Question

A ball of the mass 400gm is dropped from a height of 5m. A boy on the ground hits the ball vertically upwards with a bat with an average force of 100N so that it attains a vertical height of 20m. The time for which the ball remains in contect with the bat is [g=10m//s^(-2)]

Answer 1

Answer:

0.12sec

Explanation:velocity when ball reaches ground is (2gh)^1/2

Which gives V(initial)=10m/s

It reaches 20 m .

Hence

V^2 = U^2 + 2as

Putting values we get U= 20m/s

Therfore change in momentum is

[20-(-10)]×0.4 kg.m/s.=12 kg.m/s

Now, let time of contact be X

Then F×X= change of momentum.

100X= 12

Hence X= 12/100 sec= 0.12seconds