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High School Physics : Motion and Mechanics

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Example Questions

Example Question #61 : Linear Motion

Sam throws a 0.5kg rock off the edge of a 10m tall building at an angle of $35^{\circ}$ from the horizontal. The rock has an initial speed of $30\frac{m}{s}$ .

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

At what height above the ground will the rock change direction?

Possible Answers:

25.11m

15.11m

30.11m

6.06m

5.11m

Correct answer:

25.11m

Explanation:

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

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

We can plug in the given values and find the vertical velocity.

$30\frac{m}{s}* \sin(35^{\circ})=v_{iy}$

$30\frac{m}{s}* (0.57)=v_{iy}$

$17.21\frac{m}{s}=v_{iy}$

Remember that the vertical velocity at the highest point of a parabola is zero. Now that we know the initial and final vertical velocities, we can plug our values into an equation to solve for the maximum height.

$v_{f}^2=v_i^2+2a\Delta y$

$(0\frac{m}{s})^2=(17.21\frac{m}{s})^2+2(-9.8\frac{m}{s^2})\Delta y$

$(0\frac{m}{s})=(296.09\frac{m}{s})+(-19.6\frac{m}{s^2})\Delta y$

$-(296.09\frac{m}{s})=(-19.6\frac{m}{s^2})\Delta y$

$\frac{-296.09\frac{m}{s}}{-19.6\frac{m}{s^2}}=\Delta y$

$15.11m=\Delta y$

Remember, $\Delta y$ only tells us the CHANGE in the vertical direction. The rock started at the top of a 10m tall building, then rose an extra 15.11m .

10m+15.11m=25.11m

Its highest point is 25.11m above the ground.

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Example Question #2 : Sat Subject Test In Physics

Sam throws a 0.5kg rock off the edge of a 10m tall building at an angle of $35^{\circ}$ from the horizontal. The rock has an initial speed of $30\frac{m}{s}$ .

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

How long is the rock in the air?

Possible Answers:

4.02s

5.05s

1.12s

1.76s

2.26s

Correct answer:

4.02s

Explanation:

We first need to find the vertical component of the velocity.

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

We can plug in the given values for the angle and the initial velocity to find the vertical component.

$30\frac{m}{s}* \sin(35^{\circ})=v_{iy}$

$30\frac{m}{s}* (0.57)=v_{iy}$

$17.21\frac{m}{s}=v_{iy}$

Now we need to solve for the time that the rock travels upward. We can then add the upward travel time to the downward travel time to find the total time in the air.

Remember that the vertical velocity at the highest point of a parabola is zero. We can use that to find the time for the rock to travel upward.

v_f=v_i+at_1

$0\frac{m}{s}=17.21\frac{m}{s}+(-9.8\frac{m}{s^2})t_1$

$-17.21\frac{m}{s}=(-9.8\frac{m}{s^2}) t_1$

$\frac{-17.21\frac{m}{s}}{-9.8\frac{m}{s^2}}=t_1$

1.76s=t_1

Now let's find the time for the downward travel. We don't know the final velocity for the rock, but we CAN use the information we have been given to find the height it travels upward.

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

$(0\frac{m}{s})^2=(17.21\frac{m}{s})^2+2(-9.8\frac{m}{s^2})\Delta y$

$(0\frac{m}{s})=(296.18\frac{m}{s})+(-19.6\frac{m}{s^2})\Delta y$

$-(296.18\frac{m}{s})=(-19.6\frac{m}{s^2})\Delta y$

$\frac{-296.18\frac{m}{s}}{-19.6\frac{m}{s^2}}=\Delta y$

$15.11m=\Delta y$

Remember, $\Delta y$ only tells us the vertical CHANGE. Since the rock started at the top of a 10m building, if it rose an extra 15.11m , then at its highest point it is 25.11m above the ground.

This means that our $\Delta y$ will be -25.11m as it will be traveling down from the highest point. Using this distance, we can find the downward travel time.

