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

Options

A. `theta = cos^(-1)((pi^(2)R)/(gT^(2)))^(1//2)`
B. `theta = sin^(-1) ((pi^(2)R)/(gT^(2)))^(1//2)`
C. `theta = sin^(-1)((2gT^(2))/(piR))^(1//2)`
D. `theta = cos^(-1)((gT-1))`

Answer

C

Explanation

`T = (2piR)/v rArr v = (2piR)/T`
`H_("max") = (v^(2) sin^(2) theta)/(2g) = (2pi^(2)R^(2) sin^(2) theta)/(gT^(2)) = 4R`
`sin theta = ((2gR^(2))/(pi^(2)R))^(1//2)`
`theta = sin^(-1) [(2gT^(2))/(pi^(2)R)]^(1//2)`