JEE Mains · Physics · STD 11 - 3.2 motion in plane
A ball rolls off the top of a stairway with horizontal velocity \(\mathrm{u}\). The steps are \(0.1 \mathrm{~m}\) high and \(0.1 \mathrm{~m}\) wide. The minimum velocity \(\mathrm{u}\) with which that ball just hits the step \(5\) of the stairway will be \(\sqrt{\mathrm{x}} \mathrm{ms}^{-1}\) where \(\mathrm{x}=\) _______ [use \(\mathrm{g}=10 \mathrm{~m} / \mathrm{s}^2\) ].
- A \(1\)
- B \(2\)
- C \(3\)
- D \(4\)
Answer & Solution
Correct Answer
(B) \(2\)
Step-by-step Solution
Detailed explanation
The ball needs to just cross \(4\) steps to just hit \(5^{\text {th }}\) step Therefore, horizontal range \((R)=0.4 \mathrm{~m}\) \(\mathrm{R}=\mathrm{u} . \mathrm{t}\) Similarly, in vertical direction \( \mathrm{h}=\frac{1}{2} \mathrm{gt}^2 \)…
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