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

The correct option is C $15.4\text{m}$
$\text{Radius of curvature =}\frac{{\text{(Velocity)}}^2}{\text{Radial Acceleration}}$

From the figure, we get
Radial acceleration $=g\cos \theta$
$\therefore$ Radius of curvature $r=\frac{v^2}{g\cos \theta }$

Since, there is no acceleration along horizontal direction
$v\cos {30}^{\circ }=20\cos {60}^{\circ }$
$\Rightarrow v=\frac{20}{\sqrt{3}}\text{m/s}$

$\therefore r=\frac{{20}^2}{3\times 10\times \frac{\sqrt{3}}{2}}=15.4\text{m}$