91Ó°ÊÓ

The initial speed of the projectile is$v_0$and the projection angle is$\mathrm{θ}$.

The formula to calculate the vertical height travelled by the projectile is given by

$y=v_0\sin \mathrm{θ}t-\frac{1}{2}g{t}^2...................\left(1\right)$

Here, $y$is the vertical distance and $t$is the time taken by the object to cover the maximum range.

Substitute $0$for $y$into equation (1) and solve to calculate the required time taken by the projectile.

$\begin{array}{rcl}0& =& v_0\sin \mathrm{θ}t-\frac{1}{2}g{t}^2\\ t& =& \frac{2v_0\sin \mathrm{θ}}{g}........................\left(2\right)\end{array}$

The formula to calculate the horizontal distance $\left(R\right)$travelled by the projectile is given by

role="math" localid="1647405477077" $R=v_0\cos \mathrm{θ}t-\frac{1}{2}a{t}^2.................\left(3\right)$

Substitute the value of time from equation (2) into equation (3) and solve to obtain the horizontal distance travelled.

role="math" localid="1647405893350" $\begin{array}{rcl}R& =& v_0\cos \mathrm{θ}\left(\frac{2v_0\sin \mathrm{θ}}{g}\right)-\frac{1}{2}a{\left(\frac{2v_0\sin \mathrm{θ}}{g}\right)}^2\\ & =& \frac{{v_0}^2}{g}\left(\sin 2\mathrm{θ}-\frac{a}{g}2{\sin }^2\mathrm{θ}\right)\\ & =& \frac{{v_0}^2}{g}\left(\sin 2\mathrm{θ}-\frac{a}{g}\left(1-\cos 2\mathrm{θ}\right)\right)\\ & =& \frac{{v_0}^2}{g}\left(\sin 2\mathrm{θ}+\frac{a}{g}\cos 2\mathrm{θ}\right)-\frac{{v_0}^2}{g^2}a\\ & =& \frac{{v_0}^2}{g^2}\sqrt{a^2+g^2}\left(\frac{g}{\sqrt{a^2+g^2}}\sin 2\mathrm{θ}+\frac{a}{\sqrt{a^2+g^2}}\cos 2\mathrm{θ}\right)-\frac{{v_0}^2}{g^2}a.............\left(4\right)\end{array}$

Substitute $\tan \mathrm{Î\pm }=\frac{a}{g}$into equation (4) and simplify to obtain the range.

$\begin{array}{rcl}R& =& \frac{{v_0}^2}{g^2}\sqrt{a^2+g^2}\left(\frac{g}{\sqrt{a^2+g^2}}\sin 2\mathrm{θ}+\frac{a}{\sqrt{a^2+g^2}}\cos 2\mathrm{θ}\right)-\frac{{v_0}^2}{g^2}a\\ & =& \frac{{v_0}^2}{g^2}\sqrt{a^2+g^2}\sin \left(2\mathrm{θ}+\mathrm{Î\pm }\right)-\frac{{v_0}^2}{g^2}a......................\left(5\right)\end{array}$

The range of the projectile is maximum when

$\sin \left(2\mathrm{θ}+\mathrm{Î\pm }\right)=1...................\left(6\right)$

Solve equation (6) to obtain the required launch angle.

$\begin{array}{rcl}\sin \left(2\mathrm{θ}+\mathrm{Î\pm }\right)& =& 1\\ & =& \sin \frac{\mathrm{Ï€}}{2}\\ 2\mathrm{θ}+\mathrm{Î\pm }& =& \frac{\mathrm{Ï€}}{2}\\ \mathrm{θ}& =& \frac{1}{2}\left(\frac{\mathrm{Ï€}}{2}-\mathrm{Î\pm }\right)\\ & =& \frac{\mathrm{Ï€}}{4}-{\tan }^{-1}\left(\frac{a}{g}\right).........................\left(7\right)\end{array}$

The acceleration$\left(a\right)$ opposing the horizontal motion of the projectile is$10\%$of$g$.

Substitute $10\%g$for $a$into equation (7) and solve to calculate the required launch angle.

$\begin{array}{rcl}\mathrm{θ}& =& \frac{\mathrm{π}}{4}-{\tan }^{-1}\left(\frac{10\%g}{g}\right)\\ & ≈& 39.3°\end{array}$