Question

A ball of mass 400 gm is dropped from a height of 5 m. A boy on the ground hits the ball vertically upwards with a ba with an average force of 100 newton so that it attains a vertical height of 20 m. The time for which the ball remains in contact with the bat is (g = 10 m/s²)(a) 0.12 s(b) 0.08 s(c) 0.04 s(d) 12 s

Answer 1

it should be the answer

Answer 2

Answer:

A) 0.12 s

Explanation:

Mass = 400gm = 0.4 kg (Given)  

h1 = 5m  (Given)  

h2 = 20 m  (Given)  

g = 10 m/s^2

Velocity of the ball after dropping from 5m =  u

Velocity of the ball after being hit by the bat = v

Thus,

u² = 2 g h1  = u = √(2*10*5) = 10 m/s  (downwards)

v² = 2 g h2 = (2 *10 *20)   = v = 20 m/s  (upwards)

Change of momentum = impulse = m(v+u) = 12 kg-m/s

Force = 100 N = 12 / t

T = time of contact = 0.12 sec

Thus, the time for which the ball will remain in contact with the bat is 0.12 seconds