Q 11-02-071JEE MainJEE Main 2024 (31 Jan, Shift 1)Medium
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 $\dfrac Hh$ for which the body will take the maximum time to reach the ground is ______.
Numerical value type. Enter your answer.
Answer: 2
Time to fall $H - h$: $t_1 = \sqrt{\dfrac{2(H - h)}{g}}$. After the impact the velocity is horizontal, so the remaining fall from height $h$ takes $t_2 = \sqrt{\dfrac{2h}{g}}$.
$$t = \sqrt{\frac2g}\left(\sqrt{H - h} + \sqrt h\right)$$
$\dfrac{dt}{dh} = 0$: $\dfrac{1}{2\sqrt h} = \dfrac{1}{2\sqrt{H - h}} \Rightarrow h = \dfrac H2$, so $\dfrac Hh = 2$.
Solution by Sreeraj P, M.Sc Physics