A particle projected with velocity $2\sqrt{gh}$ so that it just clears two walls of equal height $h$ which are a distance $2h$ from each other.  The time interval for which the particle travels between this two walls is

Initial velocity $u=2\sqrt{gh}$

Horizontal velocity$=u\cos \theta =2\cos \theta \sqrt{gh}$

 $\therefore$ Hence to cover the interwall distance $t=\frac{2h}{2\cos \theta \sqrt{gh}}$

 $t=\frac{1}{\cos \theta }\sqrt{\frac{h}{g}}-\left(1\right)$

Vertical velocity at the top of it's wall 

$\begin{array}{l}{V}^2=u^2+2as\\ V^2=4gh{\sin }^2\theta -2gh\\ V^2=2gh\left[2{\sin }^2\theta -1\right]\\ V=\sqrt{2gh\left[2{\sin }^2\theta -1\right]}\\ \therefore t=\frac{2V}{g}=\frac{2\sqrt{2gh\left[2{\sin }^2\theta -1\right]}}{g}--\left(2\right)\end{array}$

Fromeq(1) & (2)

$\begin{array}{l}2\sqrt{\frac{2h}{g}\left(2{\sin }^2\theta -1\right)}=\frac{1}{\cos \theta }\sqrt{\frac{h}{g}}\\ 8{\cos }^2\left(2{\sin }^2\theta -1\right)=1\\ 8{\cos }^2\left[1-2{\cos }^2\theta \right]=1\\ -2{\cos }^4\theta +{\cos }^2\theta -\frac{1}{8}=0\\ \therefore \cos \theta =0.5\\ \theta ={60}^{\circ }\\ \text{ From eq }\left(1\right)t=2\sqrt{\frac{h}{g}}\end{array}$