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MHT CET 2026 (16 Apr, Shift 1) · Motion Under Gravity

Motion Under Gravity question from MHT CET 2026 (16 Apr, Shift 1)

Q9MHT CET 2026 · 16 APR · SHIFT 1Medium

Motion Under Gravity

A ball is released from height 'h' which makes perfectly elastic collision with ground. The frequency of periodic vibratory motion is (g=acceleration due to gravity)

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Answer: (a)

Time to fall height $h$ : $t = \sqrt{\frac{2h}{g}}$
Elastic bounce: it rises back to $h$ in the same time, so period $T = 2\sqrt{\frac{2h}{g}}$
$f = \frac{1}{T} = \frac{1}{2}\sqrt{\frac{g}{2h}}$
So the answer is (a) $\frac{1}{2}\sqrt{\frac{g}{2h}}$ .

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