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| <h2>Solutions of Savchenko Problems in Physics</h2> | | <h2>Solutions of Savchenko Problems in Physics</h2> |
| <p class="author"> | | <p class="author"> |
| Aliaksandr Melnichenka <br/> | | Aliaksandr Melnichenka <br/> |
| October 2023 | | October 2023 |
| </p> | | </p> |
| </header> | | </header> |
| | | |
| <h3 id="back-link"><a href="../../#1.3">$\leftarrow$Back</a></h3> | | <h3 id="back-link"><a href="../../#1.3">$\leftarrow$Back</a></h3> |
| | | |
| <h3> Statement </h3> | | <h3> Statement </h3> |
| <p> | | <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? | | $1.3.27^*.$ 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> | | </p> |
| | | |
| <h3>Solution</h3> | | <h3>Solution</h3> |
| <p> | | <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>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"> | | <p class="exp"> |
| $$v_x = v \cos \alpha; \quad v_y = v \sin \alpha - gt;$$ | | $$v_x = v \cos \alpha; \quad v_y = v \sin \alpha - gt;$$ |
| $$x = vt \cos \alpha; \quad y = vt \sin \alpha - gt^2 / 2.$$ | | $$x = vt \cos \alpha; \quad y = vt \sin \alpha - gt^2 / 2.$$ |
| </p> | | </p> |
| <p>The time it takes for the stone to rise to the maximum height $2R$ is found as</p> | | <p>The time it takes for the stone to rise to the maximum height $2R$ is found as</p> |
| <p class="exp"> | | <p class="exp"> |
| $$ t_1 = \frac{v_0 \sin \alpha}{g} $$ | | $$ t_1 = \frac{v_0 \sin \alpha}{g} $$ |
| </p> | | </p> |
| <p>The maximum height of the stone lift along the vertical axis should be equal to $y_{max} = 2R$, therefore</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"> | | <p class="exp"> |
| $$ \frac{v_0^2 \sin^2 \alpha}{2g} = 2R $$ | | $$ \frac{v_0^2 \sin^2 \alpha}{2g} = 2R $$ |
| </p> | | </p> |
| <p>Determine the value of the initial throw speed</p> | | <p>Determine the value of the initial throw speed</p> |
| <p class="exp"> | | <p class="exp"> |
| $$ v_0 = \sqrt{\frac{4gR}{\sin^2 \alpha}} $$ | | $$ v_0 = \sqrt{\frac{4gR}{\sin^2 \alpha}} $$ |
| </p> | | </p> |
| <p>The angle $\alpha$ at which the stone should be thrown is determined from the initial conditions</p> | | <p>The angle $\alpha$ at which the stone should be thrown is determined from the initial conditions</p> |
| <p class="exp"> | | <p class="exp"> |
| $$ v_0t_1 \cos \alpha = \frac{v_0^2 \sin \alpha \cos \alpha}{g}$$ | | $$ v_0t_1 \cos \alpha = \frac{v_0^2 \sin \alpha \cos \alpha}{g}$$ |
| $$ 2R = \frac{v_0^2 \sin \alpha \cos \alpha}{g}$$ | | $$ 2R = \frac{v_0^2 \sin \alpha \cos \alpha}{g}$$ |
| $$\tan \alpha = 2; \quad \alpha = \text{arctg} \;2 \approx 63^\circ$$ | | $$\tan \alpha = 2; \quad \alpha = \text{arctg} \;2 \approx 63^\circ$$ |
| </p> | | </p> |
| <p>Substituting into the formula for $v_0$</p> | | <p>Substituting into the formula for $v_0$</p> |
| <p class="exp"> | | <p class="exp"> |
| $$ \fbox{$v_0 = \sqrt{\frac{4gR}{\sin^2 63^\circ}} = \sqrt{5Rg}$} $$ | | $$ \fbox{$v_0 = \sqrt{\frac{4gR}{\sin^2 63^\circ}} = \sqrt{5Rg}$} $$ |
| </p> | | </p> |
| </p> | | </p> |
| | | |
| <h4>Answer</h4> | | <h4>Answer</h4> |
| <p> | | <p> |
| $$v = \sqrt{5gR}$$ | | $$v = \sqrt{5gR}$$ |
| </p> | | </p> |
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