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+ <title>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.</title>
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+ <h2>Solutions of Savchenko Problems in Physics</h2>
+ <p class="author">
+ Aliaksandr Melnichenka <br/>
+ October 2023
+ </p>
+ </header>
+
+ <h3 id="back-link"><a href="../../#1.3">$\leftarrow$Back</a></h3>
+
+ <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>
+<figure>
+<img src="https://savchenkosolutions.com/1/1.3.28/graph.png"
+loading="lazy" width="350" />
+<figcaption>
+Graphs of $y(x)$ for three angles
+</figcaption>
+</figure>
+</center>
+<p>At the initial moment of time, the normal component of $\vec{g}$, perpendicular to $\vec{v}$:</p>
+<p class="exp">
+$$g_\perp = g \cos \alpha$$
+</p>
+<p>Then the radius of curvature at the initial point is found as </p>
+<p class="exp">
+$$ \fbox{$R = \frac{v^2}{g \cos \alpha}$} $$
+</p>
+<p>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</p>
+<p class="exp">
+$$g_\perp = g$$
+$$v = v_0 \cos \alpha$$
+</p>
+<p>Similarly, the radius of curvature at this point is found as </p>
+<p class="exp">
+$$ \fbox{$R = \frac{v^2 \cos^2 \alpha}{g}$} $$
+</p>
+ </p>
+
+ <h4>Answer</h4>
+ <p>
+ $$27.5\text{ and }42.4\text{ km; }18.3\text{ and }52\text{ km; }0.2\text{ and }73.4\text{ km}$$
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