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

Tutorialspoint

Simply Easy Learning

Updated on: 10-Oct-2022

39 Views

Kickstart Your Career

Get certified by completing the course

Get Started