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7. MOTION
hard
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 $+g$.
Option A
Option B
Option C
Option D
Solution
We know for upward motion, $v^{2}=u^{2}-2 g h$ or $h=\frac{u^{2}-v^{2}}{2 g}$
But at highest point $v=0$
Therefore, $h=u^{2} / 2 g$
For first ball, $h_{1}=u_{1}^{2} / 2 g$
And for second ball, $h_{2}=u_{2}^{2} / 2 g$
Thus, $\frac{h_{1}}{h_{2}}=\frac{u_{1}^{2} / 2 g}{u_{2}^{2} / 2 g}=\frac{u_{1}^{2}}{u_{2}^{2}}$ or $h_{1}: h_{2}=u_{1}^{2}: u_{2}^{2}$
Standard 9
Science
