A projectile is given an initial velocity of $(2\hat{i}+4\hat{j})$$\text{m/s}$, where $\hat{i}$ is along the ground and $\hat{j}$ is along the vertical. If $g=10{\text{m/s}}^2$, then the equation of its trajectory is

The correct option is B $y=2x-\frac{5}{4}{x}^2$
Initial velocity = $2\hat{i}+4\hat{j}$
Magnitude of initial velocity = $u=\sqrt{2^2+4^2}=\sqrt{20}\text{m/s}$
If $\theta$ be the angle of projectile,
$\tan \theta =\frac{u_y}{u_x}=\frac{4}{2}=2$
Equation of trajectory is given by
$y=x\tan \theta -\frac{g{x}^2}{2u^2}(1+{\tan }^2\theta )$
$y=x\times 2-\frac{10x^2}{2\times 20}(1+4)$
$\therefore y=2x-\frac{5}{4}{x}^2$