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The 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.

  

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The 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) θ = \(cos^{-1}\Big(\frac{gT^2}{\pi^2R}\Big)^{\frac{1}{2}}\)

(2) θ = \(cos^{-1}\Big(\frac{\pi^2R}{gT^2}\Big)^{\frac{1}{2}}\)

(3) θ = \(sin^{-1}\Big(\frac{\pi^2R}{gT^2}\Big)^{\frac{1}{2}}\)

(4) θ = \(sin^{-1}\Big(\frac{2gT^2}{\pi^2R}\Big)^{\frac{1}{2}}\)

This topic was modified 5 months ago by admin
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Correct answer: (4) θ = \(sin^{-1}\Big(\frac{2gT^2}{\pi^2R}\Big)^{\frac{1}{2}}\)

Explanation:

T = \(\frac{2 \pi R}{v}\) ⇒ v = \(\frac{2 \pi R}{T}\)  .....(1)

Hmax = \(\frac{v^2 sin^2 \theta}{2g}\)

= \(\frac{2 \pi^2 R^2sin^2 \theta}{gT^2}\) = 4R

sin θ = \(\Big(\frac{2gT^2}{\pi^2R}\Big)^{\frac{1}{2}}\)

θ = \(sin^{-1}\Big(\frac{2gT^2}{\pi^2R}\Big)^{\frac{1}{2}}\)

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