## Solution:

### Part A

We are given the range of the projectile motion. The range is 32.0 km. We also know that for the projectile to reach its maximum distance, it should be fired at 45°. So from the formula of range,

$\displaystyle \text{R}=\frac{\text{v}_0^2\:\sin 2\theta _0}{\text{g}}$

we can say that $\sin 2\theta _0=\sin \left(2\times 45^{\circ} \right)=\sin 90^{\circ} =1$. So, we have

$\displaystyle \text{R}=\frac{\text{v}_0^2}{\text{g}}$

We can solve for v0 in terms of the other variables. That is

$\displaystyle \text{v}_0=\sqrt{\text{gR}}$

Substituting the given values, we have

$\displaystyle \text{v}_0=\sqrt{\left(9.81\:\text{m/s}^2\right)\left(32\times 10^3\:\text{m}\right)}$

$\displaystyle \text{v}_0=560.29\:\text{m/s}$        ◀

### Part B

We are solving for the maximum height here, which happened at the mid-flight of the projectile. The vertical velocity at this point is zero. Considering all this, the formula for the maximum height is derived to be

$\displaystyle \text{h}_{\text{max}}=\frac{\text{v}_{0_y}^2}{2\text{g}}$

The initial vertical velocity, v0y, is calculated as

$\displaystyle \text{v}_{\text{0y}}=\text{v}_0\sin \theta _0=\left(560.29\:\text{m/s}\right)\sin 45^{\circ} =396.18\:\text{m/s}$

Therefore, the maximum height is

$\displaystyle \text{h}_{\text{max}}=\frac{\left(396.18\:\text{m/s}\right)^2}{2\left(9.81\:\text{m/s}^2\right)}$

$\displaystyle \text{h}_{\text{max}}=8000\:\text{m}$       ◀

### Part C

Consider the following figure

A right triangle is formed with the legs, the horizontal distance and the radius of the earth, and the hypotenuse is the sum of the radius of the earth and the distance d, which is the unknown in this problem. Using Pythagorean Theorem, and converting all units to meters, we have

$\displaystyle \text{R}^2+\left(32.0\times 10^3\:\text{m}\right)^2=\left(\text{R}+\text{d}\right)^2$

$\displaystyle \left(6.37\times 10^6\:\text{m}\right)^2+\left(32.0\times 10^3\:\text{m}\right)^2=\left(6.37\times 10^6+\text{d}\right)^2$

$\displaystyle \text{d}=\sqrt{\left(6.37\times \:10^6\:\right)^2+\left(32.0\times 10^3\:\right)^2}-6.37\times \:10^6$

$\displaystyle \text{d}=80.37\:\text{m}$       ◀

This error is not significant because it is only about 1% of the maximum height computed in Part B.

## Solution:

To verify the given values in the figure, we need to solve for individual ranges for the given initial angles. To do this, we shall use the formula

$\displaystyle \text{R}=\frac{\text{v}_{\text{0}}^2 \sin 2\theta _{\text{0}}}{\text{g}}$

When the initial angle is 15°, the range is

$\displaystyle \text{R}=\frac{\left(50\:\text{m/s}\right)^2\:\sin \left(2\times 15^{\circ} \right)}{9.81\:\text{m/s}^2}=127.42\:\text{m}$        ◀

When the initial angle is 45°, the range is

$\displaystyle \text{R}=\frac{\left(50\:\text{m/s}\right)^2\:\sin \left(2\times 45^{\circ} \right)}{9.81\:\text{m/s}^2}=254.84\:\text{m}$        ◀

When the initial angle is 75°, the range is

$\displaystyle \text{R}=\frac{\left(50\:\text{m/s}\right)^2\:\sin \left(2\times 75^{\circ} \right)}{9.81\:\text{m/s}^2}=127.42\:\text{m}$        ◀

Based from the result of the calculations, we can say that the numbers in the figure is verified. The very small differences are only due to round-off errors.

