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Question

(A) Let the body fall from height $H$. It hits the inclined plane at a height $h$ from the ground,meaning it has fallen a distance of $(H-h)$.The time

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$2$

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(A) Let the body fall from height $H$. It hits the inclined plane at a height $h$ from the ground,meaning it has fallen a distance of $(H-h)$.The time taken to fall the distance $(H-h)$ is $t_1 = \sqrt{\frac{2(H-h)}{g}}$.At this point,the impact is perfectly elastic and the velocity becomes horizontal. The body then falls from height $h$ with an initial vertical velocity of $0$. The time taken to fall this remaining distance $h$ is $t_2 = \sqrt{\frac{2h}{g}}$.The total time of flight $T$ is $T = t_1 + t_2 = \sqrt{\frac{2(H-h)}{g}} + \sqrt{\frac{2h}{g}}$.To find the value of $h$ for which $T$ is maximum,we differentiate $T$ with respect to $h$ and set it to $0$:$\frac{dT}{dh} = \sqrt{\frac{2}{g}} \left( \frac{1}{2\sqrt{H-h}} \cdot (-1) + \frac{1}{2\sqrt{h}} \right) = 0$.This implies $\frac{1}{\sqrt{h}} = \frac{1}{\sqrt{H-h}}$,which gives $h = H - h$,or $2h = H$.Therefore,$\frac{H}{h} = 2$.

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