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Topic: Uniform circular motion· Kinematics

A particle moving in a circle of radius R with a uniform speed takes a time T to complete one revolution.If this particle were projected with the same speed at an angle 'θ' to the horizontal, the maximum height attained by it equals 4R. The angle of…

A particle moving in a circle of radius R with a uniform speed takes a time T to complete one revolution.

If this particle were projected with the same speed at an angle 'θ' to the horizontal, the maximum height attained by it equals 4R. The angle of projection, θ, is then given by :

1. \(\theta = \cos^{-1}\left(\frac{gT^2}{\pi^2R}\right)^{1/2}\)

2. \(\theta = \cos^{-1}\left(\frac{\pi^2R}{gT^2}\right)^{1/2}\)

3. \(\theta = \sin^{-1}\left(\frac{\pi^2R}{gT^2}\right)^{1/2}\)

4. \(\theta = \sin^{-1}\left(\frac{2gT^2}{\pi^2R}\right)^{1/2}\)

1 answer

Best answer

Answer: 4. \(\theta = \sin^{-1}\left(\frac{2gT^2}{\pi^2R}\right)^{1/2}\)

Key idea: speed \(v=\dfrac{2\pi R}T\), and the maximum height is \(H=\dfrac{v^2\sin^2\theta}{2g}\).

\(4R=\dfrac{4\pi^2R^2}{T^2}\cdot\dfrac{\sin^2\theta}{2g}\Rightarrow\sin^2\theta=\dfrac{8gRT^2}{4\pi^2R^2}=\dfrac{2gT^2}{\pi^2R}\)

\(\theta=\sin^{-1}\left(\dfrac{2gT^2}{\pi^2R}\right)^{1/2}\). Option 4.

Doubtly AI · 30 Oct 2026 AI-generated · not yet verified Report
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