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

$-25.11m=0\frac{m}{s}t+\frac{1}{2}(-9.8\frac{m}{s^2})t^2$

$-25.11m=\frac{1}{2}(-9.8\frac{m}{s^2})t^2$

$-25.11m=(-4.9\frac{m}{s^2})t^2$

$\frac{-25.11m}{-4.9\frac{m}{s^2}}=t^2$

5.12s^2=t^2

$\sqrt{5.12s^2}=\sqrt{t^2}$

2.26s=t_2

Add together the time for upward travel and downward travel to find the total flight time.

$t_{total}=t_1+t_2$

$t_{total}=1.76s+2.26s$

$t_{total}=4.02s$

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Example Question #3 : Sat Subject Test In Physics

Sam throws a 0.5kg rock off the edge of a 10m tall building at an angle of $35^{\circ}$ from the horizontal. The rock has an initial speed of $30\frac{m}{s}$ .

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

What is the horizontal distance that the rock travels?

Possible Answers:

98.77m

25.11m

160.52m

24.57m

4.66m

Correct answer:

98.77m

Explanation:

We first need to find the horizontal component of the initial velocity.

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

We can plug in the given values for the angle and initial velocity and solve.

$30\frac{m}{s}* \cos(35^{\circ})=v_{x}$

$30\frac{m}{s}* (0.82)=v_{x}$

$24.57\frac{m}{s}=v_{x}$

The only force acting on the rock during flight is gravity; there are no forces in the horizontal direction, meaning that the horizontal velocity will remain constant. We can set up a simple equation to find the relationship between distance traveled and the velocity.

$\Delta x =v_xt$

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

We need to solve for the time that the rock travels upward. We can then add the upward travel time to the downward travel time to find the total time in the air.

Remember that the vertical velocity at the highest point of a parabola is zero. We can use that to find the time for the rock to travel upward.

v_f=v_i+at_1

$0\frac{m}{s}=17.21\frac{m}{s}+(-9.8\frac{m}{s^2})t_1$

$-17.21\frac{m}{s}=(-9.8\frac{m}{s^2}) t_1$

$\frac{-17.21\frac{m}{s}}{-9.8\frac{m}{s^2}}=t_1$

1.76s=t_1

Now let's find the time for the downward travel. We don't know the final velocity for the rock, but we CAN use the information we have been given to find the height it travels upward.

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

$(0\frac{m}{s})^2=(17.21\frac{m}{s})^2+2(-9.8\frac{m}{s^2})\Delta y$

$(0\frac{m}{s})=(296.18\frac{m}{s})+(-19.6\frac{m}{s^2})\Delta y$

$-(296.18\frac{m}{s})=(-19.6\frac{m}{s^2})\Delta y$

$\frac{-296.18\frac{m}{s}}{-19.6\frac{m}{s^2}}=\Delta y$

$15.11m=\Delta y$

Remember, $\Delta y$ only tells us the vertical CHANGE. Since the rock started at the top of a 10m building, if it rose an extra 15.11m , then at its highest point it is 25.11m above the ground.

This means that our $\Delta y$ will be -25.11m as it will be traveling down from the highest point. Using this distance, we can find the downward travel time.

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

$-25.11m=0\frac{m}{s}t+\frac{1}{2}(-9.8\frac{m}{s^2})t^2$

$-25.11m=\frac{1}{2}(-9.8\frac{m}{s^2})t^2$

$-25.11m=(-4.9\frac{m}{s^2})t^2$

$\frac{-25.11m}{-4.9\frac{m}{s^2}}=t^2$

5.12s^2=t^2

$\sqrt{5.12s^2}=\sqrt{t^2}$

2.26s=t_2

Add together the time for upward travel and downward travel to find the total flight time.

$t_{total}=t_1+t_2$

$t_{total}=1.76s+2.26s$

$t_{total}=4.02s$

Now that we've finally found our time, we can plug that back into the equation from the beginning of the problem, along with our horizontal velocity, to solve for the final distance.

