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/s²)

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