A body falling on the ground from a height of | vedclass

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Question

Let $v_{1}$ and $v_{2}$ be the velocities of the body just before and just after the collision.

$\mathrm{KE}_{1}=\frac{1}{2} \mathrm{mv}_{1}^{2}=\m

Vedclass · Accepted Answer

$2$

Vedclass · Answer

Let $v_{1}$ and $v_{2}$ be the velocities of the body just before and just after the collision.

$\mathrm{KE}_{1}=\frac{1}{2} \mathrm{mv}_{1}^{2}=\mathrm{mgh}_{1}$            $..(1)$

$\mathrm{KE}_{2}=\frac{1}{2} \mathrm{mv}_{2}^{2}=\mathrm{mgh}_{2}$       $...(2)$

from $(1) \&(2) \Rightarrow \frac{v_{1}^{2}}{v_{2}^{2}}=\frac{h_{1}}{h_{2}}=\frac{10}{2.5}=4$

$\Rightarrow \frac{v_{1}}{v_{2}}=2$

A body falling on the ground from $a$ height of $10\, m$, rebounds to a height $2.5\, m,$ then the ratio of the velocities of the body just before and after the collision will be

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