A tennis ball is dropped on to the floor from a height of 9.8 m. It… | JEE Main 2023 (29 Jan, Shift 1)

Q18JEE Main 2023 · 29 JAN · SHIFT 1Medium

Motion Under Gravity

A tennis ball is dropped on to the floor from a height of 9.8 m. It rebounds to a height 5.0 m. Ball comes in contact with the floor for 0.2s. The average acceleration during contact is ___________ ms $-2$ .

(Given g = 10 ms $-2$ )

Numerical answer type. Work it out, then open the answer.

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Answer: 120

Speed on hitting: $v_1 = \sqrt{2gh_1} = \sqrt{2(10)(9.8)} = 14\text{ m/s}$ (down)
Speed on rebound: $v_2 = \sqrt{2(10)(5)} = 10\text{ m/s}$ (up)
$a = \frac{\Delta v}{\Delta t} = \frac{10 - (-14)}{0.2} = \frac{24}{0.2}$
$a = 120\text{ m s}^{-2}$

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