A ball rolls off the top of a stairway horizontally with a velocity of $4.5{\text{ms}}^{-1}$. Each step is $0.2\text{m}$ high and $0.3\text{m}$ wide. If $g$ is $10{\text{ms}}^{-2}$, then the ball will strike the $n^{th}$ step where $n$ is equal to

The correct option is B $9$
Total height by 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
$h=u_{y}t+\frac{1}{2}g{t}^2$
$u_y=0$
$\begin{array}{}\Rightarrow 0.2n=\frac{1}{2}\times 10\times t^2=5t^2& \text{(1)}\end{array}$
Horizontal distance travelled by the ball before striking the $n^{th}$ step
$R=u_{x}t=0.3\times n=0.3n$
$\begin{array}{}\Rightarrow 0.3n=4.5t& \text{(2)}\end{array}$
Dividing (i) by (ii), we get
$\frac{2}{3}=\frac{50}{45}t$
$\Rightarrow t=0.6\text{s}$
Therefore , $0.3n=4.5\times 0.6=2.7$
$\Rightarrow n=\frac{2.7}{0.3}=9$

Alternative soln :
Horizontal distance covered by the ball
$X=nw$ where $w$ is the width of each step.
Vertical distance covered by the ball
$Y=nh$ where $h$ is the width of each step.
Horizontal distance covered by a body falling from height $h$ with initial horizontal velocity $u$
$X=u\sqrt{\frac{2Y}{g}}=u\sqrt{\frac{2nh}{g}}$
$\Rightarrow nw=u\times \sqrt{\frac{2nh}{g}}$
Solving, we get
$n=\frac{2u^{2}h}{w^{2}g}=\frac{2\times {4.5}^2\times 0.2}{{0.3}^2\times 10}=9$