A ball rolls off the top of a stairway with horizontal velocity u ....

The ball needs to just cross $4$ steps to just hit ${5}^{th}$ step

Therefore, horizontal range

$\begin{cases}R & = & 4 \times 0 . 1m \\ & = & 0 . 4m\end{cases}$

The formula to calculate the horizontal range is given by

$R = ut . . . \left(1\right)$

Similarly, the vertical height covered by the ball is given by

$h = \dfrac{1}{2}g{t}^{2} . . . \left(2\right)$

From equations (1) and (2), it follows that

$4 \times 0 . 1 = \dfrac{1}{2}g{t}^{2} ⇒ 0 . 4 = \dfrac{1}{2}g{\left(\dfrac{0 . 4}{u}\right)}^{2} ⇒ {u}^{2} = 2 ⇒ u = \sqrt{2}m{s}^{- 1}$

Therefore, $x = 2$.