{"id":27253,"date":"2021-09-25T13:00:36","date_gmt":"2021-09-25T13:00:36","guid":{"rendered":"http:\/\/toposuranos.com\/material\/?p=27253"},"modified":"2024-09-16T11:51:50","modified_gmt":"2024-09-16T11:51:50","slug":"comment-calculer-le-mouvement-dun-projectile","status":"publish","type":"post","link":"https:\/\/toposuranos.com\/material\/fr\/comment-calculer-le-mouvement-dun-projectile\/","title":{"rendered":"Mouvement d&#8217;un projectile"},"content":{"rendered":"<div style=\"background-color:#F3F3F3; padding:20px;\">\n<center><\/p>\n<h1>Mouvement d&#8217;un projectile<\/h1>\n<p><em><strong>R\u00e9sum\u00e9 :<\/strong><br \/>\nDans cette le\u00e7on, nous examinerons tous les aspects cin\u00e9matiques du mouvement d&#8217;un projectile, un sujet crucial en physique qui \u00e9tend notre \u00e9tude pr\u00e9c\u00e9dente sur le mouvement uniform\u00e9ment acc\u00e9l\u00e9r\u00e9. Nous aborderons comment, en supprimant la contrainte sur la direction du mouvement, nous rencontrons des trajectoires paraboliques typiques des projectiles. Nous \u00e9tudierons comment les vitesses initiales dans n&#8217;importe quelle direction, combin\u00e9es avec l&#8217;acc\u00e9l\u00e9ration due \u00e0 la gravit\u00e9, fa\u00e7onnent ces mouvements.<br \/>\n.<\/em><\/p>\n<p><\/center><\/p>\n<p style=\"text-align:center;\"><strong>OBJECTIFS D&#8217;APPRENTISSAGE :<\/strong><br \/>\n\u00c0 la fin de cette le\u00e7on, l&#8217;\u00e9tudiant sera capable de :<\/p>\n<ol>\n<li><strong>Se souvenir<\/strong> des \u00e9quations fondamentales du mouvement parabolique et des d\u00e9finitions li\u00e9es au lancement de projectiles (telles que la vitesse initiale, l&#8217;angle de lancement, l&#8217;acc\u00e9l\u00e9ration de la gravit\u00e9).<\/li>\n<li><strong>Interpr\u00e9ter<\/strong> graphiquement la trajectoire d&#8217;un projectile.<\/li>\n<li><strong>Expliquer<\/strong> comment les diff\u00e9rentes phases du mouvement (mont\u00e9e, point culminant, descente) sont li\u00e9es aux \u00e9quations cin\u00e9matiques.<\/li>\n<li><strong>R\u00e9soudre<\/strong> des probl\u00e8mes impliquant le calcul de la hauteur maximale, de la port\u00e9e horizontale et du temps total de vol d&#8217;un projectile en utilisant les \u00e9quations du mouvement parabolique.<\/li>\n<li><strong>D\u00e9composer<\/strong> les \u00e9quations de mouvement d&#8217;un projectile pour comprendre comment chaque composant (vitesse initiale, angle de lancement, acc\u00e9l\u00e9ration due \u00e0 la gravit\u00e9) affecte la trajectoire globale.<\/li>\n<\/ol>\n<p><center><\/p>\n<p><strong>TABLE DES MATI\u00c8RES<\/strong><br \/>\n<a href=\"#1\"><strong>Introduction<\/strong><\/a><br \/>\n<a href=\"#2\"><strong>D\u00e9veloppement du mouvement du projectile<\/strong><\/a><br \/>\n<a href=\"#3\">Comment d\u00e9terminer la hauteur maximale atteinte par un projectile ?<\/a><br \/>\n<a href=\"#4\">Comment d\u00e9terminer la port\u00e9e du mouvement des projectiles ?<\/a><br \/>\n<a href=\"#5\">Quel angle de lancement maximise la port\u00e9e du projectile ?