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.
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
$\mathrm{u}=\sqrt{10 \mathrm{~g}} ; \mathrm{v}=0 \mathrm{~m} / \mathrm{s} ; \mathrm{s}=0.1 \mathrm{~m}$
$\mathrm{v}^{2}=\mathrm{u}^{2}+2 \mathrm{as}$
$0^{2}=(\sqrt{10 \mathrm{~g}})^{2}+2(a)(0.1)$
$a=-490 \mathrm{~m} / \mathrm{s}^{2}$
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