A projectile is thrown with a velocity of $50m{s}^{-1}$ at an angle of ${53}^o$ with the horizontal. The equation of the trajectory is given by (Take $g=10m/s^2$)

The correct option is A $180y=240x-x^2$
Equation of trajectory of the projectile is given by,
$y=xtan\theta -\frac{g{x}^2}{2u^{2}co{s}^2\theta }$
Given, $u=50m{s}^{-1}\theta ={53}^{\circ }$

$\Rightarrow tan\left({53}^{\circ }\right)=\frac{4}{3}\text{and}cos\left({53}^{\circ }\right)=\frac{3}{5}$
So, $y=\frac{4x}{3}-\frac{10x^2}{2\times {50}^2\times (\frac{3}{5}{)}^2}=\frac{4x}{3}-\frac{10x^2}{1800}$
$\Rightarrow 180y=240x-x^2$