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Q.
A ball of mass $0.15\, kg$ is dropped from $a$ height $10\, m$, strikes the ground and rebounds to the same height. The magnitude of impulse imparted to the ball is $\left( g =10\, m / s ^{2}\right)$ nearly :
Velocity while hitting the ground
$v =\sqrt{2 \times 10 \times 10}=10 \sqrt{2}$
As it goes to the same height it will return with same speed.
So change in velocity $v-(-v)=2 v$
Change in momentum or impulse
$=2 \,mv $
$=2 \times 0.15 \times 10 \sqrt{2}$
$=3 \sqrt{2}=4.2\, kg\,m / s$