Question:
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 $+g$ ).
Solution:
For Ist Stone
$u=u_{1}, \quad a=-g, v=0$ (Velocity at the highest point $\left.=0\right)$
$\mathrm{S}=h_{1}$
Using $v^{2}-u^{2}=2 a S$, we get
$0-u_{1}^{2}=-2 g h_{1}$ or $b_{1}=\frac{u_{1}^{2}}{2 g}$
For 2nd Stone
$u=u_{2}, a=-g, v=0$ and $\mathrm{S}=b_{2}$
Again using $v^{2}-u^{2}=2 a \mathrm{~S}$, we get
$0-u_{1}^{2}=-2 g h_{2}$ or $h_{2}=\frac{u_{2}^{2}}{2 g}$
$\therefore \quad \frac{h_{1}}{h_{2}}=\frac{u_{1}^{2}}{u_{2}^{2}}$