A particle is projected from a point $\mathrm{O}$ with velocity $u$ at an angle of $60^{\circ}$ with the…

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
  1. $\frac{u}{3}$
  2. $\frac{\mathrm{u}}{2}$
  3. $\frac{2 u}{3}$
  4. $\frac{\mathrm{u}}{\sqrt{3}}$

Solution


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

Asked in: JEE Main 2005

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