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| + | <meta name="description" content="A spherical tank standing on the ground has a radius of R. What is the lowest speed at which a rock thrown from the ground can fly over the reservoir just by touching its top?"> | ||
| + | <meta name="author" content="Aliaksandr Melnichenka"> | ||
| + | <meta name="date" content="2023-10" scheme="YYYY-MM"> | ||
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| + | <header style="text-align:center;"> | ||
| + | <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> | ||
| + | $1.3.27^*.$ A spherical tank standing on the ground has a radius of $R$. What is the lowest speed at which a rock thrown from the ground can fly over the reservoir just by touching its top? | ||
| + | </p> | ||
| + | |||
| + | <h3>Solution</h3> | ||
| + | <p> | ||
| + | <p>The stone must be thrown at an angle $\alpha$ to the horizon, satisfying the equations obtained in <a href="../1.3.6" target="_blank">1.3.6</a>:</p> | ||
| + | <p class="exp"> | ||
| + | $$v_x = v \cos \alpha; \quad v_y = v \sin \alpha - gt;$$ | ||
| + | $$x = vt \cos \alpha; \quad y = vt \sin \alpha - gt^2 / 2.$$ | ||
| + | </p> | ||
| + | <p>The time it takes for the stone to rise to the maximum height $2R$ is found as</p> | ||
| + | <p class="exp"> | ||
| + | $$ t_1 = \frac{v_0 \sin \alpha}{g} $$ | ||
| + | </p> | ||
| + | <p>The maximum height of the stone lift along the vertical axis should be equal to $y_{max} = 2R$, therefore</p> | ||
| + | <p class="exp"> | ||
| + | $$ \frac{v_0^2 \sin^2 \alpha}{2g} = 2R $$ | ||
| + | </p> | ||
| + | <p>Determine the value of the initial throw speed</p> | ||
| + | <p class="exp"> | ||
| + | $$ v_0 = \sqrt{\frac{4gR}{\sin^2 \alpha}} $$ | ||
| + | </p> | ||
| + | <p>The angle $\alpha$ at which the stone should be thrown is determined from the initial conditions</p> | ||
| + | <p class="exp"> | ||
| + | $$ v_0t_1 \cos \alpha = \frac{v_0^2 \sin \alpha \cos \alpha}{g}$$ | ||
| + | $$ 2R = \frac{v_0^2 \sin \alpha \cos \alpha}{g}$$ | ||
| + | $$\tan \alpha = 2; \quad \alpha = \text{arctg} \;2 \approx 63^\circ$$ | ||
| + | </p> | ||
| + | <p>Substituting into the formula for $v_0$</p> | ||
| + | <p class="exp"> | ||
| + | $$ \fbox{$v_0 = \sqrt{\frac{4gR}{\sin^2 63^\circ}} = \sqrt{5Rg}$} $$ | ||
| + | </p> | ||
| + | </p> | ||
| + | |||
| + | <h4>Answer</h4> | ||
| + | <p> | ||
| + | $$v = \sqrt{5gR}$$ | ||
| + | </p> | ||
| + | |||
| + | |||
| + | <footer class="row container"> | ||
| + | <br> | ||
| + | <p> | ||
| + | <small> © <strong>Savchenko Solutions</strong>, 2023-2024 <br></small> | ||
| + | </p> | ||
| + | <p> | ||
| + | <small>All rights belong to the authors. <br> Commercial use of materials - with the written permission of the authors. <br> aliaksandr@savchenkosolutions.com <br></small> | ||
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| <html lang="en"> | |||
| <head> | |||
| <meta charset="utf-8"> | |||
| <meta name="viewport" content="width=device-width, initial-scale=1.0"> | |||
| <meta http-equiv="content-language" content="en"> | |||
| <meta name="keywords" content="Savchenko Problems in Physics, Savchenko solutions, physics problems, physics olympiad preparation, IPhO, Jaan Kalda"> | |||
| <meta name="description" content="A spherical tank standing on the ground has a radius of R. What is the lowest speed at which a rock thrown from the ground can fly over the reservoir just by touching its top?"> | |||
