Skip to main content
Physics Guru is an online education and competitive examinations preparation portal founded by Manish Verma, an IIT Madras alumnus, providing learning resources for competitive examinations.

The emphasis is on conceptual understanding, analytical thinking and problem solving rather than simply memorising formulas and standard methods.

The objective is to help students understand what they are learning and develop the ability to solve problems they have not seen before.

A particle moving in a circle of radius R ....

A particle moving in a circle of radius R with a uniform speed takes 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:

(1) $\theta  = {\cos ^{ - 1}}\sqrt {\frac{{g{T^2}}}{{{\pi ^2}R}}} $
(2) $\theta  = {\cos ^{ - 1}}\sqrt {\frac{{{\pi ^2}R}}{{g{T^2}}}} $
(3) $\theta  = {\sin ^{ - 1}}\sqrt {\frac{{{\pi ^2}R}}{{g{T^2}}}} $
(4) $\theta  = {\sin ^{ - 1}}\sqrt {\frac{{2g{T^2}}}{{{\pi ^2}R}}} $

Solution

For circular motion we have, $v=\frac {2\pi R}{T}$

For projectile motion we have, max. height = $H=\frac {v^2 sin^2 \theta }{2g}=4R$

$ \Rightarrow {\left( {\frac{{2\pi R}}{T}} \right)^2}\frac{{{{\sin }^2}\theta }}{{2g}} = 4R$

$\therefore \sin \theta  = \sqrt {\frac{{2g{T^2}}}{{{\pi ^2}R}}} $ Or $\theta  = {\sin ^{ - 1}}\sqrt {\frac{{2g{T^2}}}{{{\pi ^2}R}}} $

Answer: (4)