Two particles A and B are separated by a horizontal distance x. They are projected at the same instant towards each other with speeds $u / \sqrt{3}$ and $u$ at angle of projections $30^{0}$ and $60^{0}$ respectively as shown in figure. The time after which the horizontal distance between them becomes zero is:

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$\frac{2 x}{u}$

$\frac{4 x}{u}$

$\frac{x}{2 u}$

$\frac{x}{u}$

answer is A.

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Detailed Solution

Their velocity of approach is

$v = u / \sqrt{3} \cos 30^{0} + u \cos 60^{0} = u$

$∴ t = \frac{x}{v} = \frac{x}{u}$

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