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Two stones are thrown vertically upwards simultaneously with their initial velocities $u_1$ and $u_2$ respectively. Prove that the heights reached by them would be in the ratio of $u_1^2:u_2^2$ [Assume upward acceleration is -g and downward acceleration to be +g]
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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