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

see full answer

Talk to JEE/NEET 2025 Toppers - Learn What Actually Works!

Real Strategies. Real People. Real Success Stories - Just 1 call away

An Intiative by Sri Chaitanya

$\sqrt{2}$ V

2 V

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

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

answer is B.

(Unlock A.I Detailed Solution for FREE)

Ready to Test Your Skills?

Check your Performance Today with our Free Mock Test used by Toppers!

Take Free Test

Detailed Solution

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