Updated header text

astrosander edited Sep 21, 2024, 9:21 PM

View solution History

revision #7342 parent #6414 GitHub a2d7ded ← older newer →

en/1.3.6.md +7 −5

			<!DOCTYPE html>
			<html lang="en">

			<head>
			<meta charset="utf-8">
			<meta name="viewport" content="width=device-width, initial-scale=1.0">
			<meta http-equiv="content-language" content="en">
			<meta name="keywords" content="Savchenko Problems in Physics, Savchenko solutions, physics problems, physics olympiad preparation, IPhO, Jaan Kalda">
			<meta name="description" content="The gun is fired at an angle \varphi to the horizon. Initial velocity of the projectile v. The ground surface is horizontal. Find: <br>a) the horizontal and vertical projections of velocity as a function of time; <br>b) the dependence of the x and y coordinates on time; <br>c) the trajectory equation, i.e. the dependence of y on x; <br>d) the flight time, the highest altitude and range of the projectile.">
			<meta name="author" content="Aliaksandr Melnichenka">
			<meta name="date" content="2023-10" scheme="YYYY-MM">
			<meta property="og:title" content="The gun is fired at an angle \varphi to the horizon. Initial velocity of the projectile v. The ground surface is horizontal. Find: <br>a) the horizontal and vertical projections of velocity as a function of time; <br>b) the dependence of the x and y coordinates on time; <br>c) the trajectory equation, i.e. the dependence of y on x; <br>d) the flight time, the highest altitude and range of the projectile.">
			<meta property="og:image" content="img/logo.png">
			<meta property="og:description" content="The gun is fired at an angle \varphi to the horizon. Initial velocity of the projectile v. The ground surface is horizontal. Find: <br>a) the horizontal and vertical projections of velocity as a function of time; <br>b) the dependence of the x and y coordinates on time; <br>c) the trajectory equation, i.e. the dependence of y on x; <br>d) the flight time, the highest altitude and range of the projectile.">
			<meta name="yandex-verification" content="6cfda41f74038368">
			<title>The gun is fired at an angle \varphi to the horizon. Initial velocity of the projectile v. The ground surface is horizontal. Find: <br>a) the horizontal and vertical projections of velocity as a function of time; <br>b) the dependence of the x and y coordinates on time; <br>c) the trajectory equation, i.e. the dependence of y on x; <br>d) the flight time, the highest altitude and range of the projectile.</title>
			<link rel="stylesheet" href="../../css/css-latex/style.css">
			<link rel="icon" href="../../img/logo.png" type="image/png">
			<script src="../../js/jquery-1.10.1.min.js"></script>
			<script async src="https://cdnjs.cloudflare.com/ajax/libs/mathjax/2.7.7/MathJax.js?config=TeX-MML-AM_CHTML"></script>
			<script type="text/x-mathjax-config">
			MathJax.Hub.Config({
			extensions: ['tex2jax.js'],
			jax: ['input/TeX', 'output/HTML-CSS'],
			tex2jax: {
			inlineMath: [['$', '$'], ['$', '$']],
			processEscapes: true,
			processClass: 'tex2jax',
			ignoreClass: 'html'
			},
			showProcessingMessages: false,
			messageStyle: 'none'
			});
			</script>
		@@ -35,11 +35,13 @@
			</head>
			<body style="">
			<header style="text-align:center;">
		−	~~<h2>~~Solutions of Savchenko Problems in Physics~~</h2~~>
		−	~~<p class="author"~~>
		−	Aliaksandr Melnichenka <br/>
		−	October 2023
		−	</p>
		+	<div id = "logo">
		+	<span><img src = "../../img/book.png"><span><span>Savchenko Solutions</span>
		+	</div>
		+	<p class="author">
		+	Solutions of Savchenko Problems in Physics <br/>
		+	knowledge must be free
		+	</p>
			</header>

