A body falling freely from a given height $H$ hits an inclined plane in its path at a height $h$ . As a result of this impact the direction of the velocity of the body becomes horizontal. For what value of  $(H/h)$ the body will take maximum time to reach the ground?

 Let the body is released from $A$ , it strikes the inclined plane at  $B$ and finally reaches the ground at  $C.$

The time taken by the body to travel from $A$  to  $B$  is given by  $t_{\text{AB}}=\sqrt{2(H-h)}/g.$
The direction of velocity of the body becomes horizontal after striking at  . Thus, the time taken by the body to travel from   $B$ to $C$  is  $t_{\text{BC}}=\sqrt{2h}/g.$
Thus, the total time taken by the body to reach the ground is

$t=t_{\text{AB}}+t_{\text{BC}}=\sqrt{2(H-h)}/g+\sqrt{2h}/g.$ 
The value of $h$   for maximizing $t$ is given by $\text{d}t/\text{d}h=0$ i.e., $\frac{\text{d}t}{\text{d}h}=\sqrt{\frac{2}{g}}[\frac{-1}{\sqrt{(H-h)}}+\frac{1}{\sqrt{h}}]=0.$
Solve to get  $h=H/2.$