{"id":1598,"date":"2019-10-15T10:21:40","date_gmt":"2019-10-15T08:21:40","guid":{"rendered":"https:\/\/www.mathweb.fr\/euclide\/?p=1598"},"modified":"2020-04-19T16:19:50","modified_gmt":"2020-04-19T14:19:50","slug":"saut-parabole-et-physique","status":"publish","type":"post","link":"https:\/\/www.mathweb.fr\/euclide\/2019\/10\/15\/saut-parabole-et-physique\/","title":{"rendered":"Saut, parabole et physique"},"content":{"rendered":"\n<p class=\"wp-block-paragraph\"><em>Cet article est principalement destin\u00e9 aux \u00e9l\u00e8ves de 1\u00e8re Math Sp\u00e9cialit\u00e9.<\/em><\/p>\n\n\n\n<p class=\"wp-block-paragraph\">Parlons dans cet article de math\u00e9matiques, et plus pr\u00e9cis\u00e9ment du second degr\u00e9. Alors, vous allez me dire : \u00ab\u00a0oui, mais bon ! C&rsquo;est super simple, il suffit de conna\u00eetre les formules et on sait tout faire.\u00a0\u00bb Ce n&rsquo;est pas totalement faux&#8230; mais ce n&rsquo;est pas suffisant ! Il y a beaucoup de situations qui font intervenir le second degr\u00e9, notamment ce probl\u00e8me&#8230;<\/p>\n\n\n\n<!--more-->\n\n\n\n<div id=\"ez-toc-container\" class=\"ez-toc-v2_0_85 counter-hierarchy ez-toc-counter ez-toc-white ez-toc-container-direction\">\n<div class=\"ez-toc-title-container\">\n<p class=\"ez-toc-title\" style=\"cursor:inherit\">Au menu sur cette page...<\/p>\n<span class=\"ez-toc-title-toggle\"><a href=\"#\" class=\"ez-toc-pull-right ez-toc-btn ez-toc-btn-xs ez-toc-btn-default ez-toc-toggle\" aria-label=\"Toggle Table of Content\"><span class=\"ez-toc-js-icon-con\"><span class=\"\"><span class=\"eztoc-hide\" style=\"display:none;\">Toggle<\/span><span class=\"ez-toc-icon-toggle-span\"><svg style=\"fill: #999;color:#999\" xmlns=\"http:\/\/www.w3.org\/2000\/svg\" class=\"list-377408\" width=\"20px\" height=\"20px\" viewBox=\"0 0 24 24\" fill=\"none\"><path d=\"M6 6H4v2h2V6zm14 0H8v2h12V6zM4 11h2v2H4v-2zm16 0H8v2h12v-2zM4 16h2v2H4v-2zm16 0H8v2h12v-2z\" fill=\"currentColor\"><\/path><\/svg><svg style=\"fill: #999;color:#999\" class=\"arrow-unsorted-368013\" xmlns=\"http:\/\/www.w3.org\/2000\/svg\" width=\"10px\" height=\"10px\" viewBox=\"0 0 24 24\" version=\"1.2\" baseProfile=\"tiny\"><path d=\"M18.2 9.3l-6.2-6.3-6.2 6.3c-.2.2-.3.4-.3.7s.1.5.3.7c.2.2.4.3.7.3h11c.3 0 .5-.1.7-.3.2-.2.3-.5.3-.7s-.1-.5-.3-.7zM5.8 14.7l6.2 6.3 6.2-6.3c.2-.2.3-.5.3-.7s-.1-.5-.3-.7c-.2-.2-.4-.3-.7-.3h-11c-.3 0-.5.1-.7.3-.2.2-.3.5-.3.7s.1.5.3.7z\"\/><\/svg><\/span><\/span><\/span><\/a><\/span><\/div>\n<nav><ul class='ez-toc-list ez-toc-list-level-1 ' ><li class='ez-toc-page-1 ez-toc-heading-level-2'><a class=\"ez-toc-link ez-toc-heading-1\" href=\"https:\/\/www.mathweb.fr\/euclide\/2019\/10\/15\/saut-parabole-et-physique\/#Une_histoire_de_saut%E2%80%A6\" >Une histoire de saut&#8230;<\/a><\/li><li class='ez-toc-page-1 ez-toc-heading-level-2'><a class=\"ez-toc-link