JEE Main · 2023 · 29 Jan 2023 · Shift 1 · Conservation of Momentum

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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 0.2 s. The average acceleration during contact is ______

m s^{- 2}

. [Given

g = 10 m s^{- 2}

]

Answer: 120

Solution

The ball strikes the floor at

v_{i} = 2 g h_{i} = 2 (10) (9.8) = 14

m s

^{- 1}

downward and leaves at

v_{f} = 2 g h_{f} = 2 (10) (5) = 10

m s

^{- 1}

upward. Taking upward positive, the change in velocity is

10 - (- 14) = 24

m s

^{- 1}

over the

0.2

s of contact. The average acceleration during contact is

120

m s

^{- 2}

JEE Main · 2023 · 29 Jan 2023 · Shift 1 · Integer answer

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