Question

A ball of mass 50 gram is dropped from a height of 20 metre .A boy on the ground hits the ball vertically upwards with a bat with an the average force of 200 Newton so that it attains a vertically height of 45 M .The time for which ball remain in contact with the bat is

Answer 1

First off, we should consider all the known and unknown quantities.

F = 200 N, mass = 0.05 kg, height it is dropped from = 20 m, height it goes to after being hit = 45 m.

We know, impulse given to a body is equal to the change in momentum of that body.

Velocity of the body when it reaches the ground =2gh−−−√=2gh

=2∗10∗20−−−−−−−−√=2∗10∗20

=20m/s=20m/s

Therefore initial momentum =0.05∗(−20)=0.05∗(−20)

=−1kg−m/s=−1kg−m/s

now the velocity it receives from the bat =2gh−−−√=2gh

=2∗10∗45−−−−−−−−√=2∗10∗45

=30m/s=30m/s

therefore, final momentum =1.5kg−m/s=1.5kg−m/s

change in momentum =2.5kg−m/s=2.5kg−m/s

therefore, impulse=force∗timeimpulse=force∗time

IF=tIF=t

t=2.5/200t=2.5/200

=1.25/100=1.25/100

=0.0125s