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Q.

Two stones are projected from the top of a tower in opposite directions, with the same velocity V but at 30^o & 60^o with horizontal respectively. The relative velocity of first stone relative to second stone is

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a

$\sqrt{2}$ V

b

2 V

c

$\frac{2 V}{\sqrt{3}}$

d

$\frac{V}{\sqrt{2}}$

answer is B.

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Detailed Solution

$\begin{array}{l} {\bar{V}}_{{Iwrt}_{2}} = {\bar{V}}_{1} - \bar{V_{2}} = (vcos 30 \hat{i} + vsin 30 \hat{j}) - \\ [(vcos 60 (- \hat{i}) + vsin 60 \hat{j}] \\ {\bar{V}}_{lwrt 2} = \frac{\sqrt{3} v}{2} \hat{i} + \frac{v}{2} \hat{j} + \frac{v}{2} \hat{i} - \frac{\sqrt{3} v}{2} \hat{j} \\ = \frac{v}{2} [\sqrt{3} + 1] \hat{i} + \frac{v}{2} [1 - \sqrt{3} \hat{j} \\ |{\bar{V}}_{1 wrt 2}| = \frac{v}{2} \sqrt{(\sqrt{3} + 1)^{2} + (1 - \sqrt{3})^{2}} \\ = \frac{v}{2} \sqrt{8} = \frac{v}{2} (2 \sqrt{2}) = \sqrt{2} V \end{array}$

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