A particle is projected from a point $\mathrm{O}$ with velocity $u$ at an angle of $60^{\circ}$ with the…
- $\frac{u}{3}$
- $\frac{\mathrm{u}}{2}$
- $\frac{2 u}{3}$
- $\frac{\mathrm{u}}{\sqrt{3}}$
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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