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AP Physics 1 : Motion in Two Dimensions

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Example Questions

Example Question #71 : Motion In Two Dimensions

A home run derby is being held in a stadium where the home run wall is 100m from home plate. The wall itself has a height of . The stands begin at the wall and with equal height, and rise at an angle of $30^{\circ}$ to the horizontal.

$g = 10\frac{m}{s^2}$

If a home run is hit a total distance of 120 m from an initial height of and an angle of $50^{\circ}$ to the horizontal, what is the ball's initial velocity?

Possible Answers:

$45.9\frac{m}{s}$

$14.3\frac{m}{s}$

$29.9\frac{m}{s}$

$21.2\frac{m}{s}$

$36.8\frac{m}{s}$

Correct answer:

$36.8\frac{m}{s}$

Explanation:

From the problem statement, we can use the following piecewise function to express the effective "height" at any point within the stadium:

If x < 100m then y = 0m

If $x\geq 100m$ then $y = 5m+(x-100)sin(30^{\circ})$

Before, moving on, make sure you understand the reasoning of this function. The wall begins at a distance of 100m , a height of , and rises at an angle of $30^{\circ}$ .

Using the total horizontal distance traveled, we can calculate the ball's final height using the above expression:

$y = 5m+(120m-100m)sin(30^{\circ}) = 15m$

Now we can use the following two kinematics expression:

$\Delta y = v_{y_{i}}t+\frac{1}{2}at^2$

$t = \frac{d}{v_{x_{i}}}$

Where d is the horizontal distance traveled. Replacing out component velocities, we get:

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

$t = \frac{d}{v_icos(\theta)}$

Substituting our expression for time into the other expression, we get:

$\Delta y = dtan(\theta)+\frac{1}{2}a\left ( \frac{d}{v_icos(\theta)}\right )^2$

Simplifying, we get:

$\Delta y = dtan(\theta)+\frac{ad^2}{2v_i^2cos^2(\theta)}$

Rearranging:

$\frac{ad^2}{2v_i^2cos^2(\theta)} = dtan(\theta)-\Delta y$

And again:

$\frac{ad^2}{dtan(\theta)-\Delta y} = 2v_i^2cos^2(\theta)$

One last time:

$v_i = \sqrt{\frac{ad^2}{2cos^2(\theta)\left (dtan(\theta)-\Delta y \right )}}$

It's nice and messy, but we have all of these values, so time to plug and chug:

$v_i = \sqrt{\frac{\left ( 10\frac{m}{s^2}\right )(120m)^2}{2cos^2(50^{\circ})((120m)tan(50^{\circ})-14m)}}$

$v_i = 36.8\frac{m}{s}$

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Example Question #72 : Motion In Two Dimensions

$g = 10\frac{m}{s^2}$

A ball is hit with an initial vertical velocity of $25\frac{m}{s}$ from a height of . If it takes for the ball to land in the stands, what is the total horizontal distance traveled?

Possible Answers:

92m

161m

101m

132m

116m

Correct answer:

132m

Explanation:

From the problem statement, we can use the following piecewise function to express the effective "height" at any point within the stadium:

If x < 100m then y = 0m

If $x\geq 100m$ then $y = 5m+(x-100)sin(30^{\circ})$

Before, moving on, make sure you understand the reasoning of this function. The wall begins at a distance of 100m , a height of , and rises at an angle of $30^{\circ}$ .

We can use the following expression to calculate the final height of the ball:

$y_f= y_i + v_{y_{i}}t+\frac{1}{2}at^2$

Plugging in our values, we get:

$y_f = 1m + \left ( 25\frac{m}{s}\right )(4s)+\frac{1}{2}\left ( -10\frac{m}{s^2}\right )(4s)^2$

y_f = 21m

Then we can use the piecewise expression to determine to total horizontal distance traveled:

$21m = 5m+ (x-100m)sin(30^{\circ})$

$x = \frac{16m}{sin(30^{\circ})}+100m$

x = 132m

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Example Question #73 : Motion In Two Dimensions

$g = 10\frac{m}{s^2}$

A ball is hit with an initial horizontal velocity of $30\frac{m}{s}$ and takes 3.5s for the ball to hit the stands. If the ball was hit from an initial height of , at what angle to the horizontal was the ball hit at?

Possible Answers:

$32.8^{\circ}$

$57.2^{\circ}$

$45^{\circ}$

$43.8^{\circ}$

$46.2^{\circ}$

Correct answer:

$32.8^{\circ}$

Explanation:

From the problem statement, we can use the following piecewise function to express the effective "height" at any point within the stadium:

If x < 100m then y = 0m

If $x\geq 100m$ then $y = 5m+(x-100)sin(30^{\circ})$

Before, moving on, make sure you understand the reasoning of this function. The wall begins at a distance of 100m , a height of , and rises at an angle of $30^{\circ}$ .