ez-toc-heading-2\" href=\"https:\/\/www.mathweb.fr\/euclide\/2019\/10\/15\/saut-parabole-et-physique\/#%E2%80%A6_et_de_physique\" >&#8230; et de physique<\/a><\/li><li class='ez-toc-page-1 ez-toc-heading-level-2'><a class=\"ez-toc-link ez-toc-heading-3\" href=\"https:\/\/www.mathweb.fr\/euclide\/2019\/10\/15\/saut-parabole-et-physique\/#Avec_Python\" >Avec Python<\/a><\/li><li class='ez-toc-page-1 ez-toc-heading-level-2'><a class=\"ez-toc-link ez-toc-heading-4\" href=\"https:\/\/www.mathweb.fr\/euclide\/2019\/10\/15\/saut-parabole-et-physique\/#Avec_Xcas\" >Avec Xcas<\/a><\/li><\/ul><\/nav><\/div>\n<h2 class=\"wp-block-heading\"><span class=\"ez-toc-section\" id=\"Une_histoire_de_saut%E2%80%A6\"><\/span>Une histoire de saut&#8230;<span class=\"ez-toc-section-end\"><\/span><\/h2>\n\n\n\n<p class=\"wp-block-paragraph\">Consid\u00e9rons la situation suivante : une structure est sch\u00e9matis\u00e9e ci-dessous. Une personne se met au point A et doit sauter afin d&rsquo;arriver au point B (les longueurs sont exprim\u00e9es en m\u00e8tres).<\/p>\n\n\n\n<div class=\"wp-block-image\"><figure class=\"aligncenter\"><a href=\"https:\/\/www.mathweb.fr\/euclide\/wp-content\/uploads\/2019\/10\/Saut.gif\" data-fancybox=\"gallery\"><img loading=\"lazy\" decoding=\"async\" width=\"472\" height=\"354\" src=\"https:\/\/www.mathweb.fr\/euclide\/wp-content\/uploads\/2019\/10\/Saut.gif\" alt=\"\" class=\"wp-image-1599\"\/><\/a><\/figure><\/div>\n\n\n\n<p class=\"wp-block-paragraph\">Comment d\u00e9terminer la distance OB ?<\/p>\n\n\n\n<h2 class=\"wp-block-heading\"><span class=\"ez-toc-section\" id=\"%E2%80%A6_et_de_physique\"><\/span>&#8230; et de physique<span class=\"ez-toc-section-end\"><\/span><\/h2>\n\n\n\n<p class=\"wp-block-paragraph\">D\u00e8s qu&rsquo;il s&rsquo;agit de lancer, que ce soit un ballon ou tata G\u00e9raldine, c&rsquo;est Newton qui nous aide \u00e0 l&rsquo;aide de ses lois qui (permettez-moi de sauter les d\u00e9tails) nous dit que dans notre situation, la parabole repr\u00e9sentant la trajectoire de la personne qui saute admet pour \u00e9quation (dans le rep\u00e8re d&rsquo;axes (OI) et (OJ) d&rsquo;unit\u00e9s 1 m\u00e8tre) : $$y = -\\frac{g}{2v_0^2\\cos^2(\\alpha)}(x-1)^2+\\tan(\\alpha)(x-1) + y_0$$ o\u00f9:<\/p>\n\n\n\n<ul class=\"wp-block-list\"><li>\\(g\\) est l&rsquo;acc\u00e9l\u00e9ration de la pesanteur. Sur Terre, on consid\u00e8re que \\( g\\approx 10\\text{ m}\\cdot\\text{s}^{-2}\\);<\/li><li>\\(v_0\\) est la vitesse initiale \u00e0 laquelle se lance la personne;<\/li><li>\\(\\alpha\\) est l&rsquo;angle que forme la tangente \u00e0 la parabole avec l&rsquo;horizontale;<\/li><li>\\(y_0\\) est la hauteur \u00e0 partir de laquelle la personne se lance.<\/li><\/ul>\n\n\n\n<p class=\"wp-block-paragraph\">On obtient alors, dans notre situation, apr\u00e8s avoir remplac\u00e9 les variables par leur valeur:$$y=\\frac{1}{49}\\left[-20x^2 + \\big( 40+49\\sqrt3 \\big)x + 225-49\\sqrt3 \\right].