A body falling freely from a given height $H$ hits an inclined plane in its path at a height $h$. As

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?

Professor PC Thomas Classes · Answer

(a) : $P Q R$ is an inclined plane. The body falls under gravity vertically from $A$ to $B$ till it strikes the inclined plane. At $B$, its velocity is horizontal. From $B$ to $C$, the body follows a parabolic path. At $C$, it reaches the ground.\ Let time for $A$ to $B=t_1$\ Let time for $B$ to $C=t_2$ \  $\because \quad S=u t+\frac{1}{2} g t^2$\ $	herefore \quad(H-h)=\left(0 	imes t_1ight)+\frac{1}{2} g t_1^2$\ or $\quad t_1=\sqrt{\frac{2(H-h)}{g}}$\ Similarly, $t_2=\sqrt{\frac{2 h}{g}}$\ $	herefore \quad$ Total time $t=t_1+t_2=\sqrt{\frac{2}{g}}[\sqrt{(H-h)}+\sqrt{h}]$\ $	herefore \quad \frac{d t}{d h}=\sqrt{\frac{2}{g}}\left[\frac{-1}{2 \sqrt{H-h}}+\frac{1}{2 \sqrt{h}}ight]$\ For $t$ to be maximum, $\frac{d t}{d h}=0$\ $	herefore \quad 0=\sqrt{\frac{2}{g}}\left[\frac{-1}{2 \sqrt{H-h}}+\frac{1}{2 \sqrt{h}}ight]$\ or $H-h=h \Rightarrow H=2 h$\ $\Rightarrow \quad \frac{h}{H}=\frac{1}{2} \quad 	herefore \quad \frac{h}{H}=\frac{1}{2}$.