JEE Mains · Physics · STD 11 - Motion in plane
- JEE Mains 2024
- Hard
- Numerical
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 \(\mathrm{g}=10 \mathrm{~m} / \mathrm{s}^2\) ].
Pick an option to check your answer.
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
\( \mathrm{h}=\frac{1}{2} \mathrm{gt}^2 \)
\( 0.4=\frac{1}{2} \mathrm{gt}^2 \)
\( 0.4=\frac{1}{2} \mathrm{~g}\left(\frac{0.4}{\mathrm{u}}\right)^2 \)
\( \mathrm{u}^2=2 \)
\( \mathrm{u}=\sqrt{2} \mathrm{~m} / \mathrm{s}\)
Therefore, \(\mathrm{x}=2\)
Therefore, horizontal range \((R)=0.4 \mathrm{~m}\)
\(\mathrm{R}=\mathrm{u} . \mathrm{t}\)
Similarly, in vertical direction
\( \mathrm{h}=\frac{1}{2} \mathrm{gt}^2 \)
\( 0.4=\frac{1}{2} \mathrm{gt}^2 \)
\( 0.4=\frac{1}{2} \mathrm{~g}\left(\frac{0.4}{\mathrm{u}}\right)^2 \)
\( \mathrm{u}^2=2 \)
\( \mathrm{u}=\sqrt{2} \mathrm{~m} / \mathrm{s}\)
Therefore, \(\mathrm{x}=2\)
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