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A body is projected at an angle $(θ)$ with the level ground with a speed u. If R is the horizontal range of the body $θ$ is equal to

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$\frac{1}{2} \cos^{- 1} (\frac{u^{2}}{Rg})$

$\frac{1}{\sqrt{2}} \cos^{- 1} (\frac{gR}{u^{2}})$

$\tan^{- 1} (\frac{u^{2}}{Rg})$

$\frac{1}{2} \sin^{- 1} (\frac{gR}{u^{2}})$

answer is D.

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Detailed Solution

$R = \frac{u^{2} \sin 2 θ}{g}$

$\sin 2 θ = \frac{gR}{U^{2}}$

$2 θ = \sin^{- 1} (\frac{gR}{U^{2}})$

$θ = \frac{1}{2} \sin^{- 1} (\frac{gR}{U^{2}})$

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