(112.36\frac{m}{s})+(-19.6\frac{m}{s^2})\Delta y$

$-(112.36\frac{m}{s})=(-19.6\frac{m}{s^2})\Delta y$

$\frac{-112.36\frac{m}{s}}{-19.6\frac{m}{s^2}}=\Delta y$

$5.73m=\Delta y$

Remember, $\Delta y$ only tells us the CHANGE in height. Since the rock started at the top of a 12m tall building, if it rose an extra 5.73m , then at its highest point it is 17.73m above the ground.

This means that our $\Delta y$ for the second equation will be -17.73m (the change in height is negative because it travels downward.) Use this total distance and the velocity at the top of the peak (zero) to solve for the time that the rock travels down.

$\Delta y =v_it+\frac{1}{2}at^2$

$-17.73m=0\frac{m}{s}t+\frac{1}{2}(-9.8\frac{m}{s^2})t^2$

$-17.73m=\frac{1}{2}(-9.8\frac{m}{s^2})t^2$

$-17.73m=(-4.9\frac{m}{s^2})t^2$

$\frac{-17.73m}{-4.9\frac{m}{s^2}}=t^2$

3.62s^2=t^2

$\sqrt{3.62s^2}=\sqrt{t^2}$

1.90s=t_2

Finally, add the two times together to find the total time in flight.

$t_{total}=t_1+t_2$

$t_{total}=1.08s+1.90s$

$t_{total}=2.98s$

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Example Question #37 : Understanding Motion In Two Dimensions

Laurence throws a 0.5kg rock off the edge of a 12m tall building at an angle of $20^{\circ}$ from the horizontal with an initial speed of $31\frac{m}{s}$ .

$g=-9.8\frac{m}{s^2}$ .

What is the total horizontal distance that the rock travels?

Possible Answers:

86.81m

52.71m

101.2m

160.2m

21.45m

Correct answer:

86.81m

Explanation:

To find the horizontal distance, we will need to use the horizontal velocity and the total time that the rock is in the air. We can easily find the horizontal velocity by using the cosine function and the initial velocity and angle.

$v_{i}* \cos(\theta)=v_{x}$

$31\frac{m}{s}* \cos(20^{\circ})=v_{x}$

$31\frac{m}{s}*(0.94)=v_{x}$

$29.13\frac{m}{s}=v_{x}$

There is no horizontal acceleration, so the velocity will be constant. This means our equation for distance will be simply:

$\Delta x =v_xt$ .

We know v_x , but now we need to find the time the rock is in the air.

Use the sine function with the initial velocity and angle to find the vertical velocity.

$v_{i}*\sin(\theta)=v_{iy}$

$31\frac{m}{s}*\sin(20^{\circ})=v_{iy}$

$31\frac{m}{s}* (0.34)=v_{iy}$

$10.6\frac{m}{s}=v_{iy}$

We need to break the rock's path into two parts. The first part is the time that the rock is rising to its maximum height, and the second part is the time that it is falling from the highest point to the ground.

v_f=v_i+at_1

$0\frac{m}{s}=10.6\frac{m}{s}+(-9.8\frac{m}{s^2})t_1$

$-10.6\frac{m}{s}=-9.8\frac{m}{s^2}\cdot t_1$

$\frac{-10.6\frac{m}{s}}{-9.8\frac{m}{s^2}}=t_1$

1.08s=t_1 .

$\Delta y =v_it+\frac{1}{2}at^2$

$v_f^2=v_i^2+2a\Delta y$

$(0\frac{m}{s})^2=(10.6\frac{m}{s})^2+2(-9.8\frac{m}{s^2})\Delta y$

$(0\frac{m}{s})=(112.36\frac{m}{s})+(-19.6\frac{m}{s^2})\Delta y$

$-(112.36\frac{m}{s})=(-19.6\frac{m}{s^2})\Delta y$

$\frac{-112.36\frac{m}{s}}{-19.6\frac{m}{s^2}}=\Delta y$

$5.73m=\Delta y$

Remember, $\Delta y$ only tells us the CHANGE in height. Since the rock started at the top of a 12m tall building, if it rose an extra 5.73m , then at its highest point it is 17.73m above the ground.