			<h3 id="back-link"><a href="../../#1.3">$\leftarrow$Back</a></h3>

			<h3> Statement </h3>
			<p>
			$1.3.6.$ The gun is fired at an angle $\varphi$ to the horizon. Initial velocity of the projectile $v.$ The ground surface is horizontal. Find: <br>a) the horizontal and vertical projections of velocity as a function of time; <br>b) the dependence of the $x$ and $y$ coordinates on time; <br>c) the trajectory equation, i.e. the dependence of $y$ on $x$; <br>d) the flight time, the highest altitude and range of the projectile.
			</p>

			<h3>Solution</h3>
			<p>
			a) Since the external force (gravitational force) acts only in the vertical direction, the horizontal component of the velocity $v_x$ remains unchanged
			$$ v_x = v \cdot \cos{\varphi}\;(1) $$
			And the vertical component depends as a function of time
			$$ v_y(t) = v \cdot \sin{\varphi} - gt\;(2) $$
			b) Based on the obtained expressions for the speed $(1)$ and $(2)$, we obtain the dependence of the horizontal and vertical coordinates on time, respectively:
			$$ y(t) = vt \cdot \sin{\varphi} - \frac{gt^2}{2}\;(3) $$

			$$ x(t) = vt \cdot \cos{\varphi}\;(4) $$
			c) From $(4)$ we express $t$:
			$$ t = \frac{x}{v \cdot \cos{\varphi}} \;(5) $$
			And we substitute in $(3)$
			$$ y(x) = x \tan \varphi-\frac{gx^2}{2v^2\cos^2\varphi}\;(6) $$
			d) The maximum height is achieved when the velocity is directed horizontally $(v_y = 0)$:
			$$ t_0 = \frac{v\sin\varphi}{g}\;(7) $$
			Substituting $(7)$ into $(3)$:
			$$ H = \frac{v^2}{2g}\operatorname{sin}^2\varphi\;(8) $$
			From the symmetry of the parabola, the time after which the body will again be on the ground$(y = 0)$, we find as:
			$$ T = 2t_0 = \frac{2v\sin\varphi}{g}\;(9) $$
			Substituting $(9)$ into $(4)$:
			$$ L=\frac{v^2}{g}\operatorname{sin}2\varphi\;(10) $$

			</p>

			<h4>Answer</h4>
			<p>
			a) $v_x= v\cos\varphi, v_y = v\sin\varphi-gt$<br>
			b) $x = (v\cos\varphi)t, y = (v\sin{\varphi})t - gt^2/2$<br>
			c) $y = x \tan \varphi-\frac{gx^2}{2v^2\cos^2\varphi} =x\operatorname{tg}\varphi-\frac{gx^2}{2v^2}(\operatorname{tg}^2\varphi+1)$<br>
			d) $T=\frac{2v}{g}\operatorname{sin}\varphi, H=\frac{v^2}{2g}\operatorname{sin}^2\varphi, L=\frac{v^2}{g}\operatorname{sin}2\varphi$<br>
			</p>


			<footer class="row container">
			<br>
			<p>
			<small> © <strong>Savchenko Solutions</strong>, 2023-2024 <br></small>
			</p>
			<p>
			<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>
			</p>
			</footer>
			</body>

