A particle of mass `m` is projected with velocity `u` at an angle `alpha` with horizontal. During the period when the particle descends from highest point to the position where
its velocity vector makes an angle `(alpha)/(2)` with horizontal. Work done by gravity force is

Options

A. `1/2 m u ^(2) tan^(2)(theta//2)`
B. `1/2 m u^(2)tan^(2)theta`
C. `1/2 m u^(2) cos^(2)theta tan^(2)(theta//2)`
D. `1/2 m u^(2)cos^(2)(theta//2)sin^(2)theta`

Answer

C

Explanation

As horizontal component of velocity does not change,
`v cos theta//2=u cos theta`
`v=(u cos theta)/(cos theta//2)`
`W_("gravity")=DeltaK=1/2 mv^(2)-1/2 m (ucos theta)^(2)=1/2 mu^(2)cos^(2) theta tan^(2)theta//2`
Hence (C) is correct.