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A body is projected at such an angle that the horizontal range is three times the greatest height. The angle of projection is:

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33^o7'

42^o8'

53^o8'

25^o8'

answer is D.

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Detailed Solution

$R = \frac{u^{2} \sin 2 θ}{g}; H = \frac{u^{2} \sin^{2} θ}{2 g} Hence, \frac{H}{R} = \frac{\sin^{2} θ}{2 \sin 2 θ} or \frac{1}{3} = \frac{1}{4} \tan θ or \tan θ = \frac{4}{3} or θ = \tan^{- 1} (\frac{4}{3}) = 53^{\circ} 8^{'}$

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