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Q.

A body is projected with a velocity u at an angle of $60^{0}$ to the horizontal. The time interval after which it will be moving in a direction of $30^{0}$ to the horizontal is:

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a

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

b

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

c

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

d

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

answer is A.

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Detailed Solution

$\begin{array}{l} \tan α = \frac{u \sin θ - g t}{u \cos θ} \\ \tan 30^{0} = \frac{u \sin 60 - g t}{u \cos 60^{0}} \\ \frac{1}{\sqrt{3}} = \frac{u (\frac{\sqrt{3}}{2}) - g t}{\frac{u}{2}} \\ \frac{u}{2 \sqrt{3}} = \frac{\sqrt{3} u}{2} - g t \\ g t = \frac{\sqrt{3} u}{2} - \frac{u}{2 \sqrt{3}} \\ g t = \frac{3 u - u}{2 \sqrt{3}} = \frac{u}{\sqrt{3}} \\ t = \frac{u}{\sqrt{3} g} \end{array}$

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