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]

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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, $h_{1} :\ h_{2} \ =\ u^{2}_{1} \ :\ u^{2}_{2}$.

Updated on 10-Oct-2022 13:28:47