JEE MAIN - Physics (2024 - 29th January Morning Shift - No. 29)
A ball rolls off the top of a stairway with horizontal velocity $$u$$. The steps are $$0.1 \mathrm{~m}$$ high and $$0.1 \mathrm{~m}$$ wide. The minimum velocity $$u$$ with which that ball just hits the step 5 of the stairway will be $$\sqrt{x} \mathrm{~ms}^{-1}$$ where $$x=$$ __________ [use $$\mathrm{g}=10 \mathrm{~m} / \mathrm{s}^2$$ ].
Answer
2
Explanation
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The ball needs to just cross 4 steps to just hit $$5^{\text {th }}$$ step
Therefore, horizontal range $$(R)=0.4 \mathrm{~m}$$
$$\mathrm{R}=\text { u.t }$$
Similarly, in vertical direction
$$\begin{aligned} & \mathrm{h}=\frac{1}{2} \mathrm{gt}^2 \\ & 0.4=\frac{1}{2} \mathrm{gt}^2 \\ & 0.4=\frac{1}{2} \mathrm{~g}\left(\frac{0.4}{\mathrm{u}}\right)^2 \\ & \mathrm{u}^2=2 \\ & \mathrm{u}=\sqrt{2} \mathrm{~m} / \mathrm{s} \end{aligned}$$
Therefore, $$\mathrm{x}=2$$
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