Question

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 $\left(\frac{h}{H}\right)$ the body will take maximum time to reach the ground?

Answer 1

Let the body covers $H-h$ distance in time $t_A$. Then using $2^{nd}$ equation of motion taking into consideration initial velocity of the object to be zero,
$(H-h)=\frac{1}{2}g{t}_A^2$
$\begin{array}{}\Rightarrow t_A=\sqrt{\frac{2(H-h)}{g}}& \text{(1)}\end{array}$
After the impact let the body covers $h$ vertical displacement in time $t_B$, hence
$h=\frac{1}{2}g{t}_B^2$
$\begin{array}{}\Rightarrow t_B=\sqrt{\frac{2h}{g}}& \text{(2)}\end{array}$
Total time$\left(T\right)$ to reach the ground,
$t_A+t_B=\sqrt{\frac{2(H-h)}{g}}+\frac{2h}{g}$
For maximum time,
$\frac{dT}{dh}=0$
$\frac{dT}{dh}=\frac{-1}{\sqrt{2(H-h)g}}+\frac{1}{2gh}=0$
$\Rightarrow 2(H-h)=2h$
$\Rightarrow \frac{h}{H}=\frac{1}{2}$