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MHT CET 2021 (23 Sep, Shift 1) · Motion Under Gravity

Motion Under Gravity question from MHT CET 2021 (23 Sep, Shift 1)

Q31MHT CET 2021 · 23 SEP · SHIFT 1Easy

Motion Under Gravity

A student is throwing balls vertically upwards such that he throws the $2$ nd ball when the $1$ st ball reaches maximum height. If he throws balls at an interval of 3 second, the maximum height of the balls is $\left(g = 10\,\frac{\text{m}}{\text{s}^2}\right)$

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Answer: (a)

The next ball is thrown when the previous one is at the top, so time to reach the top $= 3$ s.
$u = gt = 10 \times 3 = 30$ m/s
$H = \frac{u^2}{2g} = \frac{900}{20} = 45$ m
So the answer is (a) 45 m.

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