A ball is thrown vertically downwards from a height of 20 m with an initial velocity . It collides with the ground, loses 50 percent of its energy in collision and rebounds to the same height. The initial velocity is (Take ) [2015]

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Solution

Q.
A ball is thrown vertically downwards from a height of 20 m with an initial velocity $v_{0}$ . It collides with the ground, loses 50 percent of its energy in collision and rebounds to the same height. The initial velocity $v_{0}$ is (Take $g = 10 {ms}^{- 2}$ ) [2015]

1 28 ${ms}^{- 1}$

2 10 ${ms}^{- 1}$

3 14 ${ms}^{- 1}$

4 20 ${ms}^{- 1}$

Ans.
(4)

The situation is shown in the figure.

Let $v$ be the velocity of the ball with which it collides with the ground. Then according to the law of conservation of energy,

Gain in kinetic energy = Loss in potential energy

i.e. $\frac{1}{2} m v^{2} - \frac{1}{2} m v_{0}^{2} = m g h$ (where $m$ is the mass of the ball)

or $v^{2} - v_{0}^{2} = 2 g h$ ...(i)

Now, when the ball collides with the ground, 50% of its energy is lost and it rebounds to the same height $h$ .

$∴$    $\frac{50}{100} (\frac{1}{2} m v^{2}) = m g h$

$\frac{1}{4} v^{2} = g h or v^{2} = 4 g h$

Substituting this value of $v^{2}$ in equation (i), we get

$4 g h - v_{0}^{2} = 2 g h$

or    $v_{0}^{2} = 4 g h - 2 g h = 2 g h or v_{0} = \sqrt{2 g h}$

Here, $g = 10 {ms}^{- 2}$ and $h = 20 m$

$∴$    $v_{0} = \sqrt{2 (10 {ms}^{- 2}) (20 m)} = 20 {ms}^{- 1}$

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