{"id":34461,"date":"2021-09-25T00:00:00","date_gmt":"2021-09-25T00:00:00","guid":{"rendered":"https:\/\/toposuranos.com\/material\/?p=34461"},"modified":"2025-09-08T03:36:29","modified_gmt":"2025-09-08T03:36:29","slug":"missile-proiectilis","status":"publish","type":"post","link":"https:\/\/toposuranos.com\/material\/la\/missile-proiectilis\/","title":{"rendered":"Missile Proiectilis"},"content":{"rendered":"<div style=\"background-color:#F3F3F3; padding:20px;\">\n<center><\/p>\n<h1>Missile Proiectilis<\/h1>\n<p><em><strong>Summarium:<\/strong><br \/>\nIn hac lectione omnes aspectus cinematicos iactus proiectilis recognoscemus, thema grave in physica quod studium nostrum prius de motu uniformiter accelerato extendit. Exponemus quomodo, sublata restrictione de directione motus, occurramus trajectoriis parabolicis propriis proiectilium. Investigabimus quomodo velocitates initiales in qualibet directione, cum acceleratione gravitatis coniunctae, hos motus effingant.<br \/>\n.<\/em><\/p>\n<p><\/center><\/p>\n<p style=\"text-align:center;\"><strong>PROPOSITA DISCENDI:<\/strong><br \/>\nAd finem huius lectionis discipulus poterit:<\/p>\n<ol>\n<li><strong>Meminisse<\/strong> aequationes fundamentales motus parabolici et definitiones ad iactum proiectilium pertinentes (ut velocitas initialis, angulus iactus, acceleratio gravitatis).<\/li>\n<li><strong>Interpretari<\/strong> graphice trajectoriam proiectilis<\/li>\n<li