## Solution:

To verify the given values in the figure, we need to solve for individual ranges for the given initial velocities. To do this, we shall use the formula

$\displaystyle \text{R}=\frac{\text{v}_{\text{0}}^2\:\sin 2\theta _{\text{0}}}{\text{g}}$

When the initial velocity is 30 m/s, the range is

$\displaystyle \text{R}=\frac{\left(30\:\text{m/s}\right)^2\:\sin \left(2\times 45^{\circ} \right)}{9.81\:\text{m/s}^2}=91.74\:\text{m}$        ◀

When the initial velocity is 40 m/s, the range is

$\displaystyle \text{R}=\frac{\left(40\:\text{m/s}\right)^2\:\sin \left(2\times 45^{\circ} \right)}{9.81\:\text{m/s}^2}=163.10\:\text{m}$        ◀

When the initial velocity is 50 m/s, the range is

$\displaystyle \text{R}=\frac{\left(50\:\text{m/s}\right)^2\:\sin \left(2\times 45^{\circ} \right)}{9.81\:\text{m/s}^2}=254.84\:\text{m}$        ◀

Based from the results, we can say that the ranges are approximately equal. The differences are only due to round-off errors.

## Solution:

To illustrate the problem, consider the following figure:

### Part A

We are given the 7-meter range, R, and the initial velocity, vo, of the projectile. We have R=7.0 m, and vo=12.0 m/s. To solve for the angle of the initial velocity, we will use the formula for range

$\displaystyle \text{R}=\frac{\text{v}^{2}_{\text{o}}\sin 2\theta _{\text{o}}}{g}$

Solving for θo in terms of the other variables, we have

$\displaystyle \text{gR}=\text{v}_{\text{o}}^2\sin 2\theta _{\text{o}}$

$\displaystyle \sin 2\theta _{\text{o}}=\frac{\text{gR}}{\text{v}_{\text{o}}^2}$

$\displaystyle 2\theta _\text{o}=\sin ^{-1}\left(\frac{\text{gR}}{\text{v}_{\text{o}}^2}\right)$

$\displaystyle \theta _\text{o}=\frac{1}{2}\sin ^{-1}\left(\frac{\text{gR}}{\text{v}_{\text{o}}^2}\right)$

Substituting the given values, we have

$\displaystyle \theta _\text{o}=\frac{1}{2} \sin ^{-1}\left[\frac{\left(9.81\text{m/s}^2\right)\left(7.0\text{m}\right)}{\left(12.0\text{m/s}\right)^2}\right]$

$\displaystyle \theta _\text{o}=14.24^{\circ}$     ◀

### Part B

The other angle that would give the same range is actually the complement of the solved angle in Part A. The other angle,

$\displaystyle \theta _o'=90^{\circ} -14.24^{\circ} =75.76^{\circ}$     ◀

This angle is not used as often, because the time of flight will be longer. In rugby that means the defense would have a greater time to get into position to knock down or intercept the pass that has the larger angle of release.

### Part C

We can use the x-component of the motion to solve for the time of flight.

$\displaystyle \Delta \text{x}=\text{v}_\text{x}\text{t}$

We need the horizontal component of the velocity. We should be able to solve for the component since we are already given the initial velocity and the angle.

$\displaystyle \text{v}_{\text{x}}=\left(12\:\text{m/s}\right)\cos 14.24^{\circ} =11.63\:\text{m/s}$

Therefore, the total time of flight is

$\displaystyle \text{t}=\frac{\Delta \text{x}}{\text{v}_{\text{x}}}$

$\displaystyle \text{t}=\frac{7.0\:\text{m}}{11.63\:\text{m/s}}$

$\displaystyle \text{t}=0.60\:\text{s}$     ◀

## Solution:

To illustrate the problem, consider the following figure:

### Part A

We are given the range of 75-meter range, R, and the initial velocity, vo, of the projectile. We have R=75.0 m, and vo=35.0 m/s. To solve for the angle of the initial velocity, we will use the formula for range

$\displaystyle \text{R}=\frac{\text{v}^{2}_{\text{o}}\:\sin 2\theta _{\text{o}}}{g}$

