The equations of motion of a projectile are given by $x=36tm$ and $2y=96t-9.8t^{2}m$. The angle of projection with the horizontal is:

The correct option is A $si{n}^{-1}\left(\frac{4}{5}\right)$
Given : $x=36t$
and $2y=96t-9.8t^2$
or $y=48t-4.9t^2$
Let the initial velocity of projectile be u and angle of projection is $\theta$. Then, Initial horizontal component of velocity,
$u_x=ucos\theta ={\left(\frac{dx}{dt}\right)}_{t=0}=36$ ....(i)
$orusin\theta ={\left(\frac{dx}{dt}\right)}_{t=0}=48$ ...(ii)
Dividing (ii) by (i), we get
$tan\theta =\frac{48}{36}=\frac{4}{3}$
$\therefore sin\theta =\frac{4}{5}$ or $\theta =si{n}^{-1}\left(\frac{4}{5}\right)$