Question

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

Answer 1

Given $y=12x-\frac{3}{4} x^{2}$,
$u_{x}=3\,m \,s^{-1}$
$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 $x=0, v_{y}=u_{y}=12 \frac{dx}{dt}=12u_{x}=12\times3=36\,ms^{-1}$
$a_{y}=\frac{d}{dt} \left(\frac{dy}{dt}\right)=12 \frac{d^{2}x}{dt^{2}}-\frac{3}{2} \left[\left(\frac{dx}{dt}\right)^{2}+x \frac{d^{2}x}{dt^{2}}\right]$
But $\frac{d^{2}x}{dt^{2}}=a_{x}=0$
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\times3\times36}{27/ 2}=16\,m $
Alternatively: We have $y=12x-\frac{3}{4} x^{2}$ When projectile again comes to ground,$ y = 0$ and $x = R$
$0=12\,R-\frac{3}{4}R^{2}$
$\Rightarrow R=16\,m$