A particle is projected from ground with speed u and at an angle `theta` with horizontal. If at maximum height from ground, the speed of particle is `1//2` times of its initial velocity of projection, then find its maximum height attained.

Options

A. `(u^(2))/(g)`
B. `(2u^(2))/(g)`
C. `(u^(2))/(2g)`
D. `(3u^(2))/(8g)`

Answer

D

Explanation

Given at maximum height
`u cos theta =(1)/(2) u rArr cos theta = (1)/(2) :. Theta = 60^(@)`
`:. H = (u^(2) sin^(2) theta)/(2g) = (u^(2) sin^(2) 60^(@))/(2g) = (3u^(2))/(8g)`