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>
        $1.3.13^*.$ 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>a) We will use the formula for the flight path obtained in <a href="/1.3.6" target="_blank">1.3.6</a>:</p>

$$y(x) = x \tan\varphi-\frac{gx^2}{2v^2\cos^2 \varphi}$$

Let's use the hint,

$$y(x) = x \tan\varphi-\frac{gx^2}{2v^2}(1+\tan^2 \varphi)\quad(1)$$

Where do we express the angle $\varphi$:

$$\tan\varphi=\frac{v^2\pm\sqrt{v^4-2gv^2y-g^2x^2}}{gx}$$

b) Let's depict a set of trajectories, changing the angle $\varphi$ from $-\frac{\pi}{2}$ to $\frac{\pi}{2}$:

Having analyzed the domain of definition of the function $(1)$, we emphasize that the expression under the root is not negative

$$v^4-2gv^2y-g^2x^2 \geq 0$$

$$y \leq \frac{v^2}{2g}-\frac{gx^2}{2v^2}$$

c) From $(1)$, we get:

$$v_{x}v_{y}x-\frac{1}{2}gx^{2}=v_{x}^{2}y$$

We will use the formula for horizontal and vertical coordinates obtained in 1.3.6:

$$y(t) = vt \cdot \sin{\varphi} - \frac{gt^2}{2}$$

$$x(t) = vt \cdot \cos{\varphi}$$

Let $\xi =\cos{\varphi}$,

$$v_y = v \xi$$

$$v_x = v \sqrt{1-\xi^2}$$

Getting a function

$$v(\xi )=\frac{\sqrt{g}x}{\sqrt{2 \xi\sqrt{1-\xi^{2}}x-\xi^{2}y}} \quad(2)$$

The maximum of the radical value corresponds to the minimum of the function (2)

$$(\sqrt{x^{2}+y^{2}-y})/2 \rightarrow \max$$

Minimum of function $(2)$

$$\fbox{$v^{2}=\frac{gx}{\sqrt{1+(\frac{y}{x})^{2}}-\frac{y}{x}}$}$$

    </p>

    <h4>Answer</h4>
    <p>
        $$\tan\varphi=\frac{v^2\pm\sqrt{v^4-2gv^2y-g^2x^2}}{gx}.$$

$$y=\frac{v^2}{2g}-\frac{gx^2}{2v^2}$$
$$v_{\text{min}}=\sqrt{g(y+\sqrt{x^2+y^2})}$$

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