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A body is projected with a velocity u at two different angles of projection such that range is same in both the cases. Sum of the maximum vertical heights attained by the body in the two cases is equal to

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$\frac{u^{2}}{2 g}$

$\frac{2 u^{2}}{g}$

$\frac{u^{2}}{4 g}$

$u^{2} g$

answer is B.

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Detailed Solution

$y_{1} = \frac{u^{2} \sin^{2} θ}{2 g}$

$y_{2} = \frac{u^{2} \sin^{2} (90^{0} - θ)}{2 g} = \frac{u^{2} \cos^{2} θ}{2 g}$

$\Rightarrow y_{1} + y_{2} = \frac{u^{2}}{2 g}$

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