A particle is projected with a certain velocity at an angle $α$ above the horizontal from the foot of an inclined plane of inclination $30^{0}$ . If the particle strikes the plane normally then $α$ is equal to

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$30^{0} + \tan^{- 1} (\frac{\sqrt{3}}{2})$

$45^{0}$

$60^{0}$

$30^{0} + \tan^{- 1} 2 \sqrt{3}$

answer is A.

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Detailed Solution

Condition for normal sriking is

$\begin{aligned} 2 \tan (α - β) = \cot β \\ \tan (α - 30) = \frac{\cot 30^{\circ}}{2} = \frac{\sqrt{3}}{2} \\ α - 30 = \tan^{- 1} (\frac{\sqrt{3}}{2}) \\ α = 30 + \tan^{- 1} (\frac{\sqrt{3}}{2}) \end{aligned}$

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