<\/a><br \/>\n<a href=\"#6\"><strong>Exercices Propos\u00e9s<\/strong><\/a>\n<\/p>\n<p><iframe class=\"lazyload\" width=\"560\" height=\"315\" data-src=\"https:\/\/www.youtube.com\/embed\/_dXgQ_7u5GE\" title=\"YouTube video player\" frameborder=\"0\" allow=\"accelerometer; autoplay; clipboard-write; encrypted-media; gyroscope; picture-in-picture\" allowfullscreen=\"allowfullscreen\"><\/iframe><\/center><\/p>\n<\/div>\n<p><a name=\"1\"><\/a><\/p>\n<h2>Introduction<\/h2>\n<p style=\"text-align: justify; color: #000000;\">Dans les cours pr\u00e9c\u00e9dents, nous avons \u00e9tudi\u00e9 le <a href=\"https:\/\/www.youtube.com\/watch?v=P21bsNFF9Fw\" rel=\"noopener\" target=\"_blank\">mouvement rectiligne uniform\u00e9ment acc\u00e9l\u00e9r\u00e9<\/a> et avons vu ce qui se passe lorsqu&#8217;une acc\u00e9l\u00e9ration constante est appliqu\u00e9e dans la m\u00eame direction que le mouvement. Lorsque nous supprimons la contrainte sur la direction, nous obtenons un mouvement uniform\u00e9ment acc\u00e9l\u00e9r\u00e9, mais plus rectiligne. Dans ce sc\u00e9nario, le mouvement se d\u00e9veloppe le long du bras d&#8217;une parabole, et c&#8217;est ici que commence l&#8217;\u00e9tude du mouvement du projectile.<\/p>\n<p style=\"text-align: justify; color: #000000;\">Dans le mouvement d&#8217;un projectile, la vitesse initiale est donn\u00e9e dans n&#8217;importe quelle direction, tandis que l&#8217;acc\u00e9l\u00e9ration suit l&#8217;orientation typique de la gravit\u00e9. Lorsque le lancement du projectile est effectu\u00e9 directement vers le haut, nous obtenons un lancement vertical, qui est un cas de mouvement uniform\u00e9ment acc\u00e9l\u00e9r\u00e9.<\/p>\n<p><a name=\"2\"><\/a><\/p>\n<h2>D\u00e9veloppement du mouvement du projectile<\/h2>\n<p style=\"text-align: justify; color: #000000;\"><a href=\"https:\/\/www.youtube.com\/watch?v=_dXgQ_7u5GE&amp;t=172s\" target=\"_blank\" rel=\"noopener\"><strong><span style=\"color: #ff0000;\">Supposons que nous ayons un projectile<\/span><\/strong><\/a> lanc\u00e9 dans les airs depuis le sol par un canon avec une vitesse initiale <span dir=\"ltr\"><span class=\"katex-eq\" data-katex-display=\"false\">v_0<\/span><\/span> et un angle d&#8217;inclinaison <span dir=\"ltr\"><span class=\"katex-eq\" data-katex-display=\"false\">\\theta.<\/span><\/span> Le mouvement de ce projectile peut \u00eatre mod\u00e9lis\u00e9 sans probl\u00e8me en extrayant ses \u00e9quations de trajectoire \u00e0 partir des informations fournies. Ces \u00e9quations sont les suivantes :<\/p>\n<p style=\"text-align: center; color: #000000;\"><span dir=\"ltr\"><span class=\"katex-eq\" data-katex-display=\"false\">\n\\begin{array}{rl}\n\n\\vec{a}(t) &amp; = (0,-g) \\\\ \\\\\n\n\\vec{v}(t) &amp; =\\displaystyle \\int (0,-g) dt = (v_{0x}, -gt+v_{0y})\\\\ \\\\\n\n\\vec{r}(t) &amp; =\\displaystyle \\int (v_{0x}, -gt+v_{0y}) dt = \\left(v_{0x}t + x_0, -\\frac{1}{2}gt^2+v_{0y}t + y_0\\right)\n\n\\end{array}<\/span><\/span><\/p>\n<p><center><img