			</html>
		unchanged lines 52

	<!DOCTYPE html>		<!DOCTYPE html>
	<html lang="en">		<html lang="en">

	<head>		<head>
	<meta charset="utf-8">		<meta charset="utf-8">
	<meta name="viewport" content="width=device-width, initial-scale=1.0">		<meta name="viewport" content="width=device-width, initial-scale=1.0">
	<meta http-equiv="content-language" content="en">		<meta http-equiv="content-language" content="en">
	<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="The gun is fired at an angle \varphi to the horizon. Initial velocity of the projectile v. The ground surface is horizontal. Find: <br>a) the horizontal and vertical projections of velocity as a function of time; <br>b) the dependence of the x and y coordinates on time; <br>c) the trajectory equation, i.e. the dependence of y on x; <br>d) the flight time, the highest altitude and range of the projectile.">		<meta name="description" content="The gun is fired at an angle \varphi to the horizon. Initial velocity of the projectile v. The ground surface is horizontal. Find: <br>a) the horizontal and vertical projections of velocity as a function of time; <br>b) the dependence of the x and y coordinates on time; <br>c) the trajectory equation, i.e. the dependence of y on x; <br>d) the flight time, the highest altitude and range of the projectile.">
	<meta name="author" content="Aliaksandr Melnichenka">		<meta name="author" content="Aliaksandr Melnichenka">
	<meta name="date" content="2023-10" scheme="YYYY-MM">		<meta name="date" content="2023-10" scheme="YYYY-MM">
	<meta property="og:title" content="The gun is fired at an angle \varphi to the horizon. Initial velocity of the projectile v. The ground surface is horizontal. Find: <br>a) the horizontal and vertical projections of velocity as a function of time; <br>b) the dependence of the x and y coordinates on time; <br>c) the trajectory equation, i.e. the dependence of y on x; <br>d) the flight time, the highest altitude and range of the projectile.">		<meta property="og:title" content="The gun is fired at an angle \varphi to the horizon. Initial velocity of the projectile v. The ground surface is horizontal. Find: <br>a) the horizontal and vertical projections of velocity as a function of time; <br>b) the dependence of the x and y coordinates on time; <br>c) the trajectory equation, i.e. the dependence of y on x; <br>d) the flight time, the highest altitude and range of the projectile.">
	<meta property="og:image" content="img/logo.png">		<meta property="og:image" content="img/logo.png">
	<meta property="og:description" content="The gun is fired at an angle \varphi to the horizon. Initial velocity of the projectile v. The ground surface is horizontal. Find: <br>a) the horizontal and vertical projections of velocity as a function of time; <br>b) the dependence of the x and y coordinates on time; <br>c) the trajectory equation, i.e. the dependence of y on x; <br>d) the flight time, the highest altitude and range of the projectile.">		<meta property="og:description" content="The gun is fired at an angle \varphi to the horizon. Initial velocity of the projectile v. The ground surface is horizontal. Find: <br>a) the horizontal and vertical projections of velocity as a function of time; <br>b) the dependence of the x and y coordinates on time; <br>c) the trajectory equation, i.e. the dependence of y on x; <br>d) the flight time, the highest altitude and range of the projectile.">
	<meta name="yandex-verification" content="6cfda41f74038368">		<meta name="yandex-verification" content="6cfda41f74038368">
	<title>The gun is fired at an angle \varphi to the horizon. Initial velocity of the projectile v. The ground surface is horizontal. Find: <br>a) the horizontal and vertical projections of velocity as a function of time; <br>b) the dependence of the x and y coordinates on time; <br>c) the trajectory equation, i.e. the dependence of y on x; <br>d) the flight time, the highest altitude and range of the projectile.</title>		<title>The gun is fired at an angle \varphi to the horizon. Initial velocity of the projectile v. The ground surface is horizontal. Find: <br>a) the horizontal and vertical projections of velocity as a function of time; <br>b) the dependence of the x and y coordinates on time; <br>c) the trajectory equation, i.e. the dependence of y on x; <br>d) the flight time, the highest altitude and range of the projectile.</title>
	<link rel="stylesheet" href="../../css/css-latex/style.css">		<link rel="stylesheet" href="../../css/css-latex/style.css">
	<link rel="icon" href="../../img/logo.png" type="image/png">		<link rel="icon" href="../../img/logo.png" type="image/png">
	<script src="../../js/jquery-1.10.1.min.js"></script>		<script src="../../js/jquery-1.10.1.min.js"></script>
	<script async src="https://cdnjs.cloudflare.com/ajax/libs/mathjax/2.7.7/MathJax.js?config=TeX-MML-AM_CHTML"></script>		<script async src="https://cdnjs.cloudflare.com/ajax/libs/mathjax/2.7.7/MathJax.js?config=TeX-MML-AM_CHTML"></script>
	<script type="text/x-mathjax-config">		<script type="text/x-mathjax-config">
	MathJax.Hub.Config({		MathJax.Hub.Config({
	extensions: ['tex2jax.js'],		extensions: ['tex2jax.js'],
	jax: ['input/TeX', 'output/HTML-CSS'],		jax: ['input/TeX', 'output/HTML-CSS'],
	tex2jax: {		tex2jax: {
	inlineMath: [['$', '$'], ['$', '$']],		inlineMath: [['$', '$'], ['$', '$']],
	processEscapes: true,		processEscapes: true,
	processClass: 'tex2jax',		processClass: 'tex2jax',
	ignoreClass: 'html'		ignoreClass: 'html'
	},		},
	showProcessingMessages: false,		showProcessingMessages: false,
	messageStyle: 'none'		messageStyle: 'none'
	});		});
	</script>		</script>
	@@ -35,11 +35,13 @@
	</head>		</head>
	<body style="">		<body style="">
	<header style="text-align:center;">		<header style="text-align:center;">
	~~<h2>~~Solutions of Savchenko Problems in Physics~~</h2~~>		<div id = "logo">
	~~<p class="author"~~>		<span><img src = "../../img/book.png"><span><span>Savchenko Solutions</span>
	Aliaksandr Melnichenka <br/>		</div>
	October 2023		<p class="author">
	</p>		Solutions of Savchenko Problems in Physics <br/>
			knowledge must be free
			</p>
	</header>		</header>

