# Solve the following :

Question:

A ball is dropped from a height of $5 \mathrm{~m}$ onto a sandy floor and penetrates the sand up to $10 \mathrm{~cm}$ before coming to rest. Find the retardation of the ball in sand assuming it to be uniform.

Solution:

For ball in air

$\mathrm{u}=0 \mathrm{~m} / \mathrm{s} ; \mathrm{a}=\mathrm{g} ; \mathrm{s}=5 \mathrm{~m}$

$s=u t+\frac{1}{2} a t^{2}$

$5=0+\frac{1}{2}(g) t^{2}$

$t=\sqrt{\frac{10}{g}}$

$\mathrm{v}=\mathrm{u}+\mathrm{at}$

$\mathrm{V}=0+\mathrm{g} \sqrt{\frac{10}{\mathrm{~g}}}$

$\mathrm{v}=\sqrt{10 \mathrm{~g}}$.

For ball in sand

$\mathrm{u}=\sqrt{10 \mathrm{~g}} ; \mathrm{v}=0 \mathrm{~m} / \mathrm{s} ; \mathrm{s}=0.1 \mathrm{~m}$

$\mathrm{v}^{2}=\mathrm{u}^{2}+2 \mathrm{as}$

$0^{2}=(\sqrt{10 \mathrm{~g}})^{2}+2(a)(0.1)$

$a=-490 \mathrm{~m} / \mathrm{s}^{2}$