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 is $+g$).

For an object moving vertically upwards,the third equation of motion is given by $v^{2} = u^{2} + 2as$.
At the maximum height,the final velocity $v$ becomes $0$.
Substituting $v = 0$,$a = -g$,and $s = h$ into the equation: $0 = u^{2} - 2gh$.
Rearranging for height,we get $h = \frac{u^{2}}{2g}$.
For the first stone with initial velocity $u_{1}$,the maximum height reached is $h_{1} = \frac{u_{1}^{2}}{2g}$.
For the second stone with initial velocity $u_{2}$,the maximum height reached is $h_{2} = \frac{u_{2}^{2}}{2g}$.
Taking the ratio of the two heights: $\frac{h_{1}}{h_{2}} = \frac{u_{1}^{2} / 2g}{u_{2}^{2} / 2g} = \frac{u_{1}^{2}}{u_{2}^{2}}$.
Therefore,the ratio of the heights reached is $h_{1}: h_{2} = u_{1}^{2}: u_{2}^{2}$.