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

R=vo2 sin⁡2θog\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

gR=vo2 sin⁡2θosin⁡ 2θo=gRvo22θo=sin⁡−1(gRvo2)θo=12sin⁡−1(gRvo2)θo=12sin⁡−1[(9.81 m/s2)(75.0 m)(35.0 m/s)2]θo=18.46∘  (Answer)\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

hmax=voy22g\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

voy=(35.0 m/s)sin⁡18.46∘voy=11.08 m/s\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

hmax⁡=(11.08 m/s)22(9.81 m/s2)hmax⁡=6.26 m  (Answer)\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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