$\Delta y =v_it+\frac{1}{2}at^2$

$-17.73m=0\frac{m}{s}t+\frac{1}{2}(-9.8\frac{m}{s^2})t^2$

$-17.73m=\frac{1}{2}(-9.8\frac{m}{s^2})t^2$

$-17.73m=(-4.9\frac{m}{s^2})t^2$

$\frac{-17.73m}{-4.9\frac{m}{s^2}}=t^2$

3.62s^2=t^2

$\sqrt{3.62s^2}=\sqrt{t^2}$

1.90s=t_2

Finally, add the two times together to find the total time in flight.

$t_{total}=t_1+t_2$

$t_{total}=1.08s+1.90s$

$t_{total}=2.98s$

Now that we have finally found our time, we can plug that back into the equation from the beginning of the problem:

$\Delta x =v_xt$

Use the time and horizontal velocity to find the total horizontal distance.

$\Delta x =29.13\frac{m}{s}*2.98s$

$\Delta x=86.81m$

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Example Question #38 : Understanding Motion In Two Dimensions

Laurence throws a 0.5kg rock off the edge of a 12m tall building at an angle of $20^{\circ}$ from the horizontal with an initial speed of $31\frac{m}{s}$ .

$g=-9.8\frac{m}{s^2}$ .

What is the total final velocity of the rock right as it hits the ground?

Possible Answers:

$13.23\frac{m}{s}$

$68.21\frac{m}{s}$

$3.46\frac{m}{s}$

$19.69\frac{m}{s}$

$34.58\frac{m}{s}$

Correct answer:

$34.58\frac{m}{s}$

Explanation:

First, break our given velocity into its horizontal and vertical components.

We can find the horizontal velocity using the cosine function, initial velocity, and angle.

$v_{i}*\cos(\theta)=v_{x}$

$31\frac{m}{s}*\cos(20^{\circ})=v_{x}$

$31\frac{m}{s}*(0.94)=v_{x}$

$29.13\frac{m}{s}=v_{x}$

We can find the vertical velocity using the sine function, initial velocity, and angle.

$v_{i}* \sin(\theta)=v_{iy}$

$31\frac{m}{s}*\sin(20^{\circ})=v_{iy}$

$31\frac{m}{s}* (0.34)=v_{iy}$

$10.6\frac{m}{s}=v_{iy}$

The horizontal velocity is not going to change, since there is no horizontal acceleration. The vertical velocity, however, will be affected by gravity.

We know that the rock is going to travel a net displacement of -12m , as that's the distance between the rock's initial position (the top of the building) and the ground. The displacement must be negative because the rock is traveling downward. Using the appropriate motion equation, we can use the rock's initial velocity, displacement, and acceleration to find the final velocity.

$v_f^2=v_i^2+2a\Delta y$

$v_f^2=(10.6\frac{m}{s})^2+2(-9.8\frac{m}{s^2})(-12m)$

$v_f^2=(112.36\frac{m^2}{s^2})+235.2\frac{m^2}{s^2}$

$v_f^2=347.56\frac{m^2}{s^2}$

$\sqrt{v_f^2}=\sqrt{347.56\frac{m^2}{s^2}}$

$v_f=18.64\frac{m}{s}$

Keep in mind that square roots can be positive or negative. Since the rock is traveling in the downward direction, we need to take the negative root.

$v_f=-18.64\frac{m}{s}$ .

We now have both the final horizontal and final vertical velocities. Use the Pythagorean Theorem to find the total velocity from the directional velocities.

$(-18.64\frac{m}{s})^2+(29.13\frac{m}{s})^2=(v_{R})^2$

$347.56\frac{m^2}{s^2}+848.56\frac{m^2}{s^2}=(v_{R})^2$

$1196.2\frac{m^2}{s^2}=(v_{R})^2$

$\sqrt{1196.2\frac{m^2}{s^2}}=\sqrt{(v_{R})^2}$

$34.58\frac{m}{s}=v_{R}$

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Example Question #71 : Linear Motion

Laurence throws a 0.5kg rock off the edge of a 12m tall building at an angle of $20^{\circ}$ from the horizontal with an initial speed of $31\frac{m}{s}$ .

$g=-9.8\frac{m}{s^2}$ .

What is the angle that the rock will make with the ground right before it hits the ground?

