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 attains a vertical height of 45m. The time for which the ball remains in contact with the bat is (g= 10m/s^2)
Options are
A)1/20th of a second B)1/40th of a second
C)1/80th of a second D)1/120th of a second

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Answers

2014-12-19T04:41:27+05:30

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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

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