From a hose lying on the ground, water shoots at an angle of 45^\circ to the horizon with an initial velocity of 10 \frac{m}{s}. The cross-sectional area of the hose opening is 5 cm^2 . Determine the mass of the jet in the air.

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Solutions of Savchenko Problems in Physics
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    <h3 id="back-link"><a href="/#1.3">$\leftarrow$Back</a></h3>

    <h3> Statement </h3>
    <p>
        $1.3.12^*.$ From a hose lying on the ground, water shoots at an angle of $45^\circ$ to the horizon with an initial velocity of $10$ $\frac{m}{s}$. The cross-sectional area of the hose opening is $5$ $cm^2$ . Determine the mass of the jet in the air.
    </p>

    <h3>Solution</h3>
    <p>
        <p>First, let's determine the flight time of each water drop. This time is:</p>

$$t = \frac{2v_0 \cdot \sin \alpha}{g}$$

There is water in the air, flowing out of the hose in exactly time $t$. The mass of this water is easy to determine: it is the mass of water in a cylinder of height $v_0 t$ and base area $S$.

$$m = \rho S v_0 t = \rho S \frac{2v_0^2 \cdot \sin \alpha}{g}$$

$$m = \frac{2 \rho v_0^2 S \cdot \sin \alpha}{g} = 7.2 \text{ kg}.$$

    </p>

    <h4>Answer</h4>
    <p>
        $$x = \frac{2\nu ^{2}}{g(tg\beta +tg\alpha )}$$
    </p>


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