Question

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, 5, is then given by :

Answer 1

Correct option is (4) \(\sin^{-1} \left(\frac{2gT^2}{\pi ^2R}\right)^{\frac 12}\)

To complete a circular path of radius R, the time period is T.

So, speed of the particle,  v = \(\frac{2\pi R}T\)   ...(1)

Now, the particle is projected with the same speed at angle θ to horizontal.

So, maximum height,

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

Given, H = 4R

⇒ \(\frac{v^2 \sin^2\theta}{2g} = 4R\)

⇒ \(\sin^2\theta = \frac{8gR}{v^2}\)

⇒ \(\sin^2\theta = \frac{8gRT^2}{4\pi r^2R^2} = \frac{2gT^2}{\pi ^2 R}\)    [From equation (1)]

⇒ \(\theta = \sin^{-1} \left(\frac{2gT^2}{\pi ^2R}\right)^{\frac 12}\)

Answer 2

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

T = \(\frac{2\pi R}{v}\) 

⇒ v = \(\frac{2\pi R}{T}\) ...(1)