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JEE Main 2024 (29 Jan, Shift 1) · Horizontal Projectile & Inclined Plane

Horizontal Projectile & Inclined Plane question from JEE Main 2024 (29 Jan, Shift 1)

Q3JEE Main 2024 · 29 JAN · SHIFT 1Difficult

Horizontal Projectile & Inclined Plane

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

Numerical answer type. Work it out, then open the answer.

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Answer: 2

To land on step 5 the ball must just clear the edge of step 4: fall $y = 4(0.1) = 0.4\text{ m}$ with $x = 0.4\text{ m}$ .
$t = \sqrt{\frac{2y}{g}} = \sqrt{\frac{0.8}{10}} = \sqrt{0.08}\text{ s}$
$u = \frac{x}{t} = \frac{0.4}{\sqrt{0.08}} = \sqrt{\frac{0.16}{0.08}} = \sqrt{2}\text{ m/s}$
$x = 2$

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