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A ball is dropped onto the floor from a height of 10 m. It rebounds to a height of 5 m. If the ball was in contact with the floor for 0.01 s, what was its average acceleration during contact? (Take, $g = 10 {ms}^{- 2}$ )

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$3120 {ms}^{- 2}$

$2414 {ms}^{- 2}$

$4105 {ms}^{- 2}$

$1735 {ms}^{- 2}$

answer is A.

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Detailed Solution

$a_{av} = \frac{v_{f} - v_{i}}{Δ t}$ (As they are in opposite direction)

$= \frac{\sqrt{2 g h_{f}} + \sqrt{2 g h_{i}}}{Δ t} [a s t h e v e l o c i t i e s a r e i n o p p o s i t e d i r e c t i o n s] = \frac{\sqrt{2 \times 10 \times 5} + \sqrt{2 \times 10 \times 10}}{0.01}$

$= 2414.21 \approx 2414 {ms}^{- 2}$

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