## 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.

## 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 $D_E+D_N=D_N+D_E$

## Solution:

Consider the right triangle formed by the legs A, B, and R. We know that A is 18 m, B is 25 m, and we are solving for the magnitude of R. We can do this by using the Pythagorean Theorem. That is

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

$R=\sqrt{\left(18\:m\right)^2+\left(25\:m\right)^2}$

$R=\sqrt{324+625}$

$R=\sqrt{949}$

$R=30.8\:m$

So, the distance is about 30.8 meters from the starting point. To solve for the value of the unknown angle, φ, we can use the tangent function. That is

$tan\:\phi =\frac{A}{B}$

$tan\:\phi =\frac{18\:m}{25\:m}$

$\phi =tan^{-1}\left(\frac{18}{25}\right)$

$\phi =35.75^{\circ}$

Therefore, the compass angle is $35.75^{\circ} \:West\:of\:North$

## Solution:

Consider the right triangle formed by the legs A, B, and R. We know that A is 18 m, B is 25 m, and we are solving for the magnitude of R. We can do this by using the Pythagorean Theorem. That is

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

$R=\sqrt{\left(18\:m\right)^2+\left(25\:m\right)^2}$

$R=\sqrt{324+625}$

$R=\sqrt{949}$

$R=30.8\:m$

So, the distance is about 30.8 meters from the starting point. To solve for the value of the unknown angle, θ, we can use the tangent function. That is

$tan\:\theta =\frac{B}{A}$

$tan\:\theta =\frac{25\:m}{18\:m}$

$\theta =tan^{-1}\left(\frac{25}{18}\right)$

$\theta =54.25^{\circ}$

So, the compass reading, can be solved by taking the complimentary angle, $\phi$ as shown in the figure.

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

$\phi =35.75^{\circ}$

Therefore, the compass angle is $35.75^{\circ} \:West\:of\:North$

## Solution:

### North Component

The North component is given by

$S_N=123\:km\:\cdot sin\left(45^{\circ} \right)=86.97\:km$

### East Component

The east component is

$S_E=123\:km\:\cdot cos\:\left(45^{\circ} \right)=86.97\:km$

## Solution:

### Part A

The total distance traveled following Path D is

$d=\left(2\times 120\:m\right)+\left(6\times 120\:m\right)+\left(4\times 120\:m\right)+\left(1\times 120\:m\right)$

$d=1560\:m$

### Part B

Treat all motions upward and to the right positive, while downward and to the left negative.

The displacement in the x-direction is

$s_x=0+720\:m+0-120\:m=600\:m$

The displacement in the y-direction is

$s_y=-240\:m+0+480\:m+0=240\:m$

The total displacement is

$s=\sqrt{\left(s_x\right)^2+\left(s_y\right)^2}=\sqrt{\left(600\:m\right)^2+\left(240\:m\right)^2}=646.22\:m$

The direction angle from th x-axis is given by

$\theta _x=tan^{-1}\left|\frac{s_y}{s_x}\right|=tan^{-1}\left|\frac{240}{600}\right|=21.8^{\circ} \:North\:of\:East$

Therefore, the displacement is 646.22 m, directed at 21.8 degrees north of east.

## Solution:

### Part A

The total distance traveled following Path C is

$d=\left(1\times 120\:m\right)+\left(5\times 120\:m\right)+\left(2\times 120\:m\right)+\left(1\times 120\:m\right)+\left(1\times 120\:m\right)+\left(3\times 120\:m\right)$

$d=1560\:m$

### Part B

Treat all motions upward and to the right positive, while downward and to the left negative.

The displacement in the x-direction is

$s_x=0+600+0-120+0-360=120\:m$

The displacement in the y-direction is

$s_y=120+0-240+0+120+0=0\:m$

The total displacement is

$s=\sqrt{\left(s_x\right)^2+\left(s_y\right)^2}$

$s=\sqrt{\left(120\:m\right)^2+\left(0\:m\right)^2}$

$s=120\:m$

The direction angle from th x-axis is given by

$\theta _x=tan^{-1}\left|\frac{s_y}{s_x}\right|$

$\theta _x=tan^{-1}\left|\frac{0\:m}{120\:m}\right|$

$\theta _x=0^{\circ}$

Therefore, the displacement is 120 m, east.