A ball of mass 400 gm is dropped from a height of 5m. A boy on the ground hits the ball vertically upwards with a bat with an average force of 100N so that it attains a vertical height of 20m. Find the time (in s) for which the ball remains in contact with the bat.............. × 10^–2

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answer is 12.

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Detailed Solution

$u^{2} = 2 {gh}_{1} \Rightarrow u = 10 m / s downwards$

$v^{2} = 2 {gh}_{2} = 20 m / s upwards$

Change in momentum=impulse=m(v+u)
=0.4(10+20)

=0.4×30

=12 kg m/s

Force =100N=12/t

t=time of contact =0.12 sec.

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