$$<\/p>\n\n\n\n<p class=\"wp-block-paragraph\">Pour  r\u00e9soudre notre probl\u00e8me, il faut trouver les coordonn\u00e9es de B, donc de l&rsquo;intersection de la parabole avec la droite (OB). Cette derni\u00e8re a pour \u00e9quation:$$y=\\frac{1}{\\sqrt3}x.$$ En effet, c&rsquo;est une droite qui passe par l&rsquo;origine du rep\u00e8re (donc elle repr\u00e9sente une fonction lin\u00e9aire d&rsquo;\u00e9quation \\(y=ax\\)) et le coefficient directeur est l&rsquo;accroissement des ordonn\u00e9es quand on avance d&rsquo;une unit\u00e9 en abscisse. L&rsquo;angle du \u00ab\u00a0tremplin\u00a0\u00bb \u00e9tant de 30 degr\u00e9s, le coefficient directeur est \u00e9gal \u00e0  \\(a=\\tan(30^\\circ)\\). Si on ne conna\u00eet pas la valeur de \\(\\tan(30^\\circ)\\), on peut se ramener aux sinus et cosinus :$$a=\\frac {\\sin30^\\circ}{\\cos30^\\circ}=\\frac{\\sqrt3\/2}{1\/2}=\\frac{1}{\\sqrt3}.$$ On obtient finalement que l&rsquo;\u00e9quation de (OB) est:$$ \\frac{1}{\\sqrt3} x.$$<\/p>\n\n\n\n<p class=\"wp-block-paragraph\">Trouver l&rsquo;abscisse de B revient \u00e0 r\u00e9soudre alors l&rsquo;\u00e9quation:$$ \\frac{1}{\\sqrt3}x =  \\frac{1}{49}\\left[-20x^2 + \\big( 40+49\\sqrt3 \\big)x + 225-49\\sqrt3 \\right].$$Et apr\u00e8s avoir multipli\u00e9 par 49 chaque membre de cette derni\u00e8re \u00e9quation, puis tout mis du m\u00eame c\u00f4t\u00e9 par rapport au signe \u00ab\u00a0=\u00a0\u00bb, on obtient l&rsquo;\u00e9quation:$$-20\\sqrt3x^2+\\big(40\\sqrt3+98\\big)x+225\\sqrt3-147=0.$$<\/p>\n\n\n\n<p class=\"wp-block-paragraph\">L\u00e0, on est en terrain connu (et non en  terre inconnue, enfin&#8230; je l&rsquo;esp\u00e8re pour vous !). On sort l&rsquo;artillerie lourde : discriminant et calcul de racines. Le discriminant est:$$\\begin{align}\\Delta&amp;=b^2-4ac\\\\&amp;=(40\\sqrt3+98)^2-4\\times(-20\\sqrt3)\\times(225\\sqrt3-147)\\\\&amp;=68404-3920\\sqrt3&gt;0\\end{align}$$ donc les deux solutions de l&rsquo;\u00e9quations sont:$$x_1=\\frac{-b-\\sqrt\\Delta}{2a}=\\frac{-(40\\sqrt3+98)-\\sqrt{ 68404-3920\\sqrt3 }}{-40\\sqrt3}\\approx5,997$$et$$ x_2=\\frac{-b+\\sqrt\\Delta}{2a}=\\frac{-(40\\sqrt3+98)+\\sqrt{ 68404-3920\\sqrt3 }}{-40\\sqrt3}\\approx -1,168.$$ \\(x_2&lt;0\\) donc cette valeur ne peut pas convenir \u00e0 notre probl\u00e8me. Ainsi,$$x_B\\approx5,997.$$ Il en r\u00e9sulte alors que:$$y_B=\\frac{1}{\\sqrt3}x_B\\approx3,462.$$<\/p>\n\n\n\n<p class=\"wp-block-paragraph\">Il ne reste plus qu&rsquo;\u00e0 calculer maintenant la longueur OB:$$OB\\approx\\sqrt{5,997^2+3,462^2}\\approx6,92.