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

Formula used: $s=ut+\frac{1}{2}a{t}^2$
For horizontal displacement
$\left(0.3\right)n=\left(4.5\right)t\dots \left(i\right)$
{$\because$ Horizontal acceleration in projectile is zero}
For vertical displacement
$\left(0.2\right)n=\frac{1}{2}\times 10t^2=5t^2\dots \left(ii\right)$
{$\because$ initial velocity in vertical direction
in projectile is zero}
From equation $\left(i\right)$ putting the value of $t$ in equation $\left(ii\right)$
$\left(0.2\right)n=5{\left(\frac{0.3n}{4.5}\right)}^2$
$n=\frac{0.2\times 4.5\times 4.5}{5\times 0.3\times 0.3}$
$n=9$
Final Answer: $9$