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

( c) `H = (u^2 sin^2 theta)/(2 g)` or `80 = (u^2 sin^2 theta)/(2 xx 10)`
or `u^2 sin^2 theta = 1600` or `u sin theta = 40 ms^-1`
Horizontal velocity `= u cos theta = 3 xx 30 = 90 ms^-1`
`(u sin theta)/(u cos theta) = (40)/(90)` or `tan theta = (4)/(9)` or `theta = tan^-1 ((4)/(9))`.