The position coordinates of a projectile projected from ground on a certain planet (with no atmosphere) are given by $y=(4t-2t^2)m$ and $x=\left(3t\right)m$, where $t$ is in second and point of projection is taken as origin. The angle of projection of projectile with vertical is

The correct option is B ${37}^{\circ }$
Given : $y=4t-2t^2$ ......(i)
We know that vertical displacement of a projectile from the ground is given by
$y=u_{y}t-\frac{1}{2}{g}_p{t}^2$.
On comparing eqn (i) with above equation, we get $u_y=4$ and $g_p=4$
Horizontal displacement of the projectile is given by
$x=u_{x}t$
Comparing with $x=3t\Rightarrow u_x=3\tan \theta =\frac{u_y}{u_x}=\frac{4}{3}$
i.e Angle of projection is ${53}^{\circ }$ with horizontal.
$\therefore$ Angle of projection of projectile with vertical $=({90}^{\circ }-{53}^{\circ })={37}^{\circ }$
Hence, option (b) is correct.