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.

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    <h3> Statement </h3>
    <p>
        $1.3.28.$ 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.
    </p>

    <h3>Solution</h3>
    <p>
        <center>

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

$$\fbox{$R = \frac{v^2 \cos^2 \alpha}{g}$}$$

    <h4>Answer</h4>
    <p>
        $$27.5\text{ and }42.4\text{ km; }18.3\text{ and }52\text{ km; }0.2\text{ and }73.4\text{ km}$$
    </p>


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