The equation of motion of a projectile is $y=12x-\frac{3}{4}{x}^2$. The horizontal component of velocity is 3 ms^-1. what is the range of the projectile?

$\begin{array}{c}\text{ Given }y=12x-\frac{3}{4}{x}^2,u_x=3{\mathrm{ms}}^{-1}\\ v_y=\frac{dy}{dt}=12\frac{dx}{dt}-\frac{3}{2}x\frac{dx}{dt}\\ \text{ At }x=0,v_y=u_y=12\frac{dx}{dt}=12u_x=12\times 3=36{\mathrm{ms}}^{-1}\\ a_y=\frac{d}{dt}\left(\frac{dy}{dt}\right)=12\frac{d^{2}x}{d{t}^2}-\frac{3}{2}\left[{\left(\frac{dx}{dt}\right)}^2+x\frac{d^{2}x}{d{t}^2}\right]\end{array}$

$\text{ But }\frac{d^{2}x}{d{t}^2}=a_x=0.\text{ Hence }$

$a_y=-\frac{3}{2}{\left(\frac{dx}{dt}\right)}^2=-\frac{3}{2}{u}_x^2=-\frac{3}{2}\times (3{)}^2=-\frac{27}{2}{\mathrm{ms}}^{-2}$

$\text{ Range, }R=\frac{2u_x{u}_y}{a_y}=\frac{2\times 3\times 36}{27/2}=16\mathrm{m}$

Alternatively: we have $y=12x-\frac{3}{4}{x}^2$. when projectile again comes to ground, $y=0\text{ and }x=R$

$0=12R-\frac{3}{4}{R}^2\Rightarrow R=16\mathrm{m}$