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| <title>From the opening of the hose covered with a finger, two jets are shot at an angle \alpha and \beta to the horizon with the same initial velocity v. At what horizontal distance from the hole will the jets intersect?</title> | | <title>From the opening of the hose covered with a finger, two jets are shot at an angle \alpha and \beta to the horizon with the same initial velocity v. At what horizontal distance from the hole will the jets intersect?</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> |
| $1.3.11.$ From the opening of the hose covered with a finger, two jets are shot at an angle $\alpha$ and $\beta$ to the horizon with the same initial velocity $v$. At what horizontal distance from the hole will the jets intersect? | | $1.3.11.$ From the opening of the hose covered with a finger, two jets are shot at an angle $\alpha$ and $\beta$ to the horizon with the same initial velocity $v$. At what horizontal distance from the hole will the jets intersect? |
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| <figcaption> | | <figcaption> |
| For problem $1.3.11$ | | For problem $1.3.11$ |
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| <h3>Solution</h3> | | <h3>Solution</h3> |
| <p> | | <p> |
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| $$ vt_{1} \cdot \cos \alpha = vt_{2} \cdot \cos \beta $$ | | $$ vt_{1} \cdot \cos \alpha = vt_{2} \cdot \cos \beta $$ |
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| $$ vt_{1}\cdot \sin\alpha - \frac{gt_{1}^{2}}{2}=vt_{2}\cdot \sin\beta-\frac{gt_{2}^{2}}{2} $$ | | $$ vt_{1}\cdot \sin\alpha - \frac{gt_{1}^{2}}{2}=vt_{2}\cdot \sin\beta-\frac{gt_{2}^{2}}{2} $$ |
| <p>From the first equation,</p> | | <p>From the first equation,</p> |
| $$ t_{1}=t_{2} \cdot \frac{\cos \beta}{\cos \alpha} $$ | | $$ t_{1}=t_{2} \cdot \frac{\cos \beta}{\cos \alpha} $$ |
| <p>We substitute $t_{1}$ into the second equation and express $t_{2}$. Trigonometric formulas 63 will come in handy. The $t_{2}$ we have already obtained is enough to insert into $x=vt_{2}cos\beta$, where $x$ is the desired distance.</p> | | <p>We substitute $t_{1}$ into the second equation and express $t_{2}$. Trigonometric formulas 63 will come in handy. The $t_{2}$ we have already obtained is enough to insert into $x=vt_{2}cos\beta$, where $x$ is the desired distance.</p> |
| <p>Using trigonometric formulas:</p> | | <p>Using trigonometric formulas:</p> |
| $$ t_{2}= \frac{2\nu ^{2}}{g(tg\beta +tg\alpha )cos\beta } $$ | | $$ t_{2}= \frac{2\nu ^{2}}{g(tg\beta +tg\alpha )cos\beta } $$ |
| | | |
| $$ \fbox{$x = \frac{2\nu ^{2}}{g(tg\beta +tg\alpha )}$} $$ | | $$ \fbox{$x = \frac{2\nu ^{2}}{g(tg\beta +tg\alpha )}$} $$ |
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| </p> | | </p> |
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| <h4>Answer</h4> | | <h4>Answer</h4> |
| <p> | | <p> |
| $$x = \frac{2\nu ^{2}}{g(tg\beta +tg\alpha )}$$ | | $$x = \frac{2\nu ^{2}}{g(tg\beta +tg\alpha )}$$ |
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