decoding=\"async\" src=\"data:image\/gif;base64,R0lGODlhAQABAIAAAAAAAP\/\/\/yH5BAEAAAAALAAAAAABAAEAAAIBRAA7\" data-src=\"https:\/\/1.bp.blogspot.com\/-JQFKmGORl8A\/YVB9tKzayMI\/AAAAAAAAFks\/KPaXIUIcIVQ2yOAa_N9leIPgcZY6iv5zQCLcBGAsYHQ\/s0\/proyectil.PNG\" width=\"694\" height=\"390\" alt=\"Lancement de projectiles\" class=\"alignnone size-full lazyload\" \/><noscript><img decoding=\"async\" src=\"https:\/\/1.bp.blogspot.com\/-JQFKmGORl8A\/YVB9tKzayMI\/AAAAAAAAFks\/KPaXIUIcIVQ2yOAa_N9leIPgcZY6iv5zQCLcBGAsYHQ\/s0\/proyectil.PNG\" width=\"694\" height=\"390\" alt=\"Lancement de projectiles\" class=\"alignnone size-full lazyload\" \/><\/noscript><\/center><\/p>\n<p style=\"text-align: justify; color: #000000;\">O\u00f9 <span dir=\"ltr\"><span class=\"katex-eq\" data-katex-display=\"false\">\\vec{v}_{0} = (v_{0x},v_{0y})<\/span><\/span> est la vitesse initiale, <span dir=\"ltr\"><span class=\"katex-eq\" data-katex-display=\"false\">\\vec{r}_0=(x_0,y_0)<\/span><\/span> est la position initiale, et <span dir=\"ltr\"><span class=\"katex-eq\" data-katex-display=\"false\">g=9.81[m\/s^2]<\/span><\/span> est la magnitude de l&#8217;acc\u00e9l\u00e9ration de la gravit\u00e9. Maintenant, si nous observons le paragraphe pr\u00e9c\u00e9dent, nous remarquerons que la vitesse du projectile n&#8217;est pas indiqu\u00e9e directement, mais plut\u00f4t sa rapidit\u00e9 et son angle de tir. \u00c0 partir de ces informations et un peu de trigonom\u00e9trie, il est possible de d\u00e9terminer la vitesse initiale car :<\/p>\n<p style=\"text-align: center; color: #000000;\"><span dir=\"ltr\"><span class=\"katex-eq\" data-katex-display=\"false\">\n\\begin{array}{rl}\n\nv_{0x} &amp;= v_0 \\cos(\\theta) \\\\\n\nv_{0y} &amp;= v_0 \\sin(\\theta)\n\n\\end{array}<\/span><\/span><\/p>\n<p style=\"text-align: justify; color: #000000;\">O\u00f9 <span dir=\"ltr\"><span class=\"katex-eq\" data-katex-display=\"false\">v_0 = \\|\\vec{v}_0\\|<\/span><\/span> est la magnitude de la vitesse initiale. Si nous ajoutons \u00e9galement la position initiale <span dir=\"ltr\"><span class=\"katex-eq\" data-katex-display=\"false\">(x_0,y_0)=(0,0)<\/span><\/span>, les \u00e9quations de trajectoire sont exprim\u00e9es de la mani\u00e8re suivante :<\/p>\n<p style=\"text-align: center; color: #000000;\"><span dir=\"ltr\"><span class=\"katex-eq\" data-katex-display=\"false\">\n\\begin{array}{rl}\n\n\\vec{a}(t) &amp; = (0,-g) \\\\ \\\\\n\n\\vec{v}(t) &amp; =(v_{0}\\cos(\\theta), -gt+v_{0}\\sin(\\theta)\\\\ \\\\\n\n\\vec{r}(t) &amp; \\displaystyle =\\left(v_{0}\\cos(\\theta)t , -\\frac{1}{2}gt^2+v_{0}\\sin(\\theta)t \\right)\n\n\\end{array}<\/span><\/span><\/p>\n<p style=\"text-align: justify; color: #000000;\">Avec cela, nous pouvons r\u00e9pondre \u00e0 certaines questions relatives au lancement des projectiles : Jusqu&#8217;o\u00f9 ira-t-il ? Quelle hauteur atteindra-t-il ? Combien de temps prendra-t-il pour tomber ? etc.<\/p>\n<p><a name=\"3\"><\/a><\/p>\n<h3>Comment d\u00e9terminer la hauteur maximale atteinte par un projectile ?