Question

A ball of mass 50 g 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 icontact with the bat is (Take g=10ms^-2)

Answer 1

Answer on Question 38125, Physics, Mechanics The momentum of the ball right before the hit is

$p_{1}=-mv=-m\sqrt{2gh_{1}}$

The momentum just after the hit is

$p_{2}=mv=m\sqrt{2gh_{2}}$

Change of momentum is

$\Delta p=p_{2}-p_{1}=m\sqrt{2g}(\sqrt{h_{2}}+\sqrt{h_{1}})$

Hence needed time is

$t=\frac{\Delta p}{F}=\frac{m\sqrt{2g}(\sqrt{h_{2}}+\sqrt{h_{1}})}{F}=\frac{0.05\sqrt{2\cdot 10}(\sqrt{45}+\sqrt{20})}{200}=0.0125\,s$