A particle is projected from ground with an initial velocity $20 ext{ m/s}$ making an angle $60^\c

Question

A particle is projected from ground with an initial velocity $20 	ext{ m/s}$ making an angle $60^\circ$ with horizontal. If $R_1$ and $R_2$ are radius of curvatures of the particle at point of projection and highest point respectively, then find the value of $R_1/R_2$.

Professor PC Thomas Classes · Answer

$R_1 = u^2/(g\cos 60^\circ) = 2u^2/g$. At highest point, $v = u\cos 60^\circ = u/2$, normal acc is $g$. $R_2 = v^2/g = u^2/(4g)$. Ratio $R_1/R_2 = (2u^2/g)/(u^2/4g) = 8$.