The velocity of a projectile when it is at the greatest height is `(sqrt (2//5))` times its velocity when it is at half of its greatest height. Determine its angle of projection.

Options

A. `30^(@)`
B. `45^(@)`
C. `60^(@)`
D. `37^(@)`

Answer

C

Explanation

`(u cos alpha) = sqrt((2)/(5)) sqrt((u cos alpha^(2))+{(u sin alpha)^(2)-2gh))}`
Here, `h = (H)/(2) = (u^(2)sin^(2)alpha)/(4g)`
Solving this equation we get,`alpha = 60^(@)`