<\/h3>\n<p style=\"text-align: justify; color: #000000;\"><a href=\"https:\/\/www.youtube.com\/watch?v=_dXgQ_7u5GE&amp;t=398s\" target=\"_blank\" rel=\"noopener\"><strong><span style=\"color: #ff0000;\">Pour r\u00e9pondre \u00e0 cette question<\/span><\/strong><\/a> nous devons nous demander : Que se passe-t-il lorsque le projectile atteint sa hauteur maximale ? Ce qui se passe, c&#8217;est que la composante verticale de sa vitesse devient nulle, et donc :<\/p>\n<p style=\"text-align: center; color: #000000;\"><span dir=\"ltr\"><span class=\"katex-eq\" data-katex-display=\"false\"> -gt+v_{0}\\sin(\\theta) = 0 <\/span><\/span><\/p>\n<p style=\"text-align: justify; color: #000000;\">Cela \u00e9quivaut \u00e0 dire :<\/p>\n<p style=\"text-align: center; color: #000000;\"><span dir=\"ltr\"><span class=\"katex-eq\" data-katex-display=\"false\"> t = \\displaystyle \\frac{v_{0}\\sin(\\theta)}{g} <\/span><\/span><\/p>\n<p style=\"text-align: justify; color: #000000;\">Autrement dit, le projectile atteint la hauteur maximale apr\u00e8s un temps <span dir=\"ltr\"><span class=\"katex-eq\" data-katex-display=\"false\">t=v_0\\sin(\\theta)\/g<\/span><\/span> depuis le lancement. Nous appelons cela \u00abtemps de hauteur maximale\u00bb et \u00e9crivons :<\/p>\n<p style=\"text-align: center; color: #000000;\"><span dir=\"ltr\"><span class=\"katex-eq\" data-katex-display=\"false\">\\color{blue}{t_{alt.max} = \\frac{v_{0}\\sin(\\theta)}{g}}<\/span><\/span><\/p>\n<p style=\"text-align: justify; color: #000000;\">Ensuite, la hauteur maximale que peut atteindre le projectile peut \u00eatre obtenue en rempla\u00e7ant <span dir=\"ltr\"><span class=\"katex-eq\" data-katex-display=\"false\">t=t_{alt.max}<\/span><\/span> dans la composante verticale de la position du projectile, obtenant :<\/p>\n<p style=\"text-align: center; color: #000000;\"><span dir=\"ltr\"><span class=\"katex-eq\" data-katex-display=\"false\">\\begin{array}{rl}\n\ny_{alt.max} &amp; = \\displaystyle -\\frac{1}{2}gt_{alt.max}^2+v_{0}\\sin(\\theta)t_{alt.max}\\\\ \\\\\n\n&amp; =\\displaystyle-\\frac{1}{2}g \\left(\\frac{v_{0}\\sin(\\theta)}{g} \\right)^2 + v_{0}\\sin(\\theta) \\frac{v_{0}\\sin(\\theta)}{g} \\\\ \\\\\n\n&amp; =\\displaystyle-\\frac{1}{2} \\frac{v_{0}^2\\sin^2(\\theta)}{g} + \\frac{v_{0}^2\\sin^2(\\theta)}{g} \\\\ \\\\\n\n&amp; =\\displaystyle \\frac{v_{0}^2\\sin^2(\\theta)}{2g}\n\n\\end{array}<\/span><\/span><\/p>\n<p><a name=\"4\"><\/a><\/p>\n<h3>Comment d\u00e9terminer la port\u00e9e du mouvement des projectiles ?<\/h3>\n<p style=\"text-align: justify; color: #000000;\"><a href=\"https:\/\/www.youtube.com\/watch?v=_dXgQ_7u5GE&amp;t=653s\" target=\"_blank\" rel=\"noopener\"><strong><span style=\"color: #ff0000;\">Si vous voulez savoir<\/span><\/strong><\/a> la distance que le projectile parcourt jusqu&#8217;au moment o\u00f9 il touche le sol, tout ce que vous avez \u00e0 faire est de poser la question aux \u00e9quations de trajectoire associ\u00e9es au lancement des projectiles. Mais comment faire cela ? Simple : Que se passe-t-il lorsque le projectile touche le sol ? Ce qui se passe, c&#8217;est que la coordonn\u00e9e de position associ\u00e9e \u00e0 la hauteur devient nulle, c&#8217;est-\u00e0-dire :<\/p>\n<p style=\"text-align: center; color: #000000;\"><span dir=\"ltr\"><span class=\"katex-eq\" data-katex-display=\"false\">\\displaystyle -\\frac{1}{2}gt^2+v_{0}\\sin(\\theta)t = 0\n\n<\/span><\/span><\/p>\n<p style=\"text-align: justify; color: #000000;\">Ici, nous pouvons r\u00e9soudre le temps o\u00f9 le projectile touche le sol, ce qui se produit \u00e0 deux reprises : au moment du lancement et lorsqu&#8217;il tombe, car les solutions possibles de cette \u00e9quation sont :<\/p>\n<p style=\"text-align: center; color: #000000;\"><span dir=\"ltr\"><span class=\"katex-eq\" data-katex-display=\"false\">\n\\begin{array}{rl}\n\nt &amp; = 0\\\\ \\\\\n\nt &amp; = \\displaystyle \\frac{2v_0 \\sin(\\theta)}{g}\n\n\\end{array}\n\n<\/span><\/span><\/p>\n<p style=\"text-align: justify; color: #000000;\">Nous appelons le r\u00e9sultat non nul \u00abtemps de chute\u00bb et \u00e9crivons :<\/p>\n<p style=\"text-align: center; color: #000000;\"><span dir=\"ltr\"><span class=\"katex-eq\" data-katex-display=\"false\">\\color{blue}{t_{caida} = \\displaystyle \\frac{2v_0 \\sin(\\theta)}{g}}<\/span><\/span><\/p>\n<p style=\"text-align: justify; color: #000000;\">Si vous regardez plus haut, vous remarquerez que <span dir=\"ltr\"><span class=\"katex-eq\" data-katex-display=\"false\">t_{caida} = 2t_{alt.max}<\/span><\/span> car le temps que met le projectile pour atteindre sa hauteur maximale est le m\u00eame que celui qu&#8217;il met pour tomber de son point le plus \u00e9lev\u00e9. Cela indique une certaine sym\u00e9trie dans le mouvement du projectile. En fait, cette sym\u00e9trie se manifeste lorsque vous remarquez que la coordonn\u00e9e associ\u00e9e \u00e0 la hauteur a la forme d&#8217;une parabole.<\/p>\n<p style=\"text-align: justify; color: #000000;\">En connaissant le temps de chute, il est maintenant possible de calculer la distance parcourue par le projectile au moment o\u00f9 il touche le sol simplement en le substituant dans la premi\u00e8re coordonn\u00e9e de position :<\/p>\n<p style=\"text-align: center; color: #000000;\"><span dir=\"ltr\"><span class=\"katex-eq\" data-katex-display=\"false\">\\begin{array}{rl}\n\nx_{caida} &amp;= v_{0}\\cos(\\theta)t_{caida} \\\\ \\\\\n\n&amp; = \\displaystyle v_{0}\\cos(\\theta)\\frac{2v_0 \\sin(\\theta)}{g} \\\\ \\\\\n\n&amp; = \\displaystyle \\frac{v_0^2 \\sin(2\\theta)}{g} \\\\ \\\\\n\n\\end{array}<\/span><\/span><\/p>\n<p><a name=\"5\"><\/a><\/p>\n<h3>Quel angle de lancement maximise la port\u00e9e du projectile ?