Physics · Motion in Plane

JEE Main 2024 — 29 January, Shift 1 — Question 50

A ball rolls off the top of a stairway with horizontal velocity $u$ . The steps are 0.1 m high and 0.1 m wide. The minimum velocity 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 $\mathrm{g}=10 \mathrm{~m} / \mathrm{s}^{2}$ ].

Answer: 2

Numerical answer — enter this value.

Step-by-step solution

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

step

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

$\mathrm{R}=$ u.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, $\mathrm{x}=2$

Answer key and solution verified before publishing.

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Exam: JEE Main 2024
Paper: 29 January, Shift 1
Subject: Physics
Chapter: Motion in Plane
Topic: Oblique and Horizontal Projectile Motion