Translated 1.3.3-1.3.15; Fixed GUI image generator

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+ <meta name="description" content="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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+ <title>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.</title>
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+ <header style="text-align:center;">
+ <h2>Solutions of Savchenko Problems in Physics</h2>
+ <p class="author">
+ Aliaksandr Melnichenka <br/>
+ October 2023
+ </p>
+ </header>
+
+ <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>
+$$ y(x) = x \tan \varphi-\frac{gx^2}{2v^2\cos^2\varphi} $$
+<p>Let's use the <a href="#hint">hint</a>,</p>
+$$ y(x) = x \tan \varphi-\frac{gx^2}{2v^2}(1+\tan ^2 \varphi)\;(1) $$
+<p>Where do we express the angle $ \varphi$:</p>
+$$ \tan \varphi=\frac{v^2\pm\sqrt{v^4-2gv^2y-g^2x^2}}{gx} $$
+<p>b) Let's depict a set of trajectories, changing the angle $\varphi$ from $-\frac{\pi}{2}$ to $\frac{\pi}{2}$:</p>
+</p>
+<center>
+ <figure>
+ <img src="https://savchenkosolutions.com/1/1.3.13/graph.png"
+ loading="lazy" width="450" />
+ <figcaption>
+ Family of trajectories for different angles
+ </figcaption>
+ </figure>
+</center>
+<p>
+<p>Having analyzed the domain of definition of the function $(1)$, we emphasize that the expression under the root is not negative</p>
+$$ v^4-2gv^2y-g^2x^2 \geq 0 $$
+
+$$ y \leq \frac{v^2}{2g}-\frac{gx^2}{2v^2} $$
+<p>c) From $(1)$, we get:</p>
+$$ v_{x}v_{y}x-\frac{1}{2}gx^{2}=v_{x}^{2}y $$
+<p>We will use the formula for horizontal and vertical coordinates obtained in <a href="../1.3.6" target="_blank">1.3.6</a>:</p>
+$$ y(t) = vt \cdot \sin{\varphi} - \frac{gt^2}{2} $$
+
+$$ x(t) = vt \cdot \cos{\varphi} $$
+<p>Let $\xi=\cos{\varphi}$, </p>
+$$ v_y = v \xi $$
+
+$$ v_x = v \sqrt{1-\xi ^2} $$
+<p>Getting a function</p>
+$$ v(\xi)=\frac{\sqrt{g}x}{\sqrt{2 \xi \sqrt{1-\xi ^{2}}x-\xi ^{2}y}} \;(2) $$
+<p>The maximum of the radical value corresponds to the minimum of the function (2)</p>
+$$ (\sqrt{x^{2}+y^{2}-y})/2 \rightarrow \max $$
+<p>Minimum of function $(2)$</p>
+$$ \fbox{$v^{2}=\frac{gx}{\sqrt{1+(\frac{y}{x})^{2}}-\frac{y}{x}}$} $$
+
+ </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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