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2014-07-21T16:06:04+05:30

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While catching a ball, the cricketer withdraws his hands.
Let a ball of mass 'm' be moving with velocity 'u' before catching it.
Its initial momentum = m × u
When the cricketer stops the ball (v=0), 
Its final momentum = m × 0 =0
If the cricketer does not withdraws his hands, it takes time t
₁ to stop the ball, then the force exerted by the ball on the hands of the cricketer

F
₁= \frac{Change in Momentum}{Time Interval}

F
₁= \frac{0-m*u}{t} = \frac{-mu}{t}

But if the cricketer withdraws his hands it takes longer time t₂ to stop the ball.
The force exerted by ball on his hands will be

F
₂= \frac{-mu}{t}

Obviously, when the cricketer withdraws his hands, the ball will take more time to stop.
i.e., t
₂ > t₁.
Therefore, F
₂ < F₁ or force exerted on hands of the cricketer is less when he withdraws his hands.
Thus the cricketer avoids the chances of injury to his hands by withdrawing his hands while catching a ball.


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2014-07-21T16:09:34+05:30
The cricketers move his hands back to hold the ball as by doing that he speed of the ball speeds down because of which it becomes easier for the catcher to catch he ball.
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