Physics - Circular Motion Question with Solution | TestHub

The radius of curvature $\rho$ for a projectile is given by $\rho =\frac{v^2}{g\cos \alpha }$.

At the highest point, $\alpha =0$, so ${\rho }_{min}=\frac{v_x^2}{g}$.

Since $v_x=v\cos \alpha$,

we have $\rho =\frac{(v_x/\cos \alpha {)}^2}{g\cos \alpha }=\frac{v_x^2}{g{\cos }^3\alpha }=\frac{{\rho }_{min}}{{\cos }^3\alpha }$.

The final answer is $\boxed{A}$.