We can quickly determine the horizontal distance traveled by the ball:

$x = v_xt = \left ( 30\frac{m}{s}\right )(3.5s) = 105m$

Then we can use the piecewise function to calculate the final height of the ball. Since $x\geq 100m$ , we will use the second expression:

$y = 5m + (105m-100m)sin(30^{\circ}) = 7.5m$

Then we can use the following kinematics equation:

$\Delta y = v_{y_i}t+\frac{1}{2}at^2$

Rearranging, we get:

$v_{y_{i}}= \frac{\Delta y}{t}-\frac{at}{2}$

Plugging in our values, we get:

$v_{y_{i}}=\frac{7.5m-1m}{3.5s} - \frac{\left ( -10\frac{m}{s^2}\right )(3.5s)}{2}$

$v_{y_{i}}=19.4357\frac{m}{s}$

Then we can use the following expression to determine the initial angle of the ball:

$tan(\theta_i)=\frac{v_{y_{i}}}{v_{x_{i}}}$

Rearranging and plugging in our values, we get:

$\theta_i = tan^{-1}\left ( \frac{19.4357\frac{m}{s}}{30\frac{m}{s}}\right )$

$\theta_i = 32.8^{\circ}$

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Example Question #74 : Motion In Two Dimensions

$g = 10\frac{m}{s^2}$

If a ball is hit with an initial horizontal velocity of $25\frac{m}{s}$ and it takes 3.5s to land, what is the final height of the ball?

Possible Answers:

2.4m

5.1m

11.2m

7.6m

Correct answer:

Explanation:

From the problem statement, we can use the following piecewise function to express the effective "height" at any point within the stadium:

If x < 100m then y = 0m

If $x\geq 100m$ then $y = 5m+(x-100)sin(30^{\circ})$

Before, moving on, make sure you understand the reasoning of this function. The wall begins at a distance of 100m , a height of , and rises at an angle of $30^{\circ}$ .

We can calculate the total horizontal distance traveled by the ball using the following expression:

$x = v_xt = \left ( 25\frac{m}{s}\right )(3.5s)$

x = 87.5

The ball has yet to reach the home run wall, so it will land back on the ground with a final height of .

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Example Question #75 : Motion In Two Dimensions

A brick is dropped from the top floor of a building. The brick takes 10 s to reach the bottom. How tall is the building? Ignore air resistance.

Possible Answers:

435m

490m

135m

388m

670m

Correct answer:

490m

Explanation:

We know that the brick is dropped from rest $(v_{0}=0)$ at the top of the building and, ignoring air friction, we know the only relevant force is due to gravity $(a_{g} = -9.81\frac{m}{s})$ . Using the kinematic equation $\Delta x=v_{0}t+\frac{1}{2}at^{2}$ , plug in our known values for $v_{0}$ , , and , then solve to get $\Delta x = 490m$ .

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Example Question #76 : Motion In Two Dimensions

A brick is thrown at $10\frac{m}{s}$ from the top of a 100m building at an angle of $32\degree$ above horizontal. How long does it take to reach the ground below?

Possible Answers:

4.50s

1.02s

5.05 s

3.50 s

6.05s

Correct answer:

5.05 s

Explanation:

The problem can be split into two parts: when the brick is in the air above the building, and when the brick begins its descent past the top of the building once more. Since we are finding the time the brick takes to travel in the vertical direction, only the y-component of the velocity is relevant, $v_{y} = (10\frac{m}{s})sin(30\degree)=5\frac{m}{s}$ .

To find the time the brick is in the air for the first part, use the kinematic equation $v_y = v_{y,0}+a_{y}t$ . The time the brick takes to reach its peak is given by $0 = 5\frac{m}{s} - (9.81\frac{m}{s^2})t$ , so t=0.51s . Doubling this gives the time the brick takes to rise then fall again to the height from which it was thrown, so $t_{1} = 2t = 1.02s$ .

To find the time the brick takes to fall from the top of the building to the ground, use the kinematic equation $\Delta y = v_{y,0}t+\frac{1}{2}a_{y}t^2$ . We know that the initial velocity for this part, which is now pointing downward, is equal in magnitude to when the brick was first thrown, $v_{y,0} = -5\frac{m}{s}$ . The height of the building is given by $\Delta y = -100m$ . Hence our equation becomes $-100m = (-5 \frac{m}{s})t + \frac{1}{2}(-9.81 \frac{m}{s^2})t^2$ . Solving for gives $t_{2}=4.03s$ .

Adding the two times together, $t_{final} = t_{1} + t_{2} = 1.02s + 4.03s = 5.05s$ .

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Example Question #77 : Motion In Two Dimensions

A cannonball is shot at an angle. During the first the cannonball moves 100m horizontally.

How much horizontal distance has the cannonball covered after of motion?

