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A particle of mass 'm' is projected with a velocity 'u' at an angle '' with the horizontal. Work done by gravity during its descent from its highest point to, the position where its velocity vector makes an angle $\frac{α}{3}$ with the horizontal is,

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$\frac{1}{2} m u^{2} \tan^{2} \frac{α}{2}$

$\frac{1}{2} m u^{2} \cos^{2} \frac{α}{2} \sin^{2} α$

$\frac{1}{2} m u^{2} \tan^{2}$ $α$

$\frac{1}{2} m u^{2} \cos^{2} α \tan^{2} \frac{α}{2}$

answer is C.

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Detailed Solution

As horizontal Component of velocity remains constant,

work done = $Δ K E$ along vertical direction ,

$K_{y_{1}}$ = 0

$T a n \frac{α}{2} = \frac{V_{y}}{u C o s α} \Rightarrow V_{y} = u C o s α . T a n \frac{α}{2}$

$K_{y_{2}} = \frac{1}{2} m V_{y}^{2} = \frac{1}{2} m u^{2} C o s^{2} α . T a n^{2} \frac{α}{2}$

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