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<meta property="og: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.">
<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>
$1.3.13^*.$ 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>
$$ y(x) = x \tan\varphi-\frac{gx^2}{2v^2}(1+\tan ^2 \varphi)\;(1) $$
<small>All rights belong to the authors. <br> Commercial use of materials - with the written permission of the authors. <br> alex@savchenkosolutions.com <br></small>
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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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<meta property="og:image" content="img/logo.png">
<meta property="og:image" content="img/logo.png">
@@ -14,9 +14,9 @@
<meta property="og: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.">
<meta property="og: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.">
<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>
$1.3.13^*.$ 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>
<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>
$$ y(x) = x \tan\varphi-\frac{gx^2}{2v^2}(1+\tan ^2 \varphi)\;(1) $$
$$ y(x) = x \tan\varphi-\frac{gx^2}{2v^2}(1+\tan ^2 \varphi)\;(1) $$
<small>All rights belong to the authors. <br> Commercial use of materials - with the written permission of the authors. <br> alex@savchenkosolutions.com <br></small>
<small>All rights belong to the authors. <br> Commercial use of materials - with the written permission of the authors. <br> alex@savchenkosolutions.com <br></small>
