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College Physics by Openstax Chapter 3 Problem 34

An arrow is shot from a height of 1.5 m toward a cliff of height H . It is shot with a velocity of 30 m/s at an angle of 60º above the horizontal. It lands on the top edge of the cliff 4.0 s later.
(a) What is the height of the cliff?
(b) What is the maximum height reached by the arrow along its trajectory?
(c) What is the arrow’s impact speed just before hitting the cliff?

Solution:

Consider the following illustration:

An arrow shot at a height of 1.5 m towards a cliff of height H

Part A

We are required to solve for the value of H. We shall use the formula

\Delta \text{y}=\text{v}_{\text{0y}}\text{t}+\frac{1}{2}\text{at}^2

or, we can also write the formula as

 \text{y}-\text{y}_0=\text{v}_{\text{0y}}\text{t}+\frac{1}{2}\text{at}^2

Substituting the given values, we have

\begin{align*}
 \text{y}-\text{y}_0 &=\text{v}_{\text{0y}}\text{t}+\frac{1}{2}\text{at}^2 \\

\text{H}-1.5\:\text{m} & =\left(30\:\text{m/s}\:\sin 60^{\circ} \right)\left(4.0\:\text{s}\right)+\frac{1}{2}\left(-9.81\:\text{m/s}^2\right)\left(4.0\:\text{s}\right)^2 \\

\text{H}-1.5\:\text{m} & =25.44\:\text{m} \\

\text{H} & =25.44\:\text{m}+1.5\:\text{m} \\

\text{H} & =26.94\:\text{m} \ \qquad \ \color{DarkOrange} \left( \text{Answer} \right)
\end{align*}
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Part B

The maximum height of the projectile is given by the formula

\Delta \text{y}=\frac{\text{v}_{\text{0y}}^2}{2\text{g}}

or the formula can be written as

\text{y}_{\text{max}}-\text{y}_{\text{0}}=\frac{\text{v}_{\text{0y}}^2}{-2\text{g}}

Therefore, we have

\begin{align*}
\text{y}_{\text{max}}-\text{y}_{\text{0}} & =\frac{\text{v}_{\text{0y}}^2}{-2\text{g}} \\

\text{y}_{\text{max}}& =\frac{\text{v}_{\text{0y}}^2}{-2\text{g}}+\text{y}_{\text{0}} \\

\text{y}_{\text{max}}&=\frac{\left(\left(30\:\text{m/s}\right)\sin 60^{\circ} \right)^2}{-2\left(-9.81\:\text{m/s}^2\right)}+1.5\:\text{m} \\

\text{y}_{\text{max}}&=34.40\:\text{m}+1.5\:\text{m} \\

\text{y}_{\text{max}} & =35.90\:\text{m} \ \qquad \ \color{DarkOrange} \left( \text{Answer} \right) \\
\end{align*}
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Part C

To solve for the final speed, we need the vertical and horizontal components when the arrow hits the cliff.

We are going to use the formula for acceleration along the vertical to solve for the final speed in the vertical direction. That is

\text{a}=\frac{\text{v}_{\text{y}}-\text{v}_{\text{0y}}}{\text{t}}

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

\begin{align*}
\text{v}_{\text{y}}&=\text{v}_{\text{0y}}+\text{at}\\
\text{v}_{\text{y}}&=\left(30\:\text{m/s}\right)\sin 60^{\circ} +\left(-9.81\:\text{m/s}^2\right)\left(4.0\:\text{s}\right) \\
 \text{v}_{\text{y}}&=-13.25\:\text{m/s}
\end{align*}

Since we know that the horizontal component of the velocity does not change along the entire flight, we can equate the initial and final horizontal velocities. That is 

\begin{align*}
\text{v}_{\text{x}}&=\text{v}_{\text{0x}}=\left(30\:\text{m/s}\right)\cos 60^{\circ} \\
\text{v}_{\text{x}}&=15\:\text{m/s}
\end{align*}