<\/h3>\n<p style=\"text-align: justify; color: #000000;\"><a href=\"https:\/\/www.youtube.com\/watch?v=_dXgQ_7u5GE&amp;t=999s\" target=\"_blank\" rel=\"noopener\"><strong><span style=\"color: #ff0000;\">Si vous voulez savoir<\/span><\/strong><\/a> quel angle de lancement maximise la port\u00e9e du projectile, ou si vous voulez prouver que ce que vous savez est effectivement correct, tout ce que vous avez \u00e0 faire est de prendre parmi les expressions que nous avons d\u00e9montr\u00e9es celle qui vous permet de formuler la question de mani\u00e8re math\u00e9matique. Nous avons d\u00e9j\u00e0 calcul\u00e9 la distance de chute dans la section pr\u00e9c\u00e9dente, et il s&#8217;av\u00e8re que c&#8217;est une fonction de l&#8217;angle de lancement :<\/p>\n<p style=\"text-align: center; color: #000000;\"><span dir=\"ltr\"><span class=\"katex-eq\" data-katex-display=\"false\">\n\\displaystyle x_{caida} = x_{caida}(\\theta) = \\frac{v_0^2 \\sin(2\\theta)}{g}\n\n<\/span><\/span><\/p>\n<p style=\"text-align: justify; color: #000000;\">La fonction sinus a deux valeurs extr\u00eames possibles : +1 et -1, mais nous nous int\u00e9ressons \u00e0 la premi\u00e8re. Pour que <span dir=\"ltr\"><span class=\"katex-eq\" data-katex-display=\"false\">\\sin(2\\theta)=+1<\/span><\/span>, il est n\u00e9cessaire que <span dir=\"ltr\"><span class=\"katex-eq\" data-katex-display=\"false\">2\\theta = 90^o<\/span><\/span> (+<span dir=\"ltr\"><span class=\"katex-eq\" data-katex-display=\"false\">2k\\pi,<\/span><\/span> mais nous omettrons cette partie car nous n&#8217;en avons pas besoin) et donc <span dir=\"ltr\"><span class=\"katex-eq\" data-katex-display=\"false\">\\theta=45^o<\/span><\/span> est l&#8217;angle de lancement qui maximise la port\u00e9e. Ce probl\u00e8me peut \u00e9galement \u00eatre r\u00e9solu si nous le formulons comme un probl\u00e8me d&#8217;optimisation (en utilisant les <a href=\"https:\/\/www.youtube.com\/watch?v=OscTlX3raaE&#038;list=PL_C8rbeFjqAVaR_sgLJRBvMm5t6E1GxGI\" target=\"_blank\" rel=\"noopener\">outils de ce cours de calcul<\/a>), mais j&#8217;ai opt\u00e9 pour cette voie plus rapide et tout aussi illustrative.<\/p>\n<p><a name=\"6\"><\/a><\/p>\n<h2>Exercices Propos\u00e9s<\/h2>\n<ol style=\"text-align: justify; color: #000000;\">\n<li>Un projectile est lanc\u00e9 depuis le sol, avec un angle d&#8217;\u00e9l\u00e9vation de <span dir=\"ltr\"><span class=\"katex-eq\" data-katex-display=\"false\">\\theta=30^o<\/span><\/span> et une vitesse initiale de <span dir=\"ltr\"><span class=\"katex-eq\" data-katex-display=\"false\">v_0=70[km\/h].<\/span><\/span> a) Quelle est la hauteur maximale atteinte par le projectile ? b) Quelle distance parcourt le projectile jusqu&#8217;au moment o\u00f9 il touche le sol ? c) Combien de temps le projectile met-il pour tomber ?<\/li>\n<li>Un canon plac\u00e9 au sol tire une balle avec une vitesse de <span dir=\"ltr\"><span class=\"katex-eq\" data-katex-display=\"false\">90[km\/h]\uff1f<\/span><\/span> Sous quel angle d&#8217;\u00e9l\u00e9vation le canon doit-il \u00eatre ajust\u00e9 pour que la balle tombe \u00e0 une distance horizontale de <span dir=\"ltr\"><span class=\"katex-eq\" data-katex-display=\"false\">20[m]\uff1f<\/span><\/span><\/li>\n<li>Le m\u00eame canon de l&#8217;exercice pr\u00e9c\u00e9dent est maintenant plac\u00e9 \u00e0 une hauteur