A particle is projected from the ground at an angle of ${60}^{\circ }$ with horizontal with speed $u=20m/s$. The radius of curvature of the path of the particle, when its velocity makes an angle of ${30}^{\circ }$ with horizontal is
(Take $g=10m/s^2$)

The correct option is C $15.4m$
Let $v$ be velocity of particle when it makes ${30}^{\circ }$ with horizontal. Then $v\cos {30}^{\circ }=u\cos {60}^{\circ }$
$v=\frac{u\cos {60}^{\circ }}{\cos {30}^{\circ }}$
$=\frac{\left(20\right)\left(\frac{1}{2}\right)}{\left(\frac{\sqrt{3}}{2}\right)}=\frac{20}{\sqrt{3}}m/s$
Now, $g\cos {30}^{\circ }=\frac{v^2}{r}$
(centripetal acceleration where $r$ is the radius of curvature)
or $r=\frac{v^2}{g\cos {30}^{\circ }}=\frac{{\left(\frac{20}{\sqrt{3}}\right)}^2}{\left(10\right)\left(\frac{\sqrt{3}}{2}\right)}=15.4m$