Question 220 of 308

A ball rolls off the top of a stairway with horizontal velocity $u$. The steps are $0.1m$ high and $0.1m$ wide. The minimum velocity $u$ with which that ball just hits the step $5$ of the stairway will be $\sqrt{x}m{s}^{-1}$, where $x=$_______ [use $g=10m{s}^{-2}$].

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Answer: 2

The ball needs to just cross $4$ steps to just hit ${5}^{th}$ step
Therefore, horizontal range
$\begin{matrix}R & = & 4\times 0.1m \\ & = & 0.4m\end{matrix}$
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
$4\times 0.1=\dfrac{1}{2}g{t}^{2} \Rightarrow 0.4=\dfrac{1}{2}g{\left(\dfrac{0.4}{u}\right)}^{2} \Rightarrow {u}^{2}=2 \Rightarrow u=\sqrt{2}m{s}^{-1}$
Therefore, $x=2$.