A particle is thrown at time $t = 0$ with a velocity of $10 m s^{- 1}$ at an angle $60^{0}$ with the horizontal from a point on an inclined plane, making an angle of $30^{0}$ with the horizontal. The time when the velocity of the projectile becomes parallel to the incline is

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$\frac{2}{\sqrt{3}} s$

$\frac{1}{\sqrt{3}} s$

$\sqrt{3} s$

$\frac{1}{2 \sqrt{3}} s$

answer is B.

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Detailed Solution

$v_{y} = v \sin 30^{0} - g \cos 30^{0} t$
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$t = \frac{v \sin 30^{0}}{g \cos 30^{0}} = \frac{v}{g} \frac{1}{\sqrt{3}} = \frac{1}{\sqrt{3}} s$

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