A ball is dropped on the floor from a height of 10 m. It rebounds to a height

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A ball is dropped on the floor from a height of 10 m. It rebounds to a height of 2.5 m. If the ball is in contact with the floor for 0.01 sec, the average acceleration during contact is

Options:

A .

2100 m / s e c^{2}

downwards

B .

2100 m / s e c^{2}

upwards

C .

1400 m / s e c^{2}

D .

700 m / s e c^{2}

Answer: Option B
:
B

Velocity at the time of striking the floor,

$u = \sqrt{2 g h} = \sqrt{2 \times 9.8 \times 10} = 14 m / s$ (-ve since downwards)

Velocity with which it rebounds.

$v = \sqrt{2 g h_{2}} = \sqrt{2 \times 9.8 \times 2.5} = 7 m / s$ (+ve since upwards)

$∴$ Change in velocity $△ v = 7 - (- 14) = 21 m / s$

$∴$ Acceleration = $\frac{△ v}{△ t} = \frac{21}{0.01} = 2100 m / s^{2}$ (upwards since positive)

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