Solving for θo in terms of the other variables, we have

$\displaystyle \text{gR}=\text{v}_{\text{o}}^2\:\sin 2\theta _{\text{o}}$

$\displaystyle \sin \:2\theta _{\text{o}}=\frac{\text{gR}}{\text{v}_{\text{o}}^2}$

$\displaystyle 2\theta _\text{o}=\sin ^{-1}\left(\frac{\text{gR}}{\text{v}_{\text{o}}^2}\right)$

$\displaystyle \theta _\text{o}=\frac{1}{2}\sin ^{-1}\left(\frac{\text{gR}}{\text{v}_{\text{o}}^2}\right)$

Substituting the given values, we have

$\displaystyle \theta _o=\frac{1}{2}\sin ^{-1}\left[\frac{\left(9.81\:\text{m/s}^2\right)\left(75.0\:\text{m}\right)}{\left(35.0\:\text{m/s}\right)^2}\right]$

$\displaystyle \theta _o=18.46^{\circ}$     ◀

### Part B

We know that halfway, the maximum height of the projectile occurs. Also at this instant, the vertical velocity is zero. We can solve for the maximum height and compare it with the given height of 3.50 meters.

The maximum height can be computed using the formula

$\displaystyle \text{h}_{\text{max}}=\frac{\text{v}_{\text{oy}}^2}{2\text{g}}$

To compute for the maximum height, we need the initial vertical velocity, voy. Since we know the magnitude and direction of the initial velocity, we have

$\displaystyle \text{v}_{\text{oy}}=\left(35.0\:\text{m/s}\right)\sin 18.46^{\circ}$

$\displaystyle \text{v}_{\text{oy}}=11.08\:\text{m/s}$

Therefore, the maximum height is

$\displaystyle \text{h}_{\max }=\frac{\left(11.08\:\text{m/s}\right)^2}{2\left(9.81\:\text{m/s}^2\right)}$

$\displaystyle \text{h}_{\max }=6.26\:\text{m}$     ◀

We have known that the path of the arrow is above the branch of the tree. Therefore, the arrow will go through.

## Solution:

To illustrate the problem, consider the following figure:

### Part A

To determine the number of buses that the daredevil can clear, we will divide the range of the projectile path by 20 m, the length of 1 bus. That is

$\displaystyle \text{no. of bus}=\frac{\text{Range}}{\text{bus length}}$

First, we need to solve for the range.

$\displaystyle \text{Range}=\frac{\text{v}_{\text{o}}^2\:\sin 2\theta }{\text{g}}$

$\displaystyle \text{Range}=\frac{\left(40.0\:\text{m/s}\right)^2\sin \left[2\left(32^{\circ} \right)\right]}{9.81\:\text{m/s}^2}$

$\displaystyle \text{Range}=146.7\:\text{m}$

Therefore, the number of buses cleared is

$\displaystyle \text{no. of buses}=\frac{146.7\:\text{m}}{20\:\text{m}}$

$\displaystyle \text{no. of buses}=7.34\:\text{buses}$

Therefore, he can only clear 7 buses.          ◀

### Part B

He clears the last bus by 6.7 m, which seems to be a large margin of error, but since we neglected air resistance, it really isn’t that much room for error.

## Solution:

To illustrate the problem, consider the following figure:

### Part A

The problem states that the initial velocity is horizontal, this means that the initial vertical velocity is zero. We are also given the height of the building (which is a downward displacement), so we can solve for the time of flight using the formula y=voyt+1/2at2. That is,

$\displaystyle \text{y}=\text{v}_{\text{oy}}\text{t}+\frac{1}{2}\text{a}\text{t}^2$

$\displaystyle -60\:\text{m}=0+\frac{1}{2}\left(-9.81\:\text{m/s}^2\right)\text{t}^2$

$\displaystyle \:\text{t}^2=\frac{-60\:\text{m}}{-4.905\:\text{m/s}^2}$

$\displaystyle \:\text{t}^2=12.2324\:\text{s}^2$

$\displaystyle \text{t}=3.50\:\text{s}$          ◀

### Part B

To solve for the vox, we will use the formula $\displaystyle \text{v}_{\text{ox}}=\frac{\Delta \:\text{x}}{\text{t}}$.

