A particle aimed at a target, projected with an angle $15^{\circ}$ with the horizontal is short of the…
A particle aimed at a target, projected with an angle $15^{\circ}$ with the horizontal is short of the target by $10 \mathrm{~m}$. If projected with an angle of $45^{\circ}$ is away from the target by $15 \mathrm{~m}$, then the angle of projection to hit the target is
$\frac{1}{2} \sin ^{-1}\left(\frac{1}{10}\right)$
$\frac{1}{2} \sin ^{-1}\left(\frac{3}{10}\right)$
$\frac{1}{2} \sin ^{-1}\left(\frac{9}{10}\right)$
$\frac{1}{2} \sin ^{-1}\left(\frac{7}{10}\right)$
Solution
As Range $\propto \sin 2 \theta$
So, $\frac{R-10}{R+15}=\frac{\sin 2 \theta_1}{\sin 2 \theta_2}=\frac{\sin 30^{\circ}}{\sin 90^{\circ}}=\frac{1}{2}$
$\Rightarrow \quad 2 R-20=R+15 \Rightarrow R=35$
For the range to be maximum $\sin 2 \theta=\sin 90^{\circ}=1$
So, $R_{\max }=\frac{u^2}{g}$
As $R_1=\frac{u^2 \sin 2 \theta}{g} \Rightarrow 50=\frac{u^2}{g} \Rightarrow u^2=50 g$
and $35=\frac{50 g \times \sin 2 \theta}{g}[\because u=50 g]$
$\therefore \quad \sin 2 \theta=\frac{35}{50}=\frac{7}{10} \Rightarrow 2 \theta=\sin ^{-1}\left[\frac{7}{10}\right]$
$\Rightarrow \quad \theta=\frac{1}{2} \sin ^{-1}\left[\frac{7}{10}\right]$