JEE MAIN - Physics (2024 - 31st January Morning Shift - No. 25)
A body starts falling freely from height $$H$$ 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
2
Explanation
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Total time of flight $$=\mathrm{T}$$
$$T=\sqrt{\frac{2 h}{g}}+\sqrt{\frac{2(H-h)}{g}}$$
For max. time $$=\frac{\mathrm{dT}}{\mathrm{dh}}=0$$
$$\begin{aligned} & \sqrt{\frac{2}{\mathrm{~g}}}\left(\frac{-1}{2 \sqrt{\mathrm{H}-\mathrm{h}}}+\frac{1}{2 \sqrt{\mathrm{h}}}\right)=0 \\ & \sqrt{\mathrm{H}-\mathrm{h}}=\sqrt{\mathrm{h}} \\ & \mathrm{h}=\frac{\mathrm{H}}{2} \Rightarrow \frac{\mathrm{H}}{\mathrm{h}}=2 \end{aligned}$$
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