JEE Main 2024 — Motion In Two Dimensions Question with Solution

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

Therefore, horizontal range

$\begin{array}{rcl}R& =& 4\times 0.1 \mathrm{m}\\ & =& 0.4 \mathrm{m}\end{array}$

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

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

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

$$\begin{gathered} 4\times 0.1=\frac{1}{2}g{t}^2 \\ \Rightarrow 0.4=\frac{1}{2} g{\left(\frac{0.4}{u}\right)}^2 \\ \Rightarrow u^2=2 \\ \Rightarrow u=\sqrt{2} \mathrm{m} {\mathrm{s}}^{-1} \end{gathered}$$$$

Therefore, $x=2$.