A projectile is projected at an angle of ${60}^0$  with the horizontal with speed of 10 m/s. The minimum radius of curvature of the trajectory described by the projectile is

The radius of curvature is minimum at the highest point where the centripetal force = mg And speed V = $\mathrm{u} \cos \mathrm{\alpha }$

Now $\frac{{\mathrm{mV}}^2}{\mathrm{\rho }}=\mathrm{mg}$

$$\begin{gathered} \Rightarrow \frac{{\left(\mathrm{ucos\alpha }\right)}^2}{\mathrm{\rho }}=\mathrm{g} \\ \Rightarrow \mathrm{\rho }={\mathrm{u}}^2{\cos }^2\mathrm{\alpha }/\mathrm{g} \\ \Rightarrow {\mathrm{\rho }}_{\min }=\frac{{\left(\mathrm{ucos\alpha }\right)}^2}{\mathrm{g}}=\frac{{\left(10\cos {60}^0\right)}^2}{9.8}\mathrm{m}=2.55\mathrm{m}. \end{gathered}$$