<meta name="keywords" content="Savchenko Problems in Physics, Savchenko solutions, physics problems, physics olympiad preparation, IPhO, Jaan Kalda">
<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.">
<meta property="og:title" 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: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.">
<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>
$1.3.12^*.$ 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>
$$ t = \frac{2v_0 \cdot\sin\alpha}{g}$$
<p>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 $S$.</p>
$$ m = \rho S v_0 t = \rho S \frac{2v_0^2 \cdot\sin\alpha}{g}$$
$$ m = \frac{2 \rho v_0^2 S \cdot\sin\alpha}{g} = 7.2 \text{ kg}. $$
<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>
<meta name="keywords" content="Savchenko Problems in Physics, Savchenko solutions, physics problems, physics olympiad preparation, IPhO, Jaan Kalda">
<meta name="keywords" content="Savchenko Problems in Physics, Savchenko solutions, physics problems, physics olympiad preparation, IPhO, Jaan Kalda">
<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.">
<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.">
<meta property="og:title" 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:title" 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: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>
$1.3.12^*.$ 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.
$1.3.12^*.$ 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>
</p>
<h3>Solution</h3>
<h3>Solution</h3>
<p>
<p>
<p>First, let's determine the flight time of each water drop. This time is:</p>
<p>First, let's determine the flight time of each water drop. This time is:</p>
$$ t = \frac{2v_0 \cdot\sin\alpha}{g}$$
$$ t = \frac{2v_0 \cdot\sin\alpha}{g}$$
<p>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 $S$.</p>
<p>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 $S$.</p>
$$ m = \rho S v_0 t = \rho S \frac{2v_0^2 \cdot\sin\alpha}{g}$$
$$ m = \rho S v_0 t = \rho S \frac{2v_0^2 \cdot\sin\alpha}{g}$$
$$ m = \frac{2 \rho v_0^2 S \cdot\sin\alpha}{g} = 7.2 \text{ kg}. $$
$$ m = \frac{2 \rho v_0^2 S \cdot\sin\alpha}{g} = 7.2 \text{ kg}. $$
<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>
