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

Maximum height, `H=(u^2sin^2theta)/(2g)`
or `80 =(u^2 sin^2 theta)/(2xx10) " or " u^2sin^(2) theta= 1600 " or " u sin theta = 40 ms^(-1)`
Horizontal velocity `= u cos theta`
As per question, `u cos theta = " at" = 3 xx 30 = 90 ms^(-1)`
`:. (usin theta)/(u cos theta) = 40/90 " or " tan theta = 4/9 " or " theta = tan^(-1) (4/9)`