Question 277 of 308

A particle is projected from a point $\mathrm{O}$ with velocity $u$ at an angle of $60^{\circ}$ with the horizontal. When it is moving in a direction at right angles to its direction at $O$, its velocity then is given by

A$\frac{u}{3}$B$\frac{\mathrm{u}}{2}$C$\frac{2 u}{3}$D$\frac{\mathrm{u}}{\sqrt{3}}$

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Answer: (D) $\frac{\mathrm{u}}{\sqrt{3}}$

$\mathrm{u} \cos 60^{\circ}=\mathrm{v} \cos 30^{\circ}$
$$v=\frac{4}{\sqrt{3}}$$