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MHT CET 2023 (13 May, Shift 2) · Centripetal Acceleration & Force

Centripetal Acceleration & Force question from MHT CET 2023 (13 May, Shift 2)

Q25MHT CET 2023 · 13 MAY · SHIFT 2Medium

Centripetal Acceleration & Force

A stone is projected at angle $\theta$ with velocity $u$ . If it executes nearly a circular motion at its maximum point for short time, then the radius of the circular path will be ( $g =$ acceleration due to gravity)

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Answer: (b)

At the highest point the velocity is horizontal: $v = u\cos\theta$ .
Gravity is perpendicular to the velocity there, so it supplies the centripetal force: $g = \frac{v^2}{r}$ .
$r = \frac{u^2\cos^2\theta}{g}$
So the answer is (b) $\frac{u^2 \cos^2 \theta}{g}$ .

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