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JEE Main202329 Jan 2023Morning ShiftPhysicsMotion in One DimensionActual

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 ]

Correct answer

0

Step-by-step solution

The speed of ball just before collision with ground is v i 2 = 0 + 2 g h i ⇒ v i = 2 g h i = 2 × 10 × 9 . 8 = 14   m   s - 1     Downward The speed of ball just after collision is 0 = v f 2 - 2 g h f ⇒ v f = 2 g h f = 2 × 10 × 5 = 10   m   s - 1     Upward Average acceleration of ball is a → avg = Δ v → Δ t = v f - - v i t = 10 + 14 0 . 2 = 24 0 . 2 = 120   m   s - 2 .

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