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

Options

A. `(1)/(2) mu^(2) tan^(2) alpha`
B. `(1)/(2) mu^(2) cos^(2) (alpha)/(2)`
C. `(1)/(2) mu^(2) cos^(2)alpha tan^(2) (alpha)/(2)`
D. `(1)/(2) mu^(2) cos^(2) (alpha)/(2) sin^(2) alpha`

Answer

D

Explanation

Horizontal component of velocity i.e., `u cos alpha` remains unchanged. At highest point vertical component of velocity is zero . Let `v_(y)` be the vertical component of velocity at the desired instant.

`(v_(y))/(v_(x)) = tan . (alpha)/(2)`
`v_(y) = v_(x) tan. (alpha)/(2) = u cos alpha tan .(alpha)/(2)`
Now increase in kinetic energy = work done by gravity
`:.` The desired work done `W = (1)/(2) mv_(y)^(2)`
`= (1)/(2) mu^(2) cos^(2) alpha tan^(2). (alpha)/(2)`.