A ball of mass 400g dropped from a height of 5m. A boy on the ground hits the ball vertically upwards with a bat with an force of 100N so that it attains a vertically height of 20m. The time for which the ball remains in contact with the bat? (g = 10m/s^{2})

the initial momentum of the ball will be

p_{i} = mv_{i}

here,

m = mass of the ball = 400g = 0.4kg

v_{i} = initial velocity = [2gh_{i}]^{1/2}

{h_{i} = 5m}

= [2 x 10 x 5]^{1/2} = 10 m/s

so,

p_{i} = 0.4 x 10

thus,

p_{i} = 4 kg.m/s

and

the initial momentum of the ball will be

p_{f} = mv_{f}

here,

v_{f} = final velocity = [2gh_{f}]^{1/2}

{h_{f} = 20m}

= [2 x 10 x 20]^{1/2} = 20 m/s

so,

p_{f} = 0.4 x 20

thus,

p_{f} = 8 kg.m/s

now,

we know from Newton's Second Law

force = rate of change of momentum

F = dp/dt = [p_{f} - p_{i}] / dt

or

{F = 100 N}

dt = [p_{f} - p_{i}] / F

= [8 - (-4)] / 100

{negative sign because p_{i} is directed downwards; convention}

thus, contact time of ball with bat will be

**dt = 0.12 s**

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