A projectile has initially the same horizontal velocity as it would acquire if it had moved from rest with uniform acceleration of `3 m s^-2` for `0.5 min`. If the maximum height reached by it is `80 m`, then the angle of projection is `(g = 10 ms^-2)`.

Options

A. `tan^(-1)(3)`
B. `tan^(-1)(3/2)`
C. `tan^(-1)(4/9)`
D. `sin^(-1)(4/9)`

Answer

C

Explanation

`V=at,V=u cos theta,H=(u^(2)sin^(2)theta)/(2g)`