de <span dir=\"ltr\"><span class=\"katex-eq\" data-katex-display=\"false\">5[m]\uff1f<\/span><\/span> Sous quel angle d&#8217;\u00e9l\u00e9vation doit-il \u00eatre ajust\u00e9 pour que la balle tombe toujours \u00e0 une distance horizontale de <span dir=\"ltr\"><span class=\"katex-eq\" data-katex-display=\"false\">20[m]\uff1f<\/span><\/span><\/li>\n<li>Un bombardier vole \u00e0 une hauteur de <span dir=\"ltr\"><span class=\"katex-eq\" data-katex-display=\"false\">3,000[m]\uff1f<\/span><\/span> au-dessus du sol avec une vitesse de <span dir=\"ltr\"><span class=\"katex-eq\" data-katex-display=\"false\">1,500[km\/h].<\/span><\/span> S&#8217;il laisse tomber un projectile par son propre poids, quelle distance parcourra le projectile depuis le moment o\u00f9 il est l\u00e2ch\u00e9 jusqu&#8217;\u00e0 ce qu&#8217;il touche le sol ?<\/li>\n<\/ol>\n","protected":false},"excerpt":{"rendered":"<p>Mouvement d&#8217;un projectile R\u00e9sum\u00e9 : Dans cette le\u00e7on, nous examinerons tous les aspects cin\u00e9matiques du mouvement d&#8217;un projectile, un sujet crucial en physique qui \u00e9tend notre \u00e9tude pr\u00e9c\u00e9dente sur le mouvement uniform\u00e9ment acc\u00e9l\u00e9r\u00e9. Nous aborderons comment, en supprimant la contrainte sur la direction du mouvement, nous rencontrons des trajectoires paraboliques typiques des projectiles. Nous \u00e9tudierons [&hellip;]<\/p>\n","protected":false},"author":1,"featured_media":26193,"comment_status":"open","ping_status":"closed","sticky":false,"template":"","format":"standard","meta":{"iawp_total_views":109,"footnotes":""},"categories":[663,647],"tags":[],"class_list":["post-27253","post","type-post","status-publish","format-standard","has-post-thumbnail","hentry","category-fondements-de-la-mecanique","category-physique"],"yoast_head":"<!-- This site is optimized with the Yoast SEO plugin v26.7 - https:\/\/yoast.com\/wordpress\/plugins\/seo\/ -->\n<title>Mouvement d&#039;un projectile - toposuranos.com\/material<\/title>\n<meta name=\"description\" content=\"Apprenez tout sur les trajectoires paraboliques et leurs \u00e9quations, et comment la gravit\u00e9 influence le lancement des projectiles.\" \/>\n<meta name=\"robots\" content=\"index, follow, max-snippet:-1, max-image-preview:large, max-video-preview:-1\" \/>\n<link rel=\"canonical\" href=\"https:\/\/toposuranos.com\/material\/fr\/comment-calculer-le-mouvement-dun-projectile\/\" \/>\n<meta property=\"og:locale\" content=\"es_ES\" \/>\n<meta property=\"og:type\" content=\"article\" \/>\n<meta property=\"og:title\" content=\"Comment calculer le mouvement d&#039;un projectile\" \/>\n<meta property=\"og:description\" content=\"Apprenez tout sur les trajectoires paraboliques et leurs \u00e9quations, et comment la gravit\u00e9 influence le lancement des projectiles.