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

The correct option is B $y=2x-5x^2$
Given: Initial velocity, $\overrightarrow{v}=(\hat{i}+2\hat{j})\text{m/s}$

Magnitude of initial velocity, $u=\sqrt{1^2+2^2}=\sqrt{5}\text{m/s}$

Equation of trajectory of projectile, $y=x\tan \theta -\frac{g{x}^2}{2u^2{\cos }^2\theta }$

or
$y=x\tan \theta -\frac{g{x}^2}{2u^2}(1+{\tan }^2\theta )$

here, $\tan \theta =\frac{v_y}{v_x}=\frac{2}{1}=2$

so,
$y=x\times 2-\frac{10x^2}{2\times 5}(1+2^2)$

$y=2x-\frac{10x^2}{10}\left(5\right)$

$y=2x-5x^2$

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