Possible Answers:

250m

300m

150m

50m

200m

Correct answer:

200m

Explanation:

In projectile motion, there are no horizontal forces acting on the object during its flight. By Newton's Second Law we know that if no force is applied to an object, it does not accelerate.

So, the cannonball's horizontal velocity remains constant. Twice the time of flight leads to twice the horizontal distance.

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Example Question #78 : Motion In Two Dimensions

Jennifer is trying to catch a baseball. The ball is launched at $40\frac{m}{s}$ from a machine, $60^\circ$ above the horizontal. How far away from the machine should she be?

Possible Answers:

41.8m

17.5m

91.98m

141.2m

Correct answer:

141.2m

Explanation:

Determining the time for the upward velocity to equal zero:

v_i+at=v_f

$40*sin60^\circ-9.8*t=0$

t=3.53s

The ball will take the same amount of time to descend back to it's initial height, the height of Jennifer.

Determining how far horizontally the ball will travel in that time.

2t*v_x=d_x

$3.53*2*40cos 60^\circ=d_x$

141.2m

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Example Question #71 : Motion In Two Dimensions

Neglect air resistance and assume $g = 10\frac{m}{s^2}$

If a ball is hit from an initial height of and at an angle of $60^{\circ}$ to the horizontal, what is the minimal initial velocity that will result in a home run?

Possible Answers:

$34.4\frac{m}{s}$

$52.9\frac{m}{s}$

$18.9\frac{m}{s}$

$25.5\frac{m}{s}$

$44.4\frac{m}{s}$

Correct answer:

$34.4\frac{m}{s}$

Explanation:

Since we are asked for the shortest home run, the final height of the ball will be the height of the wall. We can then use the following expression:

$\Delta y = v_{y_{i}}t+\frac{1}{2}at^2$

Where,

$v_{y_{i}}=vsin(\theta)$

We can then use the following expression for time:

$t = \frac{x}{v_x}$

Where,

$v_x = vcos(\theta)$

Substituting our expression for time into the first expression, and replacing our component velocities, we get:

$\Delta y = xtan(\theta)-\frac{1}{2}a\left (\frac{x}{vcos(\theta)} \right )^2$

Rearranging, we get:

$\frac{1}{2}a\left (\frac{x}{vcos(\theta)} \right )^2 = xtan(\theta)-\Delta y$

$\frac{ax^2}{2v^2cos^2(\theta)} = xtan(\theta)-\Delta y$

Rearranging again:

$\frac{ax^2}{2cos^2(\theta)\left ( xtan(\theta)-\Delta y \right )} = v^2$

$v = \sqrt{\frac{ax^2}{2cos^2(\theta)\left (xtan(\theta)- \Delta y \right )}}$

We have all of these values, so time to plug and chug:

$v = \sqrt{\frac{\left ( 10\frac{m}{s^2}\right )(100m)^2}{2cos^2(60^{\circ})\left ( (100m)tan(60^{\circ})-4m\right )}}$

$v = 34.4\frac{m}{s}$

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Example Question #172 : Linear Motion And Momentum

Neglect air resistance and assume $g = 10\frac{m}{s^2}$

If a superhero baseball player hits a ball directly horizontal, what is the minimal initial velocity that will result in the ball hitting the home run wall?

Possible Answers:

$50\frac{m}{s}$

$70\frac{m}{s}$

$425\frac{m}{s}$

$150\frac{m}{s}$

$500\frac{m}{s}$

Correct answer:

$500\frac{m}{s}$

Explanation:

At the minimum velocity, the ball will reach a distance of 100 meters just when its height is 0 meters.

Let's begin with the following expression to determine how long it takes the ball to hit the ground:

$\Delta y = v_{y_{i}}t+\frac{1}{2}at^2$

There is no initial vertical velocity, so the expression becomes:

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

Rearranging:

$t = \sqrt{\frac{2\Delta y}{a}}$

Plugging in our values:

$t = \sqrt{\frac{2(1m)}{10\frac{m}{s^2}}} = 0.2s$

We can then calculate the initial velocity needed:

x = v_it

Rearranging:

$v_i = \frac{x}{t}$

Plugging in our values:

$v_i = \frac{100m}{0.2s}$

$v_i = 500\frac{m}{s}$

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AP Physics 1 : Motion in Two Dimensions

Study concepts, example questions & explanations for AP Physics 1

All AP Physics 1 Resources

Example Questions

Example Question #71 : Motion In Two Dimensions

Example Question #72 : Motion In Two Dimensions

Example Question #73 : Motion In Two Dimensions

Example Question #74 : Motion In Two Dimensions

Example Question #75 : Motion In Two Dimensions

Example Question #76 : Motion In Two Dimensions

Example Question #77 : Motion In Two Dimensions

Example Question #78 : Motion In Two Dimensions

Example Question #71 : Motion In Two Dimensions

Example Question #172 : Linear Motion And Momentum

All AP Physics 1 Resources

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