	<h3 id="back-link"><a href="../../#1.3">$\leftarrow$Back</a></h3>		<h3 id="back-link"><a href="../../#1.3">$\leftarrow$Back</a></h3>

	<h3> Statement </h3>		<h3> Statement </h3>
	<p>		<p>
	$1.3.6.$ The gun is fired at an angle $\varphi$ to the horizon. Initial velocity of the projectile $v.$ The ground surface is horizontal. Find: <br>a) the horizontal and vertical projections of velocity as a function of time; <br>b) the dependence of the $x$ and $y$ coordinates on time; <br>c) the trajectory equation, i.e. the dependence of $y$ on $x$; <br>d) the flight time, the highest altitude and range of the projectile.		$1.3.6.$ The gun is fired at an angle $\varphi$ to the horizon. Initial velocity of the projectile $v.$ The ground surface is horizontal. Find: <br>a) the horizontal and vertical projections of velocity as a function of time; <br>b) the dependence of the $x$ and $y$ coordinates on time; <br>c) the trajectory equation, i.e. the dependence of $y$ on $x$; <br>d) the flight time, the highest altitude and range of the projectile.
	</p>		</p>

	<h3>Solution</h3>		<h3>Solution</h3>
	<p>		<p>
	a) Since the external force (gravitational force) acts only in the vertical direction, the horizontal component of the velocity $v_x$ remains unchanged		a) Since the external force (gravitational force) acts only in the vertical direction, the horizontal component of the velocity $v_x$ remains unchanged
	$$ v_x = v \cdot \cos{\varphi}\;(1) $$		$$ v_x = v \cdot \cos{\varphi}\;(1) $$
	And the vertical component depends as a function of time		And the vertical component depends as a function of time
	$$ v_y(t) = v \cdot \sin{\varphi} - gt\;(2) $$		$$ v_y(t) = v \cdot \sin{\varphi} - gt\;(2) $$
	b) Based on the obtained expressions for the speed $(1)$ and $(2)$, we obtain the dependence of the horizontal and vertical coordinates on time, respectively:		b) Based on the obtained expressions for the speed $(1)$ and $(2)$, we obtain the dependence of the horizontal and vertical coordinates on time, respectively:
	$$ y(t) = vt \cdot \sin{\varphi} - \frac{gt^2}{2}\;(3) $$		$$ y(t) = vt \cdot \sin{\varphi} - \frac{gt^2}{2}\;(3) $$