$\Delta x =v_xt$

$\Delta x =24.57\frac{m}{s}* 4.02s$

$\Delta x=98.77m$

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Example Question #131 : High School Physics

Sam throws a 0.5kg rock off the edge of a 10m tall building at an angle of $35^{\circ}$ from the horizontal. The rock has an initial speed of $30\frac{m}{s}$ .

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

What is the net final velocity of the rock right before it hits the ground?

Possible Answers:

$14.8\frac{m}{s}$

$23.8\frac{m}{s}$

$59.2\frac{m}{s}$

$62.1\frac{m}{s}$

$33.1\frac{m}{s}$

Correct answer:

$33.1\frac{m}{s}$

Explanation:

The question gives the total initial velocity, but we will need to find the horizontal and vertical components.

To find the horizontal velocity we use the equation $v_{i}*\cos(\Theta)=v_{x}$ .

We can plug in the given values for the angle and initial velocity to solve.

$30\frac{m}{s}*\cos(35^{\circ})=v_{x}$

$30\frac{m}{s}*(0.82)=v_{x}$

$24.57\frac{m}{s}=v_{x}$

We can find the vertical velocity using the equation $v_{i}* \sin(\Theta)=v_{iy}$ .

$30\frac{m}{s}* \sin(35^{\circ})=v_{iy}$

$30\frac{m}{s}* (0.57)=v_{iy}$

$17.21\frac{m}{s}=v_{iy}$

The horizontal velocity will not change during flight because there are no forces in the horizontal direction. The vertical velocity, however, will be affected. We need to solve for the final vertical velocity, then combine the vertical and horizontal vectors to find the total final velocity.

We know that the rock is going to travel a net distance of -10m , as that is the distance between where the rock's initial and final positions. We now know the displacement, initial velocity, and acceleration, which will allow us to solve for the final velocity.

$v_f^2=v_1^2+2a\Delta y$

$v_f^2=(17.21\frac{m}{s})^2+2(-9.8\frac{m}{s^2})(-10m)$

$v_f^2=(296.18\frac{m^2}{s^2})+196\frac{m^2}{s^2}$

$v_f^2=492.18\frac{m^2}{s^2}$

$\sqrt{v_f^2}=\sqrt{492.18\frac{m^2}{s^2}}$

$v_f=22.19\frac{m}{s}$

Because the rock is traveling downward, our velocity will be negative: $v_f=-22.19\frac{m}{s}$ .

Now that we know our final velocities in both the horizontal and vertical directions, we can use the Pythagorean theorem to solve for the net velocity.

$(-22.19\frac{m}{s})^2+(24.57\frac{m}{s})^2=(v_{resultant})^2$

$492.4\frac{m^2}{s^2}+603.68\frac{m^2}{s^2}=(v_{resultant})^2$

$1096.08\frac{m^2}{s^2}=(v_{resultant})^2$

$\sqrt{1096.08\frac{m^2}{s^2}}=\sqrt{(v_{resultant})^2}$

$33.1\frac{m}{s}=v_{resultant}$

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Example Question #3 : Linear Motion

Sam throws a 0.5kg rock off the edge of a 10m tall building at an angle of $35^{\circ}$ from the horizontal. The rock has an initial speed of $30\frac{m}{s}$ .

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

At what angle to the horizontal will the rock impact the ground?

Possible Answers:

$41.99^{\circ}$

$33.5^{\circ}$

$21.1^{\circ}$

$17.1^{\circ}$

$44.3^{\circ}$

Correct answer:

$41.99^{\circ}$

Explanation:

The question gives the total initial velocity, but we will need to find the horizontal and vertical components.

To find the horizontal velocity we use the equation $v_{i}*\cos(\Theta)=v_{x}$ .

We can plug in the given values for the angle and initial velocity to solve.

