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?

Solutions of Savchenko Problems in Physics

Aliaksandr Melnichenka
October 2023

    <h3 id="back-link"><a href="/#1.3">$\leftarrow$Back</a></h3>

    <h3> Statement </h3>
    <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?

    </p>

    <h3>Solution</h3>
    <p>

$$vt_{1} \cdot \cos \alpha = vt_{2} \cdot \cos \beta$$

$$vt_{1}\cdot \sin\alpha - \frac{gt_{1}^{2}}{2}=vt_{2}\cdot \sin\beta-\frac{gt_{2}^{2}}{2}$$

From the first equation,

$$t_{1}=t_{2} \cdot \frac{\cos \beta}{\cos \alpha}$$

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.

Using trigonometric formulas:

$$t_{2}= \frac{2\nu ^{2}}{g(tg\beta +tg\alpha )cos\beta }$$

$$\fbox{$x = \frac{2\nu ^{2}}{g(tg\beta +tg\alpha )}$}$$

    </p>

    <h4>Answer</h4>
    <p>
        $$x = \frac{2\nu ^{2}}{g(tg\beta +tg\alpha )}$$
    </p>
    <p style="text-align: right; font-style: italic; font-size: 14;">
      <a href="https://www.instagram.com/_den.di__/" target="_blank">Dzikan Danila</a>
    </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>