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15 February, 21:17

A ball of mass 0.15 kg is dropped from rest from a height of 1.25 m. It rebounds from the floor to reach a height of 0.960 m. b). (Using conservation of energy on the way up) What is the velocity of the ball right after the collision?

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  1. 15 February, 21:31
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    If v be the velocity just after the rebound

    Kinetic energy will be converted into potential energy

    1/2 m v² = mgh

    v² = 2gh

    v = √ 2gh

    = √ 2 x 9.8 x. 96

    = 4.33 m / s
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