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

Correct Answer: 2

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 \( h \):

\[ \sqrt{H - h} = \sqrt{h} \implies h = \frac{H}{2} \implies \frac{H}{h} = 2 \]