Explanation

(b) Given, $y = 12x - \frac{3}{4}x^{2}$
$u_{x} = 3\ ms^{- 1}$
$v_{y} = \frac{dy}{dt} = 12\frac{dx}{dt} - \frac{3}{2}x\frac{dx}{dt}$
At $x = 0,v_{y} = u_{y} = 12\frac{dx}{dt} = 12u_{x} = 12 \times 3 = 36\ ms^{- 1}$
$a_{y} = \frac{d}{dt}\left( \frac{dy}{dt} \right) = 12\frac{d^{2}x}{dt^{2}} - \frac{3}{2}\left( \frac{dx}{dt} + x\frac{d^{2}x}{dt^{2}} \right)$
But $\frac{d^{2}x}{dt^{2}} = a_{x} = 0$, hence
$a_{y} = - \frac{3}{2}\ \frac{dx}{dt} = - \frac{3}{2}u_{x} = - \frac{3}{2} \times 3 - \frac{9}{2}ms^{- 2}$
Range $R = \frac{2u_{x}u_{y}}{a_{y}} = \frac{2 \times 3 \times 12}{9/2} = 16m$