Possible Answers:

$61.08^{\circ}$

$32.62^{\circ}$

$0.32^{\circ}$

$13.62^{\circ}$

$12.45^{\circ}$

Correct answer:

$32.62^{\circ}$

Explanation:

To find the angle, we will need to find the final horizontal and vertical velocities. First, break our given velocity into its horizontal and vertical components.

We can find the horizontal velocity using the cosine function, initial velocity, and angle.

$v_{i}*\cos(\theta)=v_{x}$

$31\frac{m}{s}*\cos(20^{\circ})=v_{x}$

$31\frac{m}{s}*(0.94)=v_{x}$

$29.13\frac{m}{s}=v_{x}$

We can find the vertical velocity using the sine function, initial velocity, and angle.

$v_{i}* \sin(\theta)=v_{iy}$

$31\frac{m}{s}*\sin(20^{\circ})=v_{iy}$

$31\frac{m}{s}* (0.34)=v_{iy}$

$10.6\frac{m}{s}=v_{iy}$

The horizontal velocity is not going to change, since there is no horizontal acceleration. The vertical velocity, however, will be affected by gravity.

$v_f^2=v_i^2+2a\Delta y$

$v_f^2=(10.6\frac{m}{s})^2+2(-9.8\frac{m}{s^2})(-12m)$

$v_f^2=(112.36\frac{m^2}{s^2})+235.2\frac{m^2}{s^2}$

$v_f^2=347.56\frac{m^2}{s^2}$

$\sqrt{v_f^2}=\sqrt{347.56\frac{m^2}{s^2}}$

$v_f=18.64\frac{m}{s}$

Now that we have the final directional velocities, we can use the tangent function to find the angle with which the rock impacts the ground.

$\tan(\theta)=\frac{\text{opposite}}{\text{adjacent}}$

$\tan({\theta})=\frac{v_y}{v_x}$ .

$\theta=\tan^{-1}(\frac{v_y}{v_x})$

$\theta=\tan^{-1}(\frac{18.64}{29.13})$

$\theta=\tan^{-1}(0.64)$

$\theta=32.62^{\circ}$

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Example Question #1 : Using Motion Equations

A potted plant falls off a window sill from 3 meters above the ground. How long does it take before the plant hits the ground?

$\small g=-9.8\frac{m}{s^2}$

Possible Answers:

0.782s

0.656 s

0.818 s

0.723 s

1.2 s

Correct answer:

0.782s

Explanation:

$\Delta y = V_i(t)+\frac{1}{2}at^2$

Using the equation above and the given values, we can solve for the time. First we need to find the change in distance.

$\Delta y=y_f-y_i=0m-3m=-3m$

The plant travels $\small -3$ meters in the downward direction. Now we can plug in the given values from the question to solve for time.

$-3m= (0\frac{m}{s})+\frac{1}{2}(-9.8\frac{m}{s^2})t^2$

$-3m = (-4.9\frac{m}{s^2})t^2$

$\frac {-3m}{-4.9\frac{m}{s^2}}=t^2$

0.612s^2=t^2

$\sqrt{0.612s^2}=\sqrt{t^2}$

0.782s=t

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Example Question #2 : Using Motion Equations

Walter throws a disc from 1.5 meters above the ground with purely horizontal motion. How long will it be before the disc hits the ground?

$\small g=-9.8\frac{m}{s^2}$

Possible Answers:

0.343 s

0.002 s

0.053 s

0.553 s

1.03 s

Correct answer:

0.553 s

Explanation:

The horizontal motion will not affect the time the disc is in the air. Time will be determined by the rate at which the disc falls: the acceleration due to gravity.

We can solve for the time using the equation below and the values given in the question.

$\Delta y=V_i(t)+\frac{1}{2}at^2$

$(0m-1.5m)=(0\frac{m}{s})(t)+\frac{1}{2}(-9.8\frac{m}{s^2})t^2$

$(-1.5m)=\frac{1}{2}(-9.8\frac{m}{s^2})t^2\boldsymbol{}$

$(-1.5m)=(-4.9\frac{m}{s^2})t^2$

$\frac{-1.5m}{-4.9\frac{m}{s^2}}=t^2$

0.306s^2=t^2

$\sqrt{0.306s^2}=\sqrt{t^2}$

0.553s=t

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Example Question #3 : Using Motion Equations

Leslie rolls a ball out of a window from 10 meters above the ground, such that the initial y-velocity is zero. What will the ball's final velocity be right before it hits the ground?

