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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 t₁ to stop the ball, then the force exerted by the ball on the hands of the cricketer

F₁=  \frac{Change in momentum}{Time Interval}

F₁=  \frac{0-m×u}{t₁}  \frac{-mu}{t₁}

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

F₂=  \frac{-mu}{t₂}

Obviously, when the cricketer withdraws his hands, the ball will take more time to stop.
i.e., t₂ > t₁.
Therefore, F₂ < 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.