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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> | | <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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| <h3 id="back-link"><a href="../../#1.3">$\leftarrow$Back</a></h3> | | <h3 id="back-link"><a href="../../#1.3">$\leftarrow$Back</a></h3> |
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| <h3> Statement </h3> | | <h3> Statement </h3> |
| <p> | | <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. | | $1.3.28.$ 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. |
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| <h3>Solution</h3> | | <h3>Solution</h3> |
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| loading="lazy" width="350" /> | | loading="lazy" width="350" /> |
| <figcaption> | | <figcaption> |
| Graphs of $y(x)$ for three angles | | Graphs of $y(x)$ for three angles |
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| <p>At the initial moment of time, the normal component of $\vec{g}$, perpendicular to $\vec{v}$:</p> | | <p>At the initial moment of time, the normal component of $\vec{g}$, perpendicular to $\vec{v}$:</p> |
| <p class="exp"> | | <p class="exp"> |
| $$g_\perp = g \cos \alpha$$ | | $$g_\perp = g \cos \alpha$$ |
| </p> | | </p> |
| <p>Then the radius of curvature at the initial point is found as </p> | | <p>Then the radius of curvature at the initial point is found as </p> |
| <p class="exp"> | | <p class="exp"> |
| $$ \fbox{$R = \frac{v^2}{g \cos \alpha}$} $$ | | $$ \fbox{$R = \frac{v^2}{g \cos \alpha}$} $$ |
| </p> | | </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>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"> | | <p class="exp"> |
| $$g_\perp = g$$ | | $$g_\perp = g$$ |
| $$v = v_0 \cos \alpha$$ | | $$v = v_0 \cos \alpha$$ |
| </p> | | </p> |
| <p>Similarly, the radius of curvature at this point is found as </p> | | <p>Similarly, the radius of curvature at this point is found as </p> |
| <p class="exp"> | | <p class="exp"> |
| $$ \fbox{$R = \frac{v^2 \cos^2 \alpha}{g}$} $$ | | $$ \fbox{$R = \frac{v^2 \cos^2 \alpha}{g}$} $$ |
| </p> | | </p> |
| </p> | | </p> |
| | | |
| <h4>Answer</h4> | | <h4>Answer</h4> |
| <p> | | <p> |
| $$27.5\text{ and }42.4\text{ km; }18.3\text{ and }52\text{ km; }0.2\text{ and }73.4\text{ km}$$ | | $$27.5\text{ and }42.4\text{ km; }18.3\text{ and }52\text{ km; }0.2\text{ and }73.4\text{ km}$$ |
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