\" \/>\n<meta property=\"og:url\" content=\"https:\/\/toposuranos.com\/material\/fr\/comment-calculer-le-mouvement-dun-projectile\/\" \/>\n<meta property=\"og:site_name\" content=\"toposuranos.com\/material\" \/>\n<meta property=\"article:publisher\" content=\"https:\/\/www.facebook.com\/groups\/toposuranos\" \/>\n<meta property=\"article:published_time\" content=\"2021-09-25T13:00:36+00:00\" \/>\n<meta property=\"article:modified_time\" content=\"2024-09-16T11:51:50+00:00\" \/>\n<meta property=\"og:image\" content=\"http:\/\/toposuranos.com\/material\/wp-content\/uploads\/2021\/09\/proyectil-1024x585.jpg\" \/>\n<meta name=\"author\" content=\"giorgio.reveco\" \/>\n<meta name=\"twitter:card\" content=\"summary_large_image\" \/>\n<meta name=\"twitter:title\" content=\"Comment calculer le mouvement d&#039;un projectile\" \/>\n<meta name=\"twitter:description\" content=\"Apprenez tout sur les trajectoires paraboliques et leurs \u00e9quations, et comment la gravit\u00e9 influence le lancement des projectiles.\" \/>\n<meta name=\"twitter:image\" content=\"https:\/\/toposuranos.com\/material\/wp-content\/uploads\/2021\/09\/proyectil.jpg\" \/>\n<meta name=\"twitter:creator\" content=\"@topuranos\" \/>\n<meta name=\"twitter:site\" content=\"@topuranos\" \/>\n<meta name=\"twitter:label1\" content=\"Escrito por\" \/>\n\t<meta name=\"twitter:data1\" content=\"giorgio.reveco\" \/>\n\t<meta name=\"twitter:label2\" content=\"Tiempo de lectura\" \/>\n\t<meta name=\"twitter:data2\" content=\"7 minutos\" \/>\n<script type=\"application\/ld+json\" class=\"yoast-schema-graph\">{\"@context\":\"https:\/\/schema.org\",\"@graph\":[{\"@type\":\"Article\",\"@id\":\"https:\/\/toposuranos.com\/material\/fr\/comment-calculer-le-mouvement-dun-projectile\/#article\",\"isPartOf\":{\"@id\":\"https:\/\/toposuranos.com\/material\/fr\/comment-calculer-le-mouvement-dun-projectile\/\"},\"author\":{\"name\":\"giorgio.reveco\",\"@id\":\"https:\/\/toposuranos.com\/material\/#\/schema\/person\/e15164361c3f9a2a02cf6c234cf7fdc1\"},\"headline\":\"Mouvement d&#8217;un projectile\",\"datePublished\":\"2021-09-25T13:00:36+00:00\",\"dateModified\":\"2024-09-16T11:51:50+00:00\",\"mainEntityOfPage\":{\"@id\":\"https:\/\/toposuranos.com\/material\/fr\/comment-calculer-le-mouvement-dun-projectile\/\"},\"wordCount\":1663,\"commentCount\":0,\"publisher\":{\"@id\":\"https:\/\/toposuranos.com\/material\/#organization\"},\"image\":{\"@id\":\"https:\/\/toposuranos.com\/material\/fr\/comment-calculer-le-mouvement-dun-projectile\/#primaryimage\"},\"thumbnailUrl\":\"https:\/\/toposuranos.com\/material\/wp-content\/uploads\/2021\/09\/proyectil.jpg\",\"articleSection\":[\"Fondements de la M\u00e9canique\",\"Physique\"],\"inLanguage\":\"es\",\"potentialAction\":[{\"@type\":\"CommentAction\",\"name\":\"Comment\",\"target\":[\"https:\/\/toposuranos.com\/material\/fr\/comment-calculer-le-mouvement-dun-projectile\/#respond\"]}]},{\"@type\":\"WebPage\",\"@id\":\"https:\/\/toposuranos.com\/material\/fr\/comment-calculer-le-mouvement-dun-projectile\/\",\"url\":\"https:\/\/toposuranos.com\/material\/fr\/comment-calculer-le-mouvement-dun-projectile\/\",\"name\":\"Mouvement d'un projectile - 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