# A particle of mass m is projected with velocity u at an angle $\alpha$ with horizontal. During the period when particle decends from highest point to the position where its velocity vector makes $\alpha/2$ angle with horizontal, work done by gravity force is

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

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.

edited Feb 10, 2014 by meena.p