$1.3.28.$ Projectiles fly out at an initial velocity of $600$ $\frac{m}{s}$ at an angle of $30^\circ$, $45^\circ$, $60^\circ$ to the horizon. Determine the radius of curvature of the projectile trajectory at their highest and starting points.
At the initial moment of time, the normal component of $\vec{g}$, perpendicular to $\vec{v}$:
$$g_\perp = g \cos \alpha$$
Then the radius of curvature at the initial point is found as
$$ \fbox{$R = \frac{v^2}{g \cos \alpha}$} $$
At the top point of the trajectory, the acceleration will coincide with the acceleration of gravity, and the speed will be equal to the horizontal
$$g_\perp = g$$ $$v = v_0 \cos \alpha$$
Similarly, the radius of curvature at this point is found as
$$ \fbox{$R = \frac{v^2 \cos^2 \alpha}{g}$} $$
$$27.5\text{ and }42.4\text{ km; }18.3\text{ and }52\text{ km; }0.2\text{ and }73.4\text{ km}$$