## 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.

## Farmer wants to Fence off his Four-Sided Plot with missing Side| Vector Addition and Subtraction| Analytical Method| College Physics| Problem 3.22

#### A farmer wants to fence off his four-sided plot of flat land. He measures the first three sides, shown as A, B, and C in the figure, and then correctly calculates the length and orientation of the fourth side D. What is his result?

SOLUTION:

We know that the sum of the 4 vectors is zero. So, Vector D is calculated as

$D=-A-B-C$

We solve for the x-component first. The x-component of vector D is

$D_x=-A_x-B_x-C_x$

$D_x=-\left(4.70\:km\right)cos\left(-7.50^{\circ} \right)-\left(2.48\:km\right)cos106^{\circ} -\left(3.02\:km\right)cos161^{\circ}$

$D_x=-1.12\:km$

We solve for the y-component.

$D_y=-A_y-B_y-C_y$

$D_y=-\left(4.70\:km\right)sin\left(-7.50^{\circ} \right)-\left(2.48\:km\right)sin106^{\circ} -\left(3.02\:km\right)sin161^{\circ}$

$D_y=-2.75\:km$

Since we already know the x and y component of vector D, we can finally solve for the distance of vector D.

$D=\sqrt{\left(D_x\right)^2+\left(D_y\right)^2}$

$D=\sqrt{\left(-1.12\:km\right)^2+\left(-2.75\:km\right)^2}$

$D=2.97\:km$

The corresponding direction of vector D is

$\theta =tan^{-1}\left|\frac{D_x}{D_y}\right|$

$\theta =tan^{-1}\left|\frac{1.12\:km}{2.75\:km}\right|$

$\theta =22.2^{\circ} \:W\:of\:S$

## Solution:

### Part A

The component along the south direction is

$D_s=Rsin\theta =\left(32.0\:km\right)sin\:35^{\circ} =18.4\:km$

The component along the west direction is

$D_w=Rcos\theta =\left(32.0\:km\right)cos35^{\circ} =26.2\:km$

### Part B

Consider the following figure with the rotated axes x’-y’.

The component along the southwest direction is

$D_{SW}=Rcos\theta '=\left(32.0\:km\right)cos10^{\circ} =31.5\:km$

The component along the northwest direction is

$D_{NW}=Rsin\theta '=\left(32.0\:km\right)cos10^{\circ} =5.56\:km$

## Solution:

Since the figure is closed, we know that the sum of the three vectors is equal to 0. So, we have

$\vec{A}+\vec{B}+\vec{C}=0$

Solving for C, we have

$\vec{C}=-\vec{A}-\vec{B}$

Next, we solve for the x and y components of Vectors A and B.

For Vector A, the x and y components are

$A_x=\left(80\:m\right)cos\left(21^{\circ} \right)=74.6864\:m$

$A_y=-\left(80\:m\right)sin\left(21^{\circ} \right)=-28.6694\:m$

For Vector B, the x and y components are

$B_x=-\left(105\:m\right)sin\:11^{\circ} =-20.0349\:m$

$B_y=\left(105\:m\right)cos\:11^{\circ} =103.0709\:m$

Solve for the components of Vector C.

$C_x=-A_x-B_x=-74.6864\:m\:-\left(-20.0349\:m\right)=-54.6515\:m$

$C_y=-A_y-B_y=-\left(-28.6694\:m\right)-\left(103.0709\:m\right)=-74.4015\:m$

Therefore, the length of Vector C is

$C=\sqrt{\left(C_x\right)^2+\left(C_y\right)^2}$

$C=\sqrt{\left(-54.6515\:m\right)^2+\left(-74.4015\:m\right)^2}$

$C=92.3167\:m$

The direction of Vector C is

$\theta =tan^{-1}\left(\frac{C_y}{C_x}\right)=tan^{-1}\left(\frac{74.4015\:m}{54.6515\:m}\right)=53.7^{\circ} \:\:South\:of\:West$

## Solution:

### Part A

Consider the following figure:

The east distance is the component in the horizontal direction.

$D_E=7.50\:km\:\cdot sin\:\left(15^{\circ} \right)$

$D_E=1.94\:km$

The north distance is the vertical component

$D_E=7.50\:km\cdot cos\left(15^{\circ} \right)$

$D_E=7.24\:km$

### Part B

Based from the figure, we can easily see that the order is reversible in the addition of vectors. We say that $D_E+D_N=D_N+D_E$