A ball of mass 50g is dropped from a height of 20m.A boy on the ground hits the ball vertically upwards with a bat with an average force of 200N so that it attain a vertical height of 45m. The time for which the ball remains in contact with bat is

From up to down

u1= velocity up = 0

V1= velocity before hitting bat

s1= distance before hitting bat

v1^2 = u1^2 + 2as

v1^2 = 0+ 2*10*45

V1= 20 m/s

F=M*A

A= 200/ 50*10^-3

=4000

U2 = velocity after hitting bat

V2= velocity at top = 0

s2 = distance covered after hitting bat = 45

u2^2= -2*10*45

u2= -30

time :

-30= 20 - 4000*t

t= (-30–20/-4000)

= 1/80th of a second 

Regards

Arun(askIITians forum expert)