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JEE Main202117 Mar 2021Evening ShiftPhysicsCenter of Mass, Momentum and CollisionActual

A rubber ball is released from a height of 5 m above the floor. It bounces back repeatedly, always rising to 81 100 of the height through which it falls. Find the average speed of the ball. (Take g = 10 m s - 2 )

Options

  1. A3 . 0   m   s - 1
  2. B3 . 5   m   s - 1
  3. C2 . 0   m   s - 1
  4. D2 . 50   m   s - 1

Correct answer

D. 2 . 50   m   s - 1

Step-by-step solution

v 0 = 2 g h v = e 2 g h = 2 g h ⇒ e = 0 . 9 S = h + 2 e 2 h + 2 e 4 h + … … … t = 2 h   g + 2 e 2 h   g + 2 e 2 2 h   g + … … … v av = s t = 2 . 5   m   s - 1

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