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 $(h/H)$ the body will take maximum time to reach the ground?

Answer 1

Answer: 0.5

(a) Range of both the particles is
$R=\frac{u^2\sin 2\theta }{g}=\frac{(49{)}^2\sin 90^{/circ}}{9.8}$
By symmetry we can say that they will collide at highest point.

Just before collision Just after collision
Let $v$ be the velocity of $Q$ just after collision.
Then, from conservation of linear momentum, we have
$20(u\cos 45^{\circ })-40(u\cos 45^{\circ })=40(v)$
$-20(u\cos 45^{\circ }$)
$\therefore v=0$
i.e. particle $Q$ comes to rest. So, particle $Q$ will fall vertically downwards and will strike just midway between A and B.
(b) Maximum height,
$H\frac{u^2{\sin }^2\theta }{2g}=\frac{(49{)}^2{\sin }^{2}45^{\circ }}{2\times 9.8}=61.25m$
Therefore, time taken by $Q$ to reach the ground,
$t=\sqrt{\frac{2H}{g}}=\sqrt{\frac{2\times 61.25}{9.8}}=3.52s$