	$$ x(t) = vt \cdot \cos{\varphi}\;(4) $$		$$ x(t) = vt \cdot \cos{\varphi}\;(4) $$
	c) From $(4)$ we express $t$:		c) From $(4)$ we express $t$:
	$$ t = \frac{x}{v \cdot \cos{\varphi}} \;(5) $$		$$ t = \frac{x}{v \cdot \cos{\varphi}} \;(5) $$
	And we substitute in $(3)$		And we substitute in $(3)$
	$$ y(x) = x \tan \varphi-\frac{gx^2}{2v^2\cos^2\varphi}\;(6) $$		$$ y(x) = x \tan \varphi-\frac{gx^2}{2v^2\cos^2\varphi}\;(6) $$
	d) The maximum height is achieved when the velocity is directed horizontally $(v_y = 0)$:		d) The maximum height is achieved when the velocity is directed horizontally $(v_y = 0)$:
	$$ t_0 = \frac{v\sin\varphi}{g}\;(7) $$		$$ t_0 = \frac{v\sin\varphi}{g}\;(7) $$
	Substituting $(7)$ into $(3)$:		Substituting $(7)$ into $(3)$:
	$$ H = \frac{v^2}{2g}\operatorname{sin}^2\varphi\;(8) $$		$$ H = \frac{v^2}{2g}\operatorname{sin}^2\varphi\;(8) $$
	From the symmetry of the parabola, the time after which the body will again be on the ground$(y = 0)$, we find as:		From the symmetry of the parabola, the time after which the body will again be on the ground$(y = 0)$, we find as:
	$$ T = 2t_0 = \frac{2v\sin\varphi}{g}\;(9) $$		$$ T = 2t_0 = \frac{2v\sin\varphi}{g}\;(9) $$
	Substituting $(9)$ into $(4)$:		Substituting $(9)$ into $(4)$:
	$$ L=\frac{v^2}{g}\operatorname{sin}2\varphi\;(10) $$		$$ L=\frac{v^2}{g}\operatorname{sin}2\varphi\;(10) $$

	</p>		</p>

	<h4>Answer</h4>		<h4>Answer</h4>
	<p>		<p>
	a) $v_x= v\cos\varphi, v_y = v\sin\varphi-gt$<br>		a) $v_x= v\cos\varphi, v_y = v\sin\varphi-gt$<br>
	b) $x = (v\cos\varphi)t, y = (v\sin{\varphi})t - gt^2/2$<br>		b) $x = (v\cos\varphi)t, y = (v\sin{\varphi})t - gt^2/2$<br>
	c) $y = x \tan \varphi-\frac{gx^2}{2v^2\cos^2\varphi} =x\operatorname{tg}\varphi-\frac{gx^2}{2v^2}(\operatorname{tg}^2\varphi+1)$<br>		c) $y = x \tan \varphi-\frac{gx^2}{2v^2\cos^2\varphi} =x\operatorname{tg}\varphi-\frac{gx^2}{2v^2}(\operatorname{tg}^2\varphi+1)$<br>
	d) $T=\frac{2v}{g}\operatorname{sin}\varphi, H=\frac{v^2}{2g}\operatorname{sin}^2\varphi, L=\frac{v^2}{g}\operatorname{sin}2\varphi$<br>		d) $T=\frac{2v}{g}\operatorname{sin}\varphi, H=\frac{v^2}{2g}\operatorname{sin}^2\varphi, L=\frac{v^2}{g}\operatorname{sin}2\varphi$<br>
	</p>		</p>


	<footer class="row container">		<footer class="row container">
	<br>		<br>
	<p>		<p>
	<small> © <strong>Savchenko Solutions</strong>, 2023-2024 <br></small>		<small> © <strong>Savchenko Solutions</strong>, 2023-2024 <br></small>
	</p>		</p>
	<p>		<p>
	<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>
	</p>		</p>
	</footer>		</footer>
	</body>		</body>

	</html>		</html>
	unchanged lines 52