Question

A particle is projected under gravity with velocity $\sqrt{2ag}$ from a point at a height $h$ above the level plane at an angle $\theta$ to it. The maximum range $R$ on the ground is $\frac{x\sqrt{a(a+h)}}{4}$. Find $x$.

Answer 1

Answer: 8

Co-ordinates of point $P$ are $(R,-h)$.
Hence, $-h=R\tan \theta$
$\Rightarrow R^2{\tan }^2\theta -4aR\tan \theta +\left(R^2-4ah\right)=0$
For $\theta$ to be real, $(4aR{)}^2\ge 4R^2\left(R^2-4ah\right)$
$\Rightarrow 4a^2\ge \left(R^2-4ah\right)$
$\Rightarrow R^2\le 4a(a+h)$
$\Rightarrow R\le 2\sqrt{a(a+h)}$
$\Rightarrow R_{\max }=2\sqrt{a(a+h)}$