A ball rolls down from the top of a staircase with horizontal velocity $36\text{m/s}$. If each step is $4\text{m}$ high and $3\text{m}$ wide, the ball just hits the edge of the $n^{th}$ step where $n=\frac{1152}{\alpha g}$ , then $\alpha$ is :
[$g$ is the acceleration due to gravity]

The correct option is A $1$
Let the horizontal direction be along $x-$ axis and vertical direction be along $y-$ axis.

Given, the ball just hits the edge of the $n^{th}$ step.

So, total horizontal distance to be covered $x=3\times n=3n$

and total vertical distance to be covered $y=4\times n=4n$

Then, using the kinematic relation, $s=ut+\frac{1}{2}a{t}^2$

Along horizontal direction :

we can write that, $x=u_{x}t+\frac{1}{2}{a}_x{t}^2$

$\Rightarrow 3n=u_{x}t$ [as $a_x=0$]

Given, $u_x=36\text{m/s}$

$\Rightarrow 3n=36t$ $\Rightarrow t=\frac{3n}{36}............\left(1\right)$

Along vertical direction :

we can write that, $y=u_{y}t+\frac{1}{2}{a}_y{t}^2$

From the data given in the question,

$4n=\frac{1}{2}g{t}^2...........\left(2\right)$

From $\left(1\right)$ and $\left(2\right)$

$4n=\frac{1}{2}g{\left(\frac{3n}{36}\right)}^2$

$\Rightarrow n=\frac{g}{8}\times \frac{n^2}{144}=\frac{n^{2}g}{1152}$

$\Rightarrow n^{2}g-1152n=0$

$\Rightarrow n=\frac{1152}{g}\text{or}n=0\text{(trivial)}$

Comparing with $n=\frac{1152}{\alpha g}$, we get
$\alpha =1$

Hence, option $\left(a\right)$ is the correct answer.