Question

A particle is projected under gravity with velocity $\sqrt{2 a g}$ 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

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