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A projectile is thrown at an angle $37^{0}$ from the vertical. The angle of elevation of the highest point of the projectile from point of projection is

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

$\tan^{- 1} (3 / 8)$

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

$\tan^{- 1} (8 / 3)$

answer is B.

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Detailed Solution

$90^{0} - θ = 37^{0}$

$θ = 53^{0}$

$\tan α = \frac{H_{\max}}{R / 2} = \frac{2 H}{R}$

$= \frac{2 u^{2} \sin^{2} θ / 2 g}{u^{2} \sin 2 θ / g} = \frac{\tan θ}{2}$

$= \frac{\tan 53^{0}}{2} = \frac{4 / 3}{2}$

$\tan α = \frac{2}{3}$

$α = \tan^{- 1} (2 / 3)$

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