$$<\/p>\n\n\n\n<h2 class=\"wp-block-heading\"><span class=\"ez-toc-section\" id=\"Avec_Python\"><\/span>Avec Python<span class=\"ez-toc-section-end\"><\/span><\/h2>\n\n\n\n<pre class=\"EnlighterJSRAW\" data-enlighter-language=\"python\" data-enlighter-theme=\"dracula\" data-enlighter-highlight=\"\" data-enlighter-linenumbers=\"\" data-enlighter-lineoffset=\"\" data-enlighter-title=\"\" data-enlighter-group=\"\">def f(x):\n    return (-20*x**2+(40+49*3**0.5)*x+225-49*3**0.5)\/49\n\ndef g(x):\n    return x\/(3**0.5)\n\nx = 1\n\nwhile abs(f(x)-g(x)) > 0.001:\n    print(x,f(x),g(x))\n    x += 0.0001\n\nprint(x)\n<\/pre>\n\n\n\n<p class=\"wp-block-paragraph\">Ce programme commence par d\u00e9finir deux fonctions <em>f<\/em> et <em>g<\/em> (qui correspondent respectivement \u00e0 la fonction du second degr\u00e9 repr\u00e9sentant la trajectoire de la personne qui saute et au tremplin). Pour ne pas faire appel au module <em>math<\/em> de Python, je me suis permis ici de calculer la racine carr\u00e9e de 3 en \u00e9crivant \u00ab\u00a03**0.5\u00a0\u00bb (qui est \u00e9quivalent \u00e0 sqrt(3), soit \u00e0 \\(\\sqrt3\\)).<\/p>\n\n\n\n<p class=\"wp-block-paragraph\">Ensuite, on fixe la valeur de <em>x<\/em> \u00e0 1 (car la personne saute \u00e0 partir d&rsquo;un point dont l&rsquo;abscisse est 1, donc inutile de commencer \u00e0 0). Je fais ensuite une boucle conditionnelle (<em>while<\/em>) : tant que la diff\u00e9rence entre <em>f(x)<\/em> et <em>g(x<\/em>) est strictement sup\u00e9rieure \u00e0 0,001, j&rsquo;ajoute \u00e0 <em>x<\/em> 0,0001. Notons que quand on \u00e9crit \u00ab\u00a0abs(f(x)-g(x))\u00a0\u00bb, cela signifie que l&rsquo;on consid\u00e8re la diff\u00e9rence sans tenir compte du signe (valeur absolue de la diff\u00e9rence : sans signe \u00ab\u00a0moins\u00a0\u00bb).<\/p>\n\n\n\n<p class=\"wp-block-paragraph\">Si on fait tourner ce programme, on a le temps de se faire chauffer un petit chocolat chaud avec un petit marshmallow, de le boire tranquillement, de faire un peu de gymnastique (le sport, c&rsquo;est important !) pour enfin voir le r\u00e9sultat : x = 5.996999999999774 (en enlevant le \u00ab\u00a0print\u00a0\u00bb de la boucle, \u00e7a doit aller un peu plus vite, mais rester devant un \u00e9cran o\u00f9 rien ne se passe pendant plusieurs secondes, je trouve \u00e7a angoissant&#8230;). Remarquez qu&rsquo;en arrondissant au milli\u00e8me, on obtient bien \\(x \\approx 5,997\\).<\/p>\n\n\n\n<h2 class=\"wp-block-heading\"><span class=\"ez-toc-section\" id=\"Avec_Xcas\"><\/span>Avec Xcas<span class=\"ez-toc-section-end\"><\/span><\/h2>\n\n\n\n<p class=\"wp-block-paragraph\"><a rel=\"noreferrer noopener\" aria-label=\"Xcas (s\u2019ouvre dans un nouvel onglet)\" href=\"https:\/\/www-fourier.ujf-grenoble.fr\/~parisse\/install_fr\" target=\"_blank\">Xcas<\/a> est un logiciel de calcul formel qui permet bien des choses en math\u00e9matiques. Il est t\u00e9l\u00e9chargeable gratuitement, donc abusez-en ! Et pour r\u00e9soudre notre probl\u00e8me (l&rsquo;\u00e9quation <em>f(x) = g(x)<\/em>), on a:<\/p>\n\n\n\n<div class=\"wp-block-image\"><figure class=\"aligncenter\"><a