The position coordinates of a projectile projected from ground on a certain planet (with no atmosphere) are given by $y=(4t-2t^2)\text{m}$ and $x=\left(3t\right)\text{m}$, where $t$ is in $\text{second}$ 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: $y=4t-2t^2.....\left(\text{i}\right)$

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 $\left(\text{i}\right)$ with above equation, we get $u_y=4$ and $g_p=4$.

Given: $x=3t$

We know that, horizontal displacement of the projectile is given by,

$x=u_{x}t$

Comparing with $x=3t$, we get

$u_x=3$

Now,

$\tan \theta =\frac{u_y}{u_x}=\frac{4}{3}$

i.e., Angle of projection $\left(\theta \right)$ is ${53}^{\circ }$ with horizontal.

$\therefore$Angle of projection of projectile with vertical,

$\varphi =({90}^{\circ }-\theta )=({90}^{\circ }-{53}^{\circ })={37}^{\circ }$

Hence, option $\left(\text{b}\right)$ is correct.