A ball rolls horizontally off the top of a stairway with a speed of $1.5{\text{ms}}^{-1}$. The steps are $0.2\text{m}$ high and $0.2\text{m}$ wide. If the ball hits the $n^{th}$ step first, then $n$ is

Total height through which the ball will fall down vertically before it strikes the $n^{th}$ step
$h=0.2\times n=0.2n$

Using second equation of motion, we get
$h=\frac{1}{2}g{t}^2$ or $0.2n=5t^2$ (i)

Horizontal distance travelled by the ball before striking the $n^{th}$ step
$R=0.2\times n=0.2n$

Now, $0.2n=u_x\times t$
$\Rightarrow 0.2n=1.5\times t$ (ii)

Solving (i) and (ii), we get n = 2.25
$\therefore$ Ball will first hit the 3rd step first