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 :

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

$\theta = {\sin ^{ - 1}}{\left( {{{2g{T^2}} \over {{\pi ^2}R}}} \right)^{1/2}}$

$\theta = {\cos ^{ - 1}}{\left( {{{g{T^2}} \over {{\pi ^2}R}}} \right)^{1/2}}$

$\theta = {\cos ^{ - 1}}{\left( {{{{\pi ^2}R} \over {g{T^2}}}} \right)^{1/2}}$

$\theta = {\sin ^{ - 1}}{\left( {{{{\pi ^2}R} \over {g{T^2}}}} \right)^{1/2}}$

Solution

Please login to view the detailed solution steps...

Go to DASH