$\displaystyle \text{v}_{\text{ox}}=\frac{100\:\text{m}}{3.50\:\text{s}}$

$\displaystyle \text{v}_{\text{ox}}=28.57\:\text{m/s}$          ◀

### Part C

To solve for the velocity as the ball hits the ground, we shall consider two points: (1) at the beginning of the flight, and (2) when the ball hits the ground.

We know that the initial velocity, voy, is zero. To solve for the final velocity, we will use the formula $\displaystyle \text{v}_{\text{f}}=\text{v}_{\text{o}}+\text{at}$

$\displaystyle \text{v}_{\text{f}}=0+\left(-9.81\:\text{m/s}^2\right)\left(3.50\:\text{s}\right)$

$\displaystyle \text{v}_{\text{f}}=-34.34\:\text{m/s}$          ◀

The negative velocity indicates that the motion is downward.

### Part D

Since we already know the horizontal and vertical components of the velocity when it hits the ground, we can find the resultant.

$\displaystyle \text{v}=\sqrt{\text{v}_{\text{x}}^2+\text{v}_{\text{y}}^2}$

$\displaystyle \text{v}=\sqrt{\left(28.57\:\text{m/s}\right)^2+\left(-34.34\:\text{m/s}\right)^2}$

$\displaystyle \text{v}=44.67\:\text{m/s}$          ◀

The direction of the velocity is

$\displaystyle \theta {\text{x}}=\tan ^{-1}\left|\frac{\text{v}_{\text{y}}}{\text{v}_{\text{x}}}\right|$

$\displaystyle \theta _{\text{x}}=\tan ^{-1}\left|\frac{-34.34}{28.57}\right|$

$\displaystyle \theta _{\text{x}}=50.24^{\circ}$ ⦪          ◀

The velocity is directed 50.24° down the x-axis.

## Solution:

To illustrate the problem, consider the following figure:

### Part A

Since the starting position has the same elevation as when it hits the ground, the speeds at these points are the same. The final speed is computed by solving the resultant of the horizontal and vertical velocities. That is

$\displaystyle \text{v}_{\text{f}}=\sqrt{\left(\text{v}_{\text{ox}}\right)^2+\left(\text{v}_{\text{oy}}\right)^2}$

$\displaystyle \text{v}_{\text{f}}=\sqrt{\left(16\:\text{m/s}\right)^2+\left(12\:\text{m/s}\right)^2}$

$\displaystyle \text{v}_{\text{f}}=\sqrt{400\:\text{(m/s)}^2}$

$\displaystyle \text{v}_{\text{f}}=20\:\text{m/s}$          ◀

### Part B

Consider the two points: (1) the starting point and (2) the highest point.

We know that at the highest point, the vertical velocity is zero. We also know that the total time of the flight is twice the time from the beginning to the top.

So, we shall use the formula $\displaystyle \text{t}=\frac{\text{v}_{\text{f}}-\text{v}_{\text{o}}}{\text{a}}$.

$\displaystyle \text{t}=2\left(\frac{\text{v}_{\text{top}}-\text{v}_{\text{o}}}{\text{a}}\right)$

$\displaystyle \text{t}=2\left(\frac{0\:\text{m/s}-12\:\text{m/s}}{-9.81\:\text{m/s}^2}\right)$

$\displaystyle \text{t}=2.45\:\text{s}$          ◀

### Part C

The maximum height attained can be calculated using the formula $\displaystyle \left(\text{v}_{\text{f}}\right)^2=\left(\text{v}_{\text{o}}\right)^2+2\text{a}\text{y}$.

