Правка раздела «Solution»
en/1.3.27.md
+1 −1
| ### Statement | |||
| $1.3.27^*.$ 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? | |||
| ### Solution | |||
| The stone must be thrown at an angle $\alpha$ to the horizon, satisfying the equations obtained in [1.3.6](../1.3.6): | |||
| $$v_x = v \cos\alpha ; \quad v_y = v \sin\alpha - gt;$$ $$x = vt \cos\alpha ; \quad y = vt \sin\alpha - gt^2 / 2.$$ | |||
| The time it takes for the stone to rise to the maximum height $2R$ is found as | |||
| $$ t_1 = \frac{v_0 \sin\alpha}{g} $$ | |||
| The maximum height of the stone lift along the vertical axis should be equal to $y_{max} = 2R$, therefore | |||
| $$ \frac{v_0^2 \sin^2 \alpha}{2g} = 2R $$ | |||
| Determine the value of the initial throw speed | |||
| $$ v_0 = \sqrt{\frac{4gR}{\sin^2 \alpha}} $$ | |||
| @@ -22,7 +22,7 @@Solution | |||
| The angle $\alpha$ at which the stone should be thrown is determined from the initial conditions | |||
| − | $$ v_0t_1 \cos\alpha = \frac{v_0^2 \sin\alpha\cos\alpha}{g}$$ $$ 2R = \frac{v_0^2 \sin\alpha\cos\alpha}{g}$$ $$\tan\alpha = 2; \quad \alpha = | ||
| + | $$ 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 = \arctan 2 \approx 63^\circ$$ | ||
| Substituting into the formula for $v_0$ | |||
| $$ \fbox{$v_0 = \sqrt{\frac{4gR}{\sin^2 63^\circ}} = \sqrt{5Rg}$} $$ | |||
| #### Answer | |||
| $$v = \sqrt{5gR}$$ | |||
| ещё строк без изменений 5 | |||
| ### Statement | ### Statement | ||
| $1.3.27^*.$ A spherical tank standing on the ground has a radius of $R$. What is the lowest speed at which a rock thrown from the ground can fly over the reservoir just by touching its top? | $1.3.27^*.$ A spherical tank standing on the ground has a radius of $R$. What is the lowest speed at which a rock thrown from the ground can fly over the reservoir just by touching its top? | ||
| ### Solution | ### Solution | ||
| The stone must be thrown at an angle $\alpha$ to the horizon, satisfying the equations obtained in [1.3.6](../1.3.6): | The stone must be thrown at an angle $\alpha$ to the horizon, satisfying the equations obtained in [1.3.6](../1.3.6): | ||
| $$v_x = v \cos\alpha ; \quad v_y = v \sin\alpha - gt;$$ $$x = vt \cos\alpha ; \quad y = vt \sin\alpha - gt^2 / 2.$$ | $$v_x = v \cos\alpha ; \quad v_y = v \sin\alpha - gt;$$ $$x = vt \cos\alpha ; \quad y = vt \sin\alpha - gt^2 / 2.$$ | ||
| The time it takes for the stone to rise to the maximum height $2R$ is found as | The time it takes for the stone to rise to the maximum height $2R$ is found as | ||
| $$ t_1 = \frac{v_0 \sin\alpha}{g} $$ | $$ t_1 = \frac{v_0 \sin\alpha}{g} $$ | ||
| The maximum height of the stone lift along the vertical axis should be equal to $y_{max} = 2R$, therefore | The maximum height of the stone lift along the vertical axis should be equal to $y_{max} = 2R$, therefore | ||
| $$ \frac{v_0^2 \sin^2 \alpha}{2g} = 2R $$ | $$ \frac{v_0^2 \sin^2 \alpha}{2g} = 2R $$ | ||
| Determine the value of the initial throw speed | Determine the value of the initial throw speed | ||
| $$ v_0 = \sqrt{\frac{4gR}{\sin^2 \alpha}} $$ | $$ v_0 = \sqrt{\frac{4gR}{\sin^2 \alpha}} $$ | ||
| @@ -22,7 +22,7 @@Solution | |||
| The angle $\alpha$ at which the stone should be thrown is determined from the initial conditions | The angle $\alpha$ at which the stone should be thrown is determined from the initial conditions | ||
| $$ 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 = |
$$ v_0t_1 \cos\alpha = \frac{v_0^2 \sin\alpha\cos\alpha}{g}$$ $$ 2R = \frac{v_0^2 \sin\alpha\cos\alpha}{g}$$ $$\tan\alpha = 2; \quad \alpha = \arctan 2 \approx 63^\circ$$ | ||
| Substituting into the formula for $v_0$ | Substituting into the formula for $v_0$ | ||
| $$ \fbox{$v_0 = \sqrt{\frac{4gR}{\sin^2 63^\circ}} = \sqrt{5Rg}$} $$ | $$ \fbox{$v_0 = \sqrt{\frac{4gR}{\sin^2 63^\circ}} = \sqrt{5Rg}$} $$ | ||
| #### Answer | #### Answer | ||
| $$v = \sqrt{5gR}$$ | $$v = \sqrt{5gR}$$ | ||
| ещё строк без изменений 5 | |||