Question

A ball of mass 50g is dropped from a height of 20m a boy on the ground hits the ball vertically upward with a bat with an average force of 200N so that it attain a vertical height of45m the time for which the ball remains in contact with the bat is

Answer 1

Velocity of the ball just before hitting the bat = v1 
     v1² = 0² + 2 g S = 2 * 10 * 20 = 400 
   v1 = - 20 m/s    ,  v1 is downwards.

Height attained by the ball after being hit :  45 m
The speed just after being hit by the bat.  =  v2 => 
         0 = v2² - 2 g S       =>  v2² = 2 * 10 * 45 = 900
     v2 = 30 m/sec    upwards

Change in velocity = v2 - v1 = 30 - (-20) = 50 m/s
 change in momentum = m v2 - m v1 = m Δv = 50 /1000 * 50 = 2.5 kg-m/sec

     force = Δp / Δt  = 200 Newtons
         Δt = 2.5 /200 = 1/80 sec  = 0.0125  sec 

Answer 2

1/80th of a second is right answer