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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> | | <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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| <h3 id="back-link"><a href="../../#1.3">$\leftarrow$Back</a></h3> | | <h3 id="back-link"><a href="../../#1.3">$\leftarrow$Back</a></h3> |
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| <h3> Statement </h3> | | <h3> Statement </h3> |
| <p> | | <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. | | $1.3.13^*.$ 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> | | </p> |
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| <h3>Solution</h3> | | <h3>Solution</h3> |
| <p> | | <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> | | <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> |
| $$ y(x) = x \tan \varphi-\frac{gx^2}{2v^2\cos^2\varphi} $$ | | $$ y(x) = x \tan \varphi-\frac{gx^2}{2v^2\cos^2\varphi} $$ |
| <p>Let's use the <a href="#hint">hint</a>,</p> | | <p>Let's use the <a href="#hint">hint</a>,</p> |
| $$ y(x) = x \tan \varphi-\frac{gx^2}{2v^2}(1+\tan ^2 \varphi)\;(1) $$ | | $$ y(x) = x \tan \varphi-\frac{gx^2}{2v^2}(1+\tan ^2 \varphi)\;(1) $$ |
| <p>Where do we express the angle $ \varphi$:</p> | | <p>Where do we express the angle $ \varphi$:</p> |
| $$ \tan \varphi=\frac{v^2\pm\sqrt{v^4-2gv^2y-g^2x^2}}{gx} $$ | | $$ \tan \varphi=\frac{v^2\pm\sqrt{v^4-2gv^2y-g^2x^2}}{gx} $$ |
| <p>b) Let's depict a set of trajectories, changing the angle $\varphi$ from $-\frac{\pi}{2}$ to $\frac{\pi}{2}$:</p> | | <p>b) Let's depict a set of trajectories, changing the angle $\varphi$ from $-\frac{\pi}{2}$ to $\frac{\pi}{2}$:</p> |
| </p> | | </p> |
| <center> | | <center> |
| <figure> | | <figure> |
| <img src="https://savchenkosolutions.com/1/1.3.13/graph.png" | | <img src="https://savchenkosolutions.com/1/1.3.13/graph.png" |
| loading="lazy" width="450" /> | | loading="lazy" width="450" /> |
| <figcaption> | | <figcaption> |
| Family of trajectories for different angles | | Family of trajectories for different angles |
| </figcaption> | | </figcaption> |
| </figure> | | </figure> |
| </center> | | </center> |
| <p> | | <p> |
| <p>Having analyzed the domain of definition of the function $(1)$, we emphasize that the expression under the root is not negative</p> | | <p>Having analyzed the domain of definition of the function $(1)$, we emphasize that the expression under the root is not negative</p> |
| $$ v^4-2gv^2y-g^2x^2 \geq 0 $$ | | $$ v^4-2gv^2y-g^2x^2 \geq 0 $$ |
| | | |
| $$ y \leq \frac{v^2}{2g}-\frac{gx^2}{2v^2} $$ | | $$ y \leq \frac{v^2}{2g}-\frac{gx^2}{2v^2} $$ |
| <p>c) From $(1)$, we get:</p> | | <p>c) From $(1)$, we get:</p> |
| $$ v_{x}v_{y}x-\frac{1}{2}gx^{2}=v_{x}^{2}y $$ | | $$ v_{x}v_{y}x-\frac{1}{2}gx^{2}=v_{x}^{2}y $$ |
| <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> | | <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> |
| $$ y(t) = vt \cdot \sin{\varphi} - \frac{gt^2}{2} $$ | | $$ y(t) = vt \cdot \sin{\varphi} - \frac{gt^2}{2} $$ |
| | | |
| $$ x(t) = vt \cdot \cos{\varphi} $$ | | $$ x(t) = vt \cdot \cos{\varphi} $$ |
| <p>Let $\xi=\cos{\varphi}$, </p> | | <p>Let $\xi=\cos{\varphi}$, </p> |
| $$ v_y = v \xi $$ | | $$ v_y = v \xi $$ |
| | | |
| $$ v_x = v \sqrt{1-\xi ^2} $$ | | $$ v_x = v \sqrt{1-\xi ^2} $$ |
| <p>Getting a function</p> | | <p>Getting a function</p> |
| $$ v(\xi)=\frac{\sqrt{g}x}{\sqrt{2 \xi \sqrt{1-\xi ^{2}}x-\xi ^{2}y}} \;(2) $$ | | $$ v(\xi)=\frac{\sqrt{g}x}{\sqrt{2 \xi \sqrt{1-\xi ^{2}}x-\xi ^{2}y}} \;(2) $$ |
| <p>The maximum of the radical value corresponds to the minimum of the function (2)</p> | | <p>The maximum of the radical value corresponds to the minimum of the function (2)</p> |
| $$ (\sqrt{x^{2}+y^{2}-y})/2 \rightarrow \max $$ | | $$ (\sqrt{x^{2}+y^{2}-y})/2 \rightarrow \max $$ |
| <p>Minimum of function $(2)$</p> | | <p>Minimum of function $(2)$</p> |
| $$ \fbox{$v^{2}=\frac{gx}{\sqrt{1+(\frac{y}{x})^{2}}-\frac{y}{x}}$} $$ | | $$ \fbox{$v^{2}=\frac{gx}{\sqrt{1+(\frac{y}{x})^{2}}-\frac{y}{x}}$} $$ |
| | | |
| </p> | | </p> |
| | | |
| <h4>Answer</h4> | | <h4>Answer</h4> |
| <p> | | <p> |
| $$\tan \varphi=\frac{v^2\pm\sqrt{v^4-2gv^2y-g^2x^2}}{gx}.$$ | | $$\tan \varphi=\frac{v^2\pm\sqrt{v^4-2gv^2y-g^2x^2}}{gx}.$$ |
| $$y=\frac{v^2}{2g}-\frac{gx^2}{2v^2}$$ | | $$y=\frac{v^2}{2g}-\frac{gx^2}{2v^2}$$ |
| $$v_{\text{min}}=\sqrt{g(y+\sqrt{x^2+y^2})}$$ | | $$v_{\text{min}}=\sqrt{g(y+\sqrt{x^2+y^2})}$$ |
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