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A body is projected from a point on ground. If just crosses a wall of height 20 m at a distance 20 m from the point of projection. If the body strikes the ground at a distance of 60 m, from the wall, the angle of projection is $[\tan 37^{o} = \frac{3}{4}]$

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$45^{o}$

$53^{o}$

$37^{o}$

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

answer is C.

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Detailed Solution

$\begin{array}{l} y = x \tan α (1 - \frac{x}{R}) \\ \Rightarrow 20 = 20 \tan α (1 - \frac{20}{80}) \Rightarrow \tan α = \frac{4}{3} \Rightarrow α = 53^{o} \end{array}$

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