Question

The horizontal range of projectile is $2\sqrt{3}$ times its maximum height. Find the angle of projection.

Answer 1

If $u$ and $\alpha$ are the initial velocity of projection and angle of projection, respectively, then
Maximum height attained = $\frac{u^{2}si{n}^2\alpha }{2g}$
Horizontal range = $\frac{2u^{2}sin\alpha cos\alpha }{g}$
According to the problem,
$\frac{2u^{2}sin\alpha cos\alpha }{g}=2\sqrt{3}\left(\frac{u^{2}si{n}^2\alpha }{2g}\right)\Rightarrow tan\alpha =\left(\frac{2}{\sqrt{3}}\right)$
$\Rightarrow \alpha =ta{n}^{-1}\left(\frac{2}{\sqrt{3}}\right)$