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: (a) θ = sin ⁻¹(2gT²π²R)1/2 (b) θ = cos ⁻¹(gT²π²R)1/2 (c) θ = cos ⁻¹(π²RgT²)1/2 (d) θ = sin ⁻¹(π²RgT²)1/2

Answer: (a). Speed in the circle: v = (2π R)/(T) H = (v²sin ²θ)/(2g) = 4R ⇒ sin ²θ = (8gR)/(v²) = (8gR T²)/(4π²R²) sin ²θ = (2gT²)/(π² R) So the answer is (a) θ = sin ⁻¹((2gT²)/(π² R))1/2.

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Time of Flight, Maximum Height & Range

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 ' $\theta$ ' to the horizontal, the maximum height attained by it equals 4R. The angle of projection, $\theta$ , is then given by:

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Answer: (a)

Speed in the circle: $v = \frac{2\pi R}{T}$
$H = \frac{v^2\sin^2\theta}{2g} = 4R \Rightarrow \sin^2\theta = \frac{8gR}{v^2} = \frac{8gR\,T^2}{4\pi^2R^2}$
$\sin^2\theta = \frac{2gT^2}{\pi^2 R}$
So the answer is (a) $\theta = \sin^{-1}\left(\frac{2gT^2}{\pi^2 R}\right)^{1/2}$ .

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