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A projectile is thrown at an angle $θ$ such that it is just able to cross a vertical wall as its highest point as shown in the figure. The angle $θ$ at which the projectile is thrown is given by

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

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

$\tan^{- 1} \sqrt{3}$

$\tan^{- 1} (\frac{1}{\sqrt{3}})$

answer is C.

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Detailed Solution

$\frac{R / 2}{H} = \frac{\sqrt{3} H}{H} = \sqrt{3}$

$o r \frac{(v_{0}^{2} \sin θ \cos θ) / g}{(v_{0}^{2} \sin^{2} θ) / 2 g} = \sqrt{3}$

$2 \cot θ = \sqrt{3}$

$\tan θ = \frac{2}{\sqrt{3}}$

$o r θ = \tan^{- 1} (\frac{2}{\sqrt{3}})$

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