Thank you for reporting, we will resolve it shortly
Q.
A particle is projected with velocity $\sqrt{2gh}$ , such that it just crosses two walls of height $ h $ and separated by $ h $ . Find the angle of projection.
For vertically upward motion of a projectile,
$y=(u \sin \alpha) t-\frac{1}{2} g t^{2}$
or $h=(u \sin \alpha) t-\frac{1}{2} g t^{2}$
or $g t^{2}-(2 u \sin \alpha) t+2 h=0$ ...(i)
$\therefore t_{1}=\frac{2 u \sin \alpha \pm \sqrt{\left(4 u^{2} \sin ^{2} \alpha\right)-8 g h}}{2 g}$
If two roots of quadratic Eq. (i) are $t_{1}, t_{2}$, then
$t=\frac{2 u \sin \alpha+\sqrt{\left(4 u^{2} \sin ^{2} \alpha-8 g h\right.}}{2 g}$
and $t_{2}=\frac{2 u \sin \alpha-\sqrt{\left(4 u^{2} \sin ^{2} \alpha\right)-8 g h}}{2 g}$
If particle crosses the walls at times $t_{1}$ and $t_{2}$ respectively, then time of flight $t$ is $t=\sqrt{t_{1} t_{2}}$
or $t^{2}=t_{1} t_{2}$
$\therefore \left(\frac{2 u \sin \alpha}{g}\right)^{2}=\frac{(2 u \sin \alpha)^{2}-\left(4 u^{2} \sin ^{2} \alpha-8 g h\right)}{4 g^{2}}$
or $\frac{4 u^{2} \sin ^{2} \alpha}{g^{2}}=\frac{8 g h}{4 g^{2}}$
or $2 u^{2} \sin ^{2} \alpha=g h$
Given, $u=\sqrt{2 g h}$
$\therefore 2(2 g h) \sin ^{2} \alpha=g h$
or $\sin ^{2} \alpha=\frac{1}{4}$
or $\sin \alpha=\frac{1}{2}$
$\therefore \alpha=30^{\circ}$