The maximum height is calculated as follows:

$\displaystyle \left(\text{v}_{\text{f}}\right)^2=\left(\text{v}_{\text{o}}\right)^2+2\text{ay}$

$\displaystyle \text{y}_{\max }=\frac{\left(\text{v}_{\text{top}}\right)^2-\left(\text{v}_{\text{o}}\right)^2}{2\text{a}}$

$\displaystyle \text{y}_{\max }=\frac{\left(0\:\text{m/s}\right)^2-\left(16\:\text{m/s}\right)^2}{2\left(-9.81\:\text{m/s}^2\right)}$

$\displaystyle \text{y}_{\max }=7.34\:\text{m}$          ◀

## Solution:

Since we do not know the exact location of the projectile after 3 seconds, consider the following arbitrary figure:

From the figure, we can solve for the components of the initial velocity.

$\displaystyle \text{v}_{\text{ox}}=\left(50\:\text{m/s}\right)\cos 30^{\circ} =43.3013\:\text{m/s}$

$\displaystyle \text{v}_{\text{oy}}=\left(50\:\text{m/s}\right)\sin 30^{\circ} =25\:\text{m/s}$

So, we are asked to solve for the values of x and y. To solve for the value of the horizontal displacement, x, we shall use the formula x=voxt. That is,

$\displaystyle \text{x}=\text{v}_{\text{ox}}\text{t}$

$\displaystyle \text{x}=\left(43.3013\:\text{m/s}\right)\left(3\:\text{s}\right)$

$\displaystyle \text{x}=129.9\:\text{m}$          ◀

To solve for the vertical displacement, y, we shall use the formula y=voyt+1/2at2. That is

$\displaystyle \text{y}=\text{v}_{\text{oy}}\text{t}+\frac{1}{2}\text{a}\text{t}^2$

$\displaystyle \text{y}=\left(25\:\text{m/s}\right)\left(3\:\text{s}\right)+\frac{1}{2}\left(-9.81\:\text{m/s}^2\right)\left(3\:\text{s}\right)^2$

$\displaystyle \text{y}=30.9\:\text{m}$          ◀

Therefore, the projectile strikes a target at a distance 129.9 meters horizontally and 30.9 meters vertically from the launching point.

## Solution:

### Part A

From the given statement, you first walk 18.0 m straight west and then 25.0 straight south. These vectors are represented by the graph shown below.

To solve for the resultant, we simply need to use the Pythagorean theorem to solve for the hypotenuse of the right triangle formed. That is,

$\displaystyle \text{R}=\sqrt{\text{A}^2+\text{B}^2}$

$\displaystyle \text{R}=\sqrt{\left(18.0\:\text{m}\right)^2+\left(25.0\:\text{m}\right)^2}$

$\displaystyle \text{R}=30.8\:\text{m}$

To solve for the angle, θ, we shall use the tangent function.

$\displaystyle \theta =\tan ^{-1}\left(\frac{\text{B}}{\text{A}}\right)$

$\displaystyle \theta =\tan ^{-1}\left(\frac{25.0\:\text{m}}{18.0\:\text{m}}\right)$

$\displaystyle \theta =54.25^{\circ}$

Thus, the value of 𝜙 can be calculated as

$\displaystyle \phi=90^{\circ} -54.25^{\circ}$

$\displaystyle \phi=35.75^{\circ}$

Therefore, the compass direction of the resultant is 35.75° West of South.

### Part B

From the statement, you walk 25.0 m north first and then 18.0 m east. This is represented by the figure shown below.

So, we have a right triangle with legs 25.0 m and 18.0 m. We are tasked to solve for the value of R, and the angle 𝜙 for the compass direction. The value of R can be solved using the Pythagorean Theorem as in Part A.

$\displaystyle \text{R}=\sqrt{\text{A}^2+\text{B}^2}$

$\displaystyle \text{R}=\sqrt{\left(18.0\:\text{m}\right)^2+\left(25.0\:\text{m}\right)^2}$

$\displaystyle \text{R}=30.8\:\text{m}$

To solve for the angle, θ, we shall use the tangent function.

$\displaystyle \theta =\tan ^{-1}\left(\frac{\text{B}}{\text{A}}\right)$

$\displaystyle \theta =\tan ^{-1}\left(\frac{18.0\:\text{m}}{25.0\:\text{m}}\right)$

$\displaystyle \theta =35.75^{\circ}$

Therefore, the compass direction of the resultant is 35.75° East of North.