Question

A body starts falling freely from height $H$ and hits an inclined plane in its path at height $h$. As a result of this perfectly elastic impact, the direction of the velocity of the body becomes horizontal. The value of $\frac{H}{h}$ for which the body will take the maximum time to reach the ground is _____.

Answer 1

Consider the total time of flight $T$ as:

$T = \sqrt{\frac{2h}{g}} + \sqrt{\frac{2(H - h)}{g}}$

To find the value of $\frac{H}{h}$ that maximizes the time, we differentiate $T$ with respect to $h$ and set it to zero:

$\frac{dT}{dh} = 0 \implies \frac{\sqrt{2}}{g} \left( -\frac{1}{2\sqrt{H - h}} + \frac{1}{2\sqrt{h}} \right) = 0$

Solving for $h$:

$\sqrt{H - h} = \sqrt{h} \implies h = \frac{H}{2} \implies \frac{H}{h} = 2$