A ball is falling freely under gravity from a height $10 m$ with an initial downward velocity $v_{0}$ . If collides with the ground loses $50 %$ of its kinetic energy in collision and then rises back to the same height. Then find the initial velocity $v_{0}$ .

10 m / s

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12 m / s

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10 \sqrt{2} m / s

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20 \sqrt{2} m / s

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Solution

The correct option is C $10 \sqrt{2} m / s$
we are going to solve this in reverse.
Minimum velocity $v_{3}$ required to raise at height $10 m$ .
from conservation of energy.

Kinetic energy = potential energy

$\frac{1}{2} m v_{3}^{2} = m g h$

$v_{3} = \sqrt{2 g h} = \sqrt{2 \times 10 \times 10} = 10 \sqrt{2} m s^{- 1}$

Striking kinetic energy is double of the rebound kinetic energy.

$K . E_{s t r i k e} = 2 \times K . E_{r e b o u n d}$

$\frac{1}{2} m v_{2}^{2} = 2 \times \frac{1}{2} m v_{3}$

$v_{2} = v_{3} \sqrt{2} = 10 \sqrt{2} \times \sqrt{2} = 20 m s^{- 1}$

Velocity of striking is $v_{2} = 20 m s^{- 1}$

Apply kinematic equation of motion

$v_{2}^{2} - v_{0}^{2} = 2 a s$

$20^{2} - v_{0}^{2} = 2 \times 10 \times 10$

$v_{0} = 10 \sqrt{2} m s^{- 1}$

Initial velocity is $10 \sqrt{2} m s^{- 1}$ .

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A ball is falling freely under gravity from a height 10 m with an initial downward velocity v0. If collides with the ground loses 50% of its kinetic energy in collision and then rises back to the same height. Then find the initial velocity v0.

A ball is falling freely under gravity from a height $10 m$ with an initial downward velocity $v_{0}$ . If collides with the ground loses $50 %$ of its kinetic energy in collision and then rises back to the same height. Then find the initial velocity $v_{0}$ .