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QJEEM24Ph26ad: JEE Main 2024 Physics - Class 11 JEEM24Ph26ad Solution: A ball rolls off the top of a stairway with horizontal velocity …

A ball rolls off the top of a stairway with horizontal velocity uu . The steps are 0.1 m0.1 \mathrm{~m} high and 0.1 m0.1 \mathrm{~m} wide. The minimum velocity uu with which that ball just hits the step 5 of the stairway will be x ms−1\sqrt{x} \mathrm{~ms}^{-1} where x=x= __________ [use g=10 m/s2\mathrm{g}=10 \mathrm{~m} / \mathrm{s}^2 ].

Step-by-Step Verified Solution

JEE Main 2024 (Online) 29th January Morning Shift Physics - Motion in a Plane Question 12 English Explanation

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

Therefore, horizontal range (R)=0.4 m(R)=0.4 \mathrm{~m}

R= u.t \mathrm{R}=\text { u.t }

Similarly, in vertical direction

amp;h=12gt2amp;0.4=12gt2amp;0.4=12 g(0.4u)2amp;u2=2amp;u=2 m/s\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, x=2\mathrm{x}=2

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