| <meta name="author" content="Aliaksandr Melnichenka"> | |||
| <meta name="date" content="2023-10" scheme="YYYY-MM"> | |||
| <meta property="og:title" content="A spherical tank standing on the ground has a radius of R. What is the lowest speed at which a rock thrown from the ground can fly over the reservoir just by touching its top?"> | |||
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| <meta property="og:description" content="A spherical tank standing on the ground has a radius of R. What is the lowest speed at which a rock thrown from the ground can fly over the reservoir just by touching its top?"> | |||
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| <title>A spherical tank standing on the ground has a radius of R. What is the lowest speed at which a rock thrown from the ground can fly over the reservoir just by touching its top?</title> | |||
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| }); | |||
| </script> | |||
| </head> | |||
| <body style=""> | |||
| <header style="text-align:center;"> | |||
| <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> | |||
| $1.3.27^*.$ A spherical tank standing on the ground has a radius of $R$. What is the lowest speed at which a rock thrown from the ground can fly over the reservoir just by touching its top? | |||
| </p> | |||
| <h3>Solution</h3> | |||
| <p> | |||
| <p>The stone must be thrown at an angle $\alpha$ to the horizon, satisfying the equations obtained in <a href="../1.3.6" target="_blank">1.3.6</a>:</p> | |||
| <p class="exp"> | |||
| $$v_x = v \cos \alpha; \quad v_y = v \sin \alpha - gt;$$ | |||
| $$x = vt \cos \alpha; \quad y = vt \sin \alpha - gt^2 / 2.$$ | |||
| </p> | |||
| <p>The time it takes for the stone to rise to the maximum height $2R$ is found as</p> | |||
| <p class="exp"> | |||
| $$ t_1 = \frac{v_0 \sin \alpha}{g} $$ | |||
| </p> | |||
| <p>The maximum height of the stone lift along the vertical axis should be equal to $y_{max} = 2R$, therefore</p> | |||
| <p class="exp"> | |||
| $$ \frac{v_0^2 \sin^2 \alpha}{2g} = 2R $$ | |||
| </p> | |||
| <p>Determine the value of the initial throw speed</p> | |||
| <p class="exp"> | |||
| $$ v_0 = \sqrt{\frac{4gR}{\sin^2 \alpha}} $$ | |||
| </p> | |||
| <p>The angle $\alpha$ at which the stone should be thrown is determined from the initial conditions</p> | |||
| <p class="exp"> | |||
| $$ v_0t_1 \cos \alpha = \frac{v_0^2 \sin \alpha \cos \alpha}{g}$$ | |||
| $$ 2R = \frac{v_0^2 \sin \alpha \cos \alpha}{g}$$ | |||
| $$\tan \alpha = 2; \quad \alpha = \text{arctg} \;2 \approx 63^\circ$$ | |||
| </p> | |||
| <p>Substituting into the formula for $v_0$</p> | |||
| <p class="exp"> | |||
| $$ \fbox{$v_0 = \sqrt{\frac{4gR}{\sin^2 63^\circ}} = \sqrt{5Rg}$} $$ | |||
| </p> | |||
| </p> | |||
| <h4>Answer</h4> | |||
| <p> | |||
| $$v = \sqrt{5gR}$$ | |||
| </p> | |||
| <footer class="row container"> | |||
| <br> | |||
| <p> | |||
| <small> © <strong>Savchenko Solutions</strong>, 2023-2024 <br></small> | |||
| </p> | |||
| <p> | |||
| <small>All rights belong to the authors. <br> Commercial use of materials - with the written permission of the authors. <br> aliaksandr@savchenkosolutions.com <br></small> | |||
| </p> | |||
| </footer> | |||
| </body> | |||
| </html> | |||