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A body is projected with a velocity u at an angle of 60° to the horizontal. The time interval after which it will be moving in a direction of 30° to the horizontal is

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\frac{\sqrt{3} u}{2g}

\frac{\sqrt{3} u}{g}

\frac{u}{\sqrt{3} g}

\frac{2u}{\sqrt{3} g}

answer is A.

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Detailed Solution

During projectile motion, angle at any instant t is such that

$\tan α = \frac{u \sin θ - gt}{u \cos θ}$

$\tan 30 = \frac{u \sin 60 - gt}{u \cos 60}$

$\tan 30 - \tan 60 = \frac{- gt}{u \cos 60}$

$\frac{1}{\sqrt{3}} - \sqrt{3} = \frac{- gt}{u \cos 60}$

$\Rightarrow \frac{- 2}{\sqrt{3}} = \frac{- gt}{u \times 1/2} \Rightarrow \frac{1}{\sqrt{3}} = \frac{gt}{u}$

$\Rightarrow t = \frac{u}{\sqrt{3} g}$

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