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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Solutions of Savchenko Problems in Physics
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    <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>
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