A particle is moving in xy plane and equation of its trajectory is $y=\left(120x-3x^2\right)$ where x and y are in metre. When the distance of the particle from x-axis is maximum, speed of the particle is

When distance from x-axis is maximum, velocity component along y-axis must be zero. 

i.e. $\frac{dy}{dt}=0$ i.e. $120-3\times 2x\frac{dx}{dt}=0\Rightarrow \frac{dx}{dt}=\frac{20}{x}$

Again, when y is maximum, $\frac{dy}{dx}=0\Rightarrow 120-6x=0\Rightarrow x=20m$

$\therefore \frac{dx}{dt}=\frac{20}{20}m/s=1m/s$