href=\"https:\/\/www.mathweb.fr\/euclide\/wp-content\/uploads\/2019\/10\/xcas.png\" data-fancybox=\"gallery\"><img loading=\"lazy\" decoding=\"async\" width=\"902\" height=\"282\" src=\"https:\/\/www.mathweb.fr\/euclide\/wp-content\/uploads\/2019\/10\/xcas.png\" alt=\"\" class=\"wp-image-1608\" srcset=\"https:\/\/www.mathweb.fr\/euclide\/wp-content\/uploads\/2019\/10\/xcas.png 902w, https:\/\/www.mathweb.fr\/euclide\/wp-content\/uploads\/2019\/10\/xcas-300x94.png 300w, https:\/\/www.mathweb.fr\/euclide\/wp-content\/uploads\/2019\/10\/xcas-600x188.png 600w, https:\/\/www.mathweb.fr\/euclide\/wp-content\/uploads\/2019\/10\/xcas-768x240.png 768w\" sizes=\"auto, (max-width: 902px) 100vw, 902px\" \/><\/a><figcaption>Capture d&rsquo;\u00e9cran des calculs sous Xcas<\/figcaption><\/figure><\/div>\n\n\n\n<p class=\"wp-block-paragraph\">La fonction \u00ab\u00a0fsolve(f(x) = g(x) , x)\u00a0\u00bb nous permet d&rsquo;obtenir les valeurs approch\u00e9es des solutions. Je prends donc la valeur qui correspond \u00e0 ce que nous cherchions (x:=5.997). Ensuite, je d\u00e9finis <em>y<\/em> comme \u00e9tant le quotient de <em>x<\/em> par \\(\\sqrt3\\) et je calcule la longueur OB.<\/p>\n","protected":false},"excerpt":{"rendered":"<p>Cet article est principalement destin\u00e9 aux \u00e9l\u00e8ves de 1\u00e8re Math Sp\u00e9cialit\u00e9. Parlons dans cet article de math\u00e9matiques, et plus pr\u00e9cis\u00e9ment du second degr\u00e9. Alors, vous allez me dire : \u00ab\u00a0oui, mais bon ! C&rsquo;est super simple, il suffit de conna\u00eetre les formules et on sait tout faire.\u00a0\u00bb Ce n&rsquo;est pas [&hellip;]<\/p>\n","protected":false},"author":1,"featured_media":0,"comment_status":"open","ping_status":"open","sticky":false,"template":"","format":"standard","meta":{"footnotes":""},"categories":[21,6,5],"tags":[144,40,143],"class_list":["post-1598","post","type-post","status-publish","format-standard","hentry","category-enseignement","category-mathematiques","category-python","tag-premiere-math-specialite","tag-second-degre","tag-xcas"],"yoast_head":"<!-- This site is optimized with the Yoast SEO plugin v27.8 - https:\/\/yoast.com\/product\/yoast-seo-wordpress\/ -->\n<title>Saut, parabole et physique - Mathweb.fr<\/title>\n<meta name=\"robots\" content=\"index, follow, max-snippet:-1, max-image-preview:large, max-video-preview:-1\" \/>\n<link rel=\"canonical\" href=\"https:\/\/www.mathweb.fr\/euclide\/2019\/10\/15\/saut-parabole-et-physique\/\" \/>\n<meta property=\"og:locale\" content=\"fr_FR\" \/>\n<meta property=\"og:type\" content=\"article\" \/>\n<meta property=\"og:title\" content=\"Saut, parabole et physique - Mathweb.fr\" \/>\n<meta property=\"og:description\" content=\"Cet article est principalement destin\u00e9 aux \u00e9l\u00e8ves de 1\u00e8re Math Sp\u00e9cialit\u00e9. 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