$30\frac{m}{s}*\cos(35^{\circ})=v_{x}$

$30\frac{m}{s}*(0.82)=v_{x}$

$24.57\frac{m}{s}=v_{x}$

We can find the vertical velocity using the equation $v_{i}* \sin(\Theta)=v_{iy}$ .

$30\frac{m}{s}* \sin(35^{\circ})=v_{iy}$

$30\frac{m}{s}* (0.57)=v_{iy}$

$17.21\frac{m}{s}=v_{iy}$

$v_f^2=v_1^2+2a\Delta y$

$v_f^2=(17.21\frac{m}{s})^2+2(-9.8\frac{m}{s^2})(-10m)$

$v_f^2=(296.18\frac{m^2}{s^2})+196\frac{m^2}{s^2}$

$v_f^2=492.18\frac{m^2}{s^2}$

$\sqrt{v_f^2}=\sqrt{492.18\frac{m^2}{s^2}}$

$v_f=22.19\frac{m}{s}$

Because the rock is traveling downward, our velocity will be negative: $v_f=-22.19\frac{m}{s}$ .

Now that we know our final velocities in both the horizontal and vertical directions, we can find the angle created between the two trajectories. The horizontal and vertical velocities can be compared using trigonometry.

$\tan(\Theta)=\frac{v_y}{v_x}$ ,

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

Plug in our values and solve for the angle.

$\tan^{-1}(\frac{22.19\frac{m}{s}}{24.57\frac{m}{s}})=\Theta$

$\tan^{-1}(0.90)=\Theta$

$41.99^{\circ}=\Theta$

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Example Question #62 : Linear Motion

Sam throws a 0.5kg rock off the edge of a 10m tall building at an angle of $35^{\circ}$ from the horizontal. The rock has an initial speed of $30\frac{m}{s}$ .

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

What is the total velocity of the rock when it reaches maximum height?

Possible Answers:

$0\frac{m}{s}$

$12.17\frac{m}{s}$

$9.8\frac{m}{s}$

$24.57\frac{m}{s}$

$-9.8\frac{m}{s}$

Correct answer:

$24.57\frac{m}{s}$

Explanation:

The question gives the total initial velocity, but we will need to find the horizontal and vertical components.

To find the horizontal velocity we use the equation $v_{i}*\cos(\Theta)=v_{x}$ .

We can plug in the given values for the angle and initial velocity to solve.

$30\frac{m}{s}*\cos(35^{\circ})=v_{x}$

$30\frac{m}{s}*(0.82)=v_{x}$

$24.57\frac{m}{s}=v_{x}$

We can find the vertical velocity using the equation $v_{i}* \sin(\Theta)=v_{iy}$ .

$30\frac{m}{s}* \sin(35^{\circ})=v_{iy}$

$30\frac{m}{s}* (0.57)=v_{iy}$

$17.21\frac{m}{s}=v_{iy}$

The question asks for the total velocity at the maximum height. At the top of the parabola, $v_y=0\frac{m}{s}$ as the direction of the motion changes from upward to downward. Yet, even though it has no vertical velocity, the horizontal velocity remains constant during flight. The only force on the rock is that of gravity, and gravity will only affect the vertical velocity. There are no horizontal forces present to alter the horizontal velocity.

Our final answer will be equal to the horizontal velocity: $24.57\frac{m}{s}$ .

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Example Question #31 : Understanding Motion In Two Dimensions

Sam throws a 0.5kg rock off the edge of a 10m tall building at an angle of $35^{\circ}$ from the horizontal. The rock has an initial speed of $30\frac{m}{s}$ .

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

What is the initial vertical velocity?

Possible Answers:

$34.21\frac{m}{s}$

$17.21\frac{m}{s}$

$2.21\frac{m}{s}$

$7.21\frac{m}{s}$

$89.21\frac{m}{s}$

Correct answer:

$17.21\frac{m}{s}$

Explanation:

The question gives the total initial velocity, but we will need to find the vertical component.

To find the vertical velocity we use the equation $v_{i}* \sin(\Theta)=v_{iy}$ .

