A ball rolls of the top of stair-way with a horizontal velocity of magnitude $1.8m{s}^{-1}.$ The steps are 0.20 m high and 0.20 m wide. Which step will the ball hit first?

$Given,x=0.20m,y=0.20m,u=1.8m{s}^{-1}$

Let the ball strike the nth step of stairs,

Vertical distance travelled $=ny=n\times 0.20=\frac{1}{2}g{t}^2$

Horizontal distance travelled, nx = ut

$ort=\frac{nx}{u}$

$\therefore ny=\frac{1}{2}g\times \frac{n^2{x}^2}{u^2}$

$orn=\frac{2u^2}{g}\frac{y}{x^2}=\frac{2\times {\left(1.8\right)}^2\times 0.20}{9.8\times {\left(0.20\right)}^2}=3.3\approx 4$