$\small g=-9.8\frac{m}{s^2}$

Possible Answers:

$14\frac{m}{s}$

$28\frac{m}{s}$

$9\frac{m}{s}$

$12\frac{m}{s}$

$13.2\frac{m}{s}$

Correct answer:

$14\frac{m}{s}$

Explanation:

We are given the initial velocity, acceleration, and the initial and final distances.

$\Delta y=y_f-y_i=0m-10m=-10m$

Using the equation below and the given values, we can solve for the final velocity.

$V_f^2=V_i^2+2a\Delta y$

$V_f^2=(0\frac{m}{s})^2+2(-9.8\frac{m}{s^2})(-10m)$

$V_f^2=196\frac{m^2}{s^2}$

$\sqrt{V_f^2}=\sqrt{196\frac{m^2}{s^2}}$

$V_f=14\frac{m}{s}$

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Example Question #4 : Using Motion Equations

A 2.9kg crate slides across a floor for before coming to rest 10m from its original position.

What is the initial velocity of the crate?

Possible Answers:

$2\frac{m}{s}$

$50\frac{m}{s}$

$4\frac{m}{s}$

$8\frac{m}{s}$

$1\frac{m}{s}$

Correct answer:

$4\frac{m}{s}$

Explanation:

The problem gives us distance, final velocity, and change in time. We can use these values in the equation below to solve for the initial velocity.

$\Delta x=\frac{(v_f+v_i)}{2}\Delta t$

Plug in our given values and solve.

$10m=\frac{(0\frac{m}{s}+v_i)}{2}5s$

Divide both sides by $\small 5$ .

$\frac{10m}{5s}=\frac{v_i}{2}$

$2\frac{m}{s}=\frac{v_i}{2}$

Multiply both sides by $\small 2$ .

$4\frac{m}{s}=v_i$

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Example Question #5 : Using Motion Equations

A 2.9kg crate slides across a floor for before coming to rest 10m from its original position.

What is the net acceleration on the crate?

Possible Answers:

$-1\frac{m}{s^2}$

$-10\frac{m}{s}$

$-0.8\frac{m}{s^2}$

$-4\frac{m}{s}$

$-2.32\frac{m}{s^2}$

Correct answer:

$-0.8\frac{m}{s^2}$

Explanation:

All of the equations regarding acceleration require an initial velocity. We will need to solve for the initial velocity using the given variables. The problem gives us distance, final velocity, and change in time. We can use these values in the equation below to solve for the initial velocity.

$\Delta x=\frac{(v_f+v_i)}{2}\Delta t$

Plug in our given values and solve.

$10m=\frac{(0\frac{m}{s}+v_i)}{2}5s$

$\frac{10m}{5s}=\frac{v_i}{2}$

$2\frac{m}{s}=\frac{v_i}{2}$

$4\frac{m}{s}=v_i$

We can use a linear motion equation to solve for the acceleration, using the velocity we just found. We now have the distance, time, and initial velocity.

$\Delta x=v_it+\frac{1}{2}at^2$

Plug in the given values to solve for acceleration.

$10m=(4\frac{m}{s}) (5s)+\frac{1}{2}a(5s)^2$

$10m=(20m)+\frac{1}{2}a(25s^2)$

$-10m=\frac{1}{2}a(25s^2)$

-20m=a(25s^2)

$\frac{-20m}{25s^2}=a$

$-0.8\frac{m}{s^2}=a$

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High School Physics : Linear Motion

Study concepts, example questions & explanations for High School Physics

All High School Physics Resources

Example Questions

Example Question #35 : Understanding Motion In Two Dimensions

Example Question #36 : Understanding Motion In Two Dimensions

Example Question #37 : Understanding Motion In Two Dimensions

Example Question #38 : Understanding Motion In Two Dimensions

Example Question #71 : Linear Motion

Example Question #1 : Using Motion Equations

Example Question #2 : Using Motion Equations

Example Question #3 : Using Motion Equations

Example Question #4 : Using Motion Equations

Example Question #5 : Using Motion Equations

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