Question

Question 24
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 ration of $u_1^2$ ; $u_2^2$ ( Assume upward acceleration is –g and downward acceleration to be +g).

Answer 1

We know for upward motion, $v^2=u^2-2gh$ or $h=\frac{u^2-v^2}{2g}$
But at highest point $v=0$
Therefore, $h=\frac{u^2}{2g}$
For first ball, $h_2=\frac{u_1^2}{2g}$
and for second ball $h_2=\frac{u_2^2}{2g}$
Thus $\frac{h_1}{h_2}=\frac{\frac{u_1^2}{2g}}{\frac{u_2^2}{2g}}=\frac{u_1^2}{u_1^2}or{h}_1:h_2=u_1^2:u_2^2$