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Motion — Science STD 9 — Question

CBSE BoardEnglish MediumSTD 9ScienceMotion3 Marks
Question
Two stones are thrown vertically upwards simultaneously with their initial velocities u1 and u2 respectively. Prove that the heights reached by them would be in the ratio of $\text{u}_1^2:\text{u}_2^1$ (Assume upward acceleration is –g and downward acceleration to be +g).
✓

Answer

We know for upward motion,  $\text{v}^2=\text{u}^2-2\text{gh}$ or $\text{h}=\frac{\text{u}^2-\text{v}^2}{2\text{g}}$
But at highest point $\text{v}=0$
Therefore, $\text{h}=\frac{\text{u}^2}{\text{g}}$
For first ball, $\text{h}_1=\text{u}_1\frac{2}{2\text{g}}$
And for second ball, $\text{h}_2=\frac{\text{u}^2_2}{2\text{g}}$
Thus, $\frac{\text{h}_1}{\text{h}_2}=\frac{\text{u}_1^2/2\text{g}}{\text{u}_2^2/2\text{g}}=\frac{\text{u}_1^1}{\text{u}_2^2}$ or $\text{h}_1:\text{h}_2=\text{u}_1^2:\text{u}_2^2$

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