JEE Mains · Physics · STD 11 - Work,energy,power and collision
- JEE Mains 2024
- Hard
- Numerical
A body starts falling freely from height \(\mathrm{H}\) hits an inclined plane in its path at height \(\mathrm{h}\). As a result of this perfectly elastic impact, the direction of the velocity of the body becomes horizontal. The value of \(\frac{\mathrm{H}}{\mathrm{h}}\) for which the body will take the maximum time to reach the ground is _______.
Pick an option to check your answer.
Answer & Solution
Correct Answer
(A) \(2\)
Step-by-step Solution
Detailed explanation
Total time of flight \(=\mathrm{T}\)
\(T=\sqrt{\frac{2 h}{g}}+\sqrt{\frac{2(H-h)}{g}}\)
For \(\max\) time \(=\frac{d T}{d h}=0\)
\(\sqrt{\frac{2}{g}}\left(\frac{-1}{2 \sqrt{\mathrm{H}-\mathrm{h}}}+\frac{1}{2 \sqrt{\mathrm{h}}}\right)=0\)
\(\sqrt{\mathrm{H}-\mathrm{h}}=\sqrt{\mathrm{h}}\)
\(\mathrm{h}=\frac{\mathrm{H}}{2} \Rightarrow \frac{\mathrm{H}}{\mathrm{h}}=2\)

\(T=\sqrt{\frac{2 h}{g}}+\sqrt{\frac{2(H-h)}{g}}\)
For \(\max\) time \(=\frac{d T}{d h}=0\)
\(\sqrt{\frac{2}{g}}\left(\frac{-1}{2 \sqrt{\mathrm{H}-\mathrm{h}}}+\frac{1}{2 \sqrt{\mathrm{h}}}\right)=0\)
\(\sqrt{\mathrm{H}-\mathrm{h}}=\sqrt{\mathrm{h}}\)
\(\mathrm{h}=\frac{\mathrm{H}}{2} \Rightarrow \frac{\mathrm{H}}{\mathrm{h}}=2\)

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