91Ó°ÊÓ

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,

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

Here , R is the horizontal range, $\mathit{θ}$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:

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

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 $u$,

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

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:

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

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

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

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

Hence, the arrow will go over the branch.