We can plug in the given values for the angle and initial velocity to solve.

$30\frac{m}{s}* \sin(35^{\circ})=v_{iy}$

$30\frac{m}{s}* (0.57)=v_{iy}$

$17.21\frac{m}{s}=v_{iy}$

Report an Error

Example Question #31 : 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 initial vertical velocity?

Possible Answers:

$24.7 \frac{m}{s}$

$31 \frac{m}{s}$

$0\frac{m}{s}$

$10.6 \frac{m}{s}$

$-9.8 \frac{m}{s}$

Correct answer:

$10.6 \frac{m}{s}$

Explanation:

The given initial velocity is at an angle, so we have to use some trigonometric functions to break it into horizontal and vertical components.

Effectively, the initial velocity becomes the hypotenuse of a right triangle with the horizontal velocity becoming the base and the vertical velocity becoming the height. To find the vertical velocity, we use the relationship between the hypotenuse and the opposite side.

$\sin{\theta}=\frac{\text{opposite}}{\text{hypotenuse}}$

$\sin{\theta}=\frac{v_y}{v_{i}}$

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

Use the given initial velocity and angle to solve for the vertical velocity.

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

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

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

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Example Question #32 : 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 initial horizontal velocity?

Possible Answers:

$31 \frac{m}{s}$

$15.5\frac{m}{s}$

$10.6\frac{m}{s}$

$9.8\frac{m}{s}$

$29.13 \frac{m}{s}$

Correct answer:

$29.13 \frac{m}{s}$

Explanation:

The given initial velocity is at an angle, so we have to use some trigonometric functions to break it into horizontal and vertical components.

Effectively, the initial velocity becomes the hypotenuse of a right triangle with the horizontal velocity becoming the base and the vertical velocity becoming the height. To find the horizontal velocity, we use the relationship between the hypotenuse and the adjacent side.

$\cos{\theta}=\frac{\text{adjacent}}{\text{hypotenuse}}$

$\cos{\theta}=\frac{v_x}{v_{i}}$

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

Use the given initial velocity and angle to find the horizontal velocity.

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

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

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

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Example Question #31 : 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 velocity at the maximum height?

Possible Answers:

$6.2\frac{m}{s}$

$29.13 \frac{m}{s}$

$10.6\frac{m}{s}$

$0\frac{m}{s}$

$31\frac{m}{s}$

Correct answer:

$29.13 \frac{m}{s}$

Explanation:

In a problem with parabolic motion, your first step should always be to break the given velocity into its horizontal and vertical components.

Use cosine to find the initial horizontal velocity.

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

We are given the initial velocity and angle.

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

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

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

Use these values with sine to find the initial vertical velocity.

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

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

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

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

The question is asking for the velocity at the maximum height or, in our terms, the top of the parabola. At the top of the parabola the vertical velocity will be zero, but the horizontal velocity will remain constant. To find the total velocity, we normally use the Pythagorean Theorem with the horizontal and vertical velocities.

$v_{R}^2=v_x^2+v_y^2$

Since we know that the vertical velocity is zero, however, we can see that the total velocity will simply be equal to the horizontal velocity.

v_R^2=v_x^2

$\sqrt{v_{R}^2}=\sqrt{(29.13\frac{m}{s})^2}$

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

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High School Physics : Motion and Mechanics

Study concepts, example questions & explanations for High School Physics

All High School Physics Resources

Example Questions

Example Question #61 : Linear Motion

Example Question #2 : Sat Subject Test In Physics

Example Question #3 : Sat Subject Test In Physics

Example Question #131 : High School Physics

Example Question #3 : Linear Motion

Example Question #62 : Linear Motion

Example Question #31 : Understanding Motion In Two Dimensions

Example Question #31 : Understanding Motion In Two Dimensions

Example Question #32 : Understanding Motion In Two Dimensions

Example Question #31 : Understanding Motion In Two Dimensions

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