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Two stones are thrown vertically upwards simultaneously with their initial velocities u1 and u2respectively. Find the ratio of the height reached by the two stones.
In this case, acceleration will be 'g' in downwards
We know that, $v^{2} \ =\ u^{2} \ - \ 2gh$ or $ h = \frac{u^{2} \ -\ v^{2}}{2g}$
But at the highest point, $v\ =\ 0$
Therefore, $h\ =\ \frac{u^{2}}{2g}$
For first ball, $h_{1} \ =\ \frac{u^{2}_{1}}{2g}$
For second ball, $h_{2} \ =\ \frac{u^{2}_{2}}{2g}$
If we divide, $h_{1}$ and $h_{2}$, then we will get,
$\frac{h_{1}}{h_{2}} \ =\ \frac{\frac{u^{2}_{1}}{2g}}{\frac{u^{2}_{2}}{2g}} \
=\ \frac{u^{2}_{1}}{u^{2}_{2}}$
Therefore, $h_{1} :\ h_{2} \ =\ u^{2}_{1} \ :\ u^{2}_{2}$.
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