\[(a)\;\frac{1}{2} mu^2\tan ^2 \alpha \quad (b)\;\frac{1}{2} mu^2 \cos ^2 \alpha \tan ^2 \alpha/2 \quad (c)\;\frac{1}{2} mu^2 \tan ^2 \alpha/2 \quad (d)\;\frac{1}{2} mu^2 \cos ^2 \alpha/2 m^2 \alpha \]

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Let $v_y$ vertical component of velocity at instant when velocity vector makes $\large\frac{\alpha}{2}$ with horizontal

$v_x $=horizontal component $=u \cos \alpha$

$\large\frac{v_y}{v_x}=$$\tan \alpha/2$

$v_y=v_x \tan \alpha/2$

$\quad= u \cos \alpha \tan \alpha/2$

Now increase in Kinetic energy = Work done by gravity

$W= \large\frac{1}{2}$$m{v_y}^2$

$\quad=\large\frac{1}{2}$$mu^2\cos ^2 \alpha \tan ^2 \alpha/2$

Hence b is the correct answer.

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