A ball of mass 50g isdropped from a height of 20m. A boy on the ground hits the ball with an average force of 200N, so that it attains a vertical height of 45m. The time for which the ball remains in contact with the bat is (g=10m/s^2)A)1/20th ofa second B)1/40th of a secondC)1/80th of a secondD)1/120th of a second

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