Therefore, the final speed is

\begin{align*}
\text{v}&=\sqrt{\text{v}_{\text{y}}^2+\text{v}_{\text{x}}^2} \\
\text{v}&=\sqrt{\left(-13.25\:\text{m/s}\right)^2+\left(15\:\text{m/s}\right)^2}\\
\text{v}&=20.01\:\text{m/s} \ \qquad \ \color{DarkOrange} \left( \text{Answer} \right)
\end{align*}
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College Physics by Openstax Chapter 3 Problem 29

An archer shoots an arrow at a 75.0 m distant target; the bull’s-eye of the target is at same height as the release height of the arrow. (a) At what angle must the arrow be released to hit the bull’s-eye if its initial speed is 35.0 m/s? In this part of the problem, explicitly show how you follow the steps involved in solving projectile motion problems. (b) There is a large tree halfway between the archer and the target with an overhanging horizontal branch 3.50 m above the release height of the arrow. Will the arrow go over or under the branch?

Solution:

To illustrate the problem, consider the following figure:

The archer and the target at 75 meter range

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

\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

\begin{align*}

\text{gR} & =\text{v}_{\text{o}}^2\:\sin 2\theta _{\text{o}} \\
\sin \:2\theta _{\text{o}} & =\frac{\text{gR}}{\text{v}_{\text{o}}^2} \\
2\theta _\text{o} & =\sin ^{-1}\left(\frac{\text{gR}}{\text{v}_{\text{o}}^2}\right) \\
\theta _\text{o} & =\frac{1}{2}\sin ^{-1}\left(\frac{\text{gR}}{\text{v}_{\text{o}}^2}\right) \\
\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] \\
\theta _o & =18.46^{\circ} \ \qquad \ {\color{DarkOrange} \left( \text{Answer} \right)}
 
\end{align*}

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

\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

\begin{align*}

\text{v}_{\text{oy}} & =\left(35.0\:\text{m/s}\right)\sin 18.46^{\circ} \\
\text{v}_{\text{oy}} & =11.08\:\text{m/s}
 
\end{align*}

Therefore, the maximum height is

\begin{align*}

\text{h}_{\max } & =\frac{\left(11.08\:\text{m/s}\right)^2}{2\left(9.81\:\text{m/s}^2\right)} \\
\text{h}_{\max } & =6.26\:\text{m} \ \qquad \ {\color{DarkOrange} \left( \text{Answer} \right)}

 
\end{align*}

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

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College Physics by Openstax Chapter 3 Problem 1

Find the following for path A in Figure 3.52:
(a) The total distance traveled, and
(b) The magnitude and direction of the displacement from start to finish.

Figure 3.54 The various lines represent paths taken by different people walking in a city. All blocks are 120 m on a side
Figure 3.52 The various lines represent paths taken by different people walking in a city. All blocks are 120 m on a side

Solution:

Part A

The total distance traveled is 

\begin{align*}

\text{d} & =\left(3\times 120\ \text{m}\right)+\left(1\times 120\:\text{m}\right) \\
\text{d} & =480\:\text{m}  \ \qquad \ {\color{DarkOrange} \left( \text{Answer} \right)}\\

\end{align*}

Part B

The magnitude of the displacement is 

\begin{align*}

\text{s }& =\sqrt{\left( s_x \right)^{2\:}+\left( s_y \right)^2} \\
\text{s }& = \sqrt{\left(1\times 120\:\text{m}\right)^2+\left(3\times 120\:\text{m}\right)^2} \\
\text{s }& = 379\ \text{m} \ \qquad \ {\color{DarkOrange} \left( \text{Answer} \right)}

\end{align*}

The direction is

\begin{align*}

 \theta & = \tan^{-1}\left(\frac{s_x}{s_y}\right) \\
\theta & = \tan^{-1}\left(\frac{1\times 120\:\text{m}}{3\times 120 \ \text{m}}\right) \\
\theta & =71.6^{\circ} ,\:\text{E of N} \ \qquad \ {\color{DarkOrange} \left( \text{Answer} \right)}

\end{align*}
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