Two balls of same mass are dropped from the same height h , on to the floor . the first ball bounces to a height h/4, after the collection & the second ball to a height h/16. the impulse applied by the first & second ball on the floor are ;${I_1}$ & ;and ${I_2}$ respectively . then

When the ball is dropped from height h its velocity ${v}_{1}=\sqrt{2gh}$

Now velocity of the ball after the bounce, where height $h=\dfrac{h}{4}$,velocity${v}_{1a}=\sqrt{2g\dfrac{h}{4}}$

Now impulse ${I}_{1}={F}_{1}t=mat=m({v}_{1}-{v}_{1a})=m(\sqrt { 2gh } -\sqrt { 2g\dfrac { h }{ 4 } } )=m({ v }_{ 1 }-\dfrac { 1 }{ 2 } { v }_{ 1 })=\dfrac { 1 }{ 2 } m{ v }_{ 1 }$

Similarly for the ball when reaches a height of $\dfrac{h}{4}$ after bounce,velocity${v}_{2a}=\sqrt{2g\dfrac{h}{16}}$ and velocity of the ball before

bounce${v}_{2}=\sqrt{2gh}$

Similarly the impulse of the second ball${I}_{2}=\dfrac{3}{4}{v}_{2}m$

Given mass of both the ball are same.

We can compare both the impulses as

$\dfrac{4{I}_{2}}{3{v}_{2}}=\dfrac{2{I}_{2}}{{v}_{1}}$

Now as both ${v}_{2}={v}_{1}=\sqrt{2gh}$

$3{I}_{1}=2{I}_{2}$