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The time of flight of a projectile is related to its horizontal range by the equation ${gT}^{2} = 2 R .$ The angle of projection is

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$30^{\circ}$

$\tan^{- 1} (2)$

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

$45^{\circ}$

answer is D.

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Detailed Solution

$g T^{2} = 2 R$

$g (\frac{2 u \sin θ}{g}) = 2 (\frac{u^{2} \sin 2 θ}{g})$

$g (\frac{4 u^{2} \sin^{2} θ}{g^{2}}) = 2 (\frac{u^{2} \sin 2 θ}{g})$

$g (\frac{4 u^{2} \sin^{2} θ}{g^{2}}) = 2 (\frac{u^{2} 2 \sin θ \cos θ}{g})$

$\frac{\sin θ}{\cos θ} = 1$

$\tan θ = 1$

$θ = 45^{\circ}$

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