Question

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

Answer 1

Answer: 16

Given $y=12x-\frac{3}{4}{x}^2$,
$u_x=3m{s}^{-1}$
$v_y=\frac{dy}{dt}=12\frac{dx}{dt}-\frac{3}{2}x\frac{dx}{dt}$
At $x=0,v_y=u_y=12\frac{dx}{dt}-\frac{3}{2}x\frac{dx}{dt}$
At $x=0,v_y=u_y=12\frac{dx}{dt}=12u_x=12\times 3=36m{s}^{-1}$
$a_y=\frac{d}{dt}\left(\frac{dy}{dt}\right)=12\frac{d^{2}x}{d{t}^2}-\frac{3}{2}\left[{\left(\frac{dx}{dt}\right)}^2+x\frac{d^{2}x}{d{t}^2}\right]$
But $\frac{d^{2}x}{d{t}^2}=a_x=0$
Hence $a_y=-\frac{3}{2}{\left(\frac{dx}{dt}\right)}^2=-\frac{3}{2}{u}^{2}x=-\frac{3}{2}\times {\left(3\right)}^2=-\frac{27}{2}m{s}^{-2}$
Range, $R=\frac{2u_x{u}_y}{a_y}=\frac{2\times 3\times 36}{27/2}=16m$
Alternatively: We have $y=12x-\frac{3}{4}{x}^2$ When projectile again comes to ground,$y=0$ and $x=R$
$0=12R-\frac{3}{4}{R}^2$
$\Rightarrow R=16m$