$A$ projectile has initially the same horizontal velocity as it would acquire if it had moved from rest with uniform acceleration of $3 \, ms^{-2}$ for $0.5 \, minutes$. If the maximum height reached by it is $80 \, m$,then the angle of projection is (Take $g = 10 \, ms^{-2}$)
- A$\tan^{-1}(3)$
- B$\tan^{-1}(3/2)$
- C$\tan^{-1}(4/9)$
- D$\sin^{-1}(4/9)$
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