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?

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.

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 $18.4°$.

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.