Question

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

Answer 1

Given $y=12x-\frac{3}{4}{x}^2,u_x=3m/s$
$v_y=\frac{dy}{dt}=12\frac{dx}{dt}-\frac{3}{2}x\frac{dx}{dt}$
At$x=0,v_y=u_y=12\frac{dx}{dt}=12u_x=12\times 3=36m/s$
$a_y=\frac{d}{dt}\left(\frac{dy}{dt}\right)=12\frac{d^{2}x}{d{t}^2}-\frac{3}{2}(\frac{dx}{dt}\frac{dx}{dt}+x\frac{d^{2}x}{d{t}^2})$
but $\frac{d^{2}x}{d{t}^2}=a_x=0$, hence
$a_y=-\frac{3}{2}\frac{dx}{dt}\frac{dx}{dt}=-\frac{3}{2}{u}_x^2=-\frac{3}{2}\times 9=-\frac{27}{2}m/s^2$
Range $R=\frac{2u_x{u}_y}{a_y}=\frac{2\times 3\times 36}{27/2}=16m$