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At certain height a shell at rest explodes into equal fragments. One of the fragments receives a horizontal velocity u. The time interval after which, the velocity vectors will be inclined at $120^{0}$ to each other is

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$\frac{υ}{\sqrt{3} g}$

$\frac{2 υ}{\sqrt{3} g}$

$\frac{υ}{2 \sqrt{3} g}$

$\frac{\sqrt{3} υ}{g}$

answer is A.

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Detailed Solution

$t = \frac{2 \sqrt{u_{1} u_{2}}}{g} cotθ / 2 = \frac{\sqrt{u . u}}{g} \cot 60 = \frac{u}{\sqrt{3 g}}$

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