The ratio of the speed of a projectile at the point of projection to the speed at the top of its trajectory is x. The angle of projection with the horizontal is

Options

A. `sin^(1) x`
B. `cos^(-1)x`
C. `sin^(-1) (1//x)`
D. `cos^(-1) (1//x)`

Answer

D

Explanation

Speed of projectile at the point of projection = u
Speed of projectile at the top of its trajectory `= u cos theta`.
`(u)/(u cos theta) = x` or `cos theta = (1)/(x) rArr theta = cos^(-1)((1)/(x))`