Question

An archer shoots an arrow at a $\mathbf{75}\mathbf{.}\mathbf{0}\mathbf{\text{m}}$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 $\mathbf{35}\mathbf{.}\mathbf{0}\mathbf{\text{m}}\mathbf{/}\mathbf{\text{s}}$? 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$\mathbf{3}\mathbf{.}\mathbf{50}\mathbf{\text{m}}$ above the release height of the arrow. Will the arrow go over or under the branch?

Short answer

(a) The angle at which the arrow must be released to hit the bull’s eye is $18.4°$.

(b) The arrow will go over the branch.

Step-by-step solution

Projectile motion

When a body is shot at a particular beginning velocity at an angle to the horizontal, the acceleration due to gravity acts on the body along the path, and the body follows a parabolic path known as a trajectory. This motion is referred to as projectile motion.

(a) Angle of projection

The horizontal range of the projectile motion is,

$R=\frac{u^2\sin \left(2\theta \right)}{g}$

Here , R is the horizontal range, $\mathit{\theta }$is the angle of projection of the arrow, and g is the acceleration due to gravity.

Rearranging the above equation in order to get an expression for the angle of projection:

$\begin{array}{rcl}\sin \left(2\theta \right)& =& \frac{Rg}{u^2}\\ 2\theta & =& {\sin }^{-1}\left(\frac{Rg}{u^2}\right)\\ \theta & =& \frac{1}{2}{\sin }^{-1}\left(\frac{Rg}{u^2}\right)\\ & & \end{array}$

Substitute $75.0\text{m}$forR , data-custom-editor="chemistry" $9.8\text{m}/{\text{s}}^2$for $g$, and $35.0\text{m/s}$for $u$,

$\begin{array}{rcl}\theta & =& \frac{1}{2}{\sin }^{-1}\left[\frac{\left(75.0\text{m}\right)\times \left(9.8{\text{m/s}}^{\text{2}}\right)}{{\left(35.0\text{m/s}\right)}^2}\right]\\ & =& 18.4°\\ & & \end{array}$

Hence, the angle at which the arrow must be released to hit the bull’s eye is $18.4°$.

(b) Maximum height attained

The maximum height attained by the arrow is:

$\mathit{H}\mathbf{=}\frac{{\mathbf{u}}^{\mathbf{2}}\mathbf{s}\mathbf{i}{\mathbf{n}}^{\mathbf{2}}\mathbf{\theta }}{\mathbf{2}\mathbf{g}}$

Substitute $9.8\text{m}/{\text{s}}^2$for $g$, and $35.0\text{m/s}$for $u$, and$18.4°$for $\theta$,

$\begin{array}{rcl}H& =& \frac{{\left(35.0\text{m/s}\right)}^2\times {\sin }^2\left(18.4°\right)}{2\times \left(9.8{\text{m/s}}^{\text{2}}\right)}\\ & =& 6.23\text{m}\\ & & \end{array}$

Since the height of the branch of the tree is $3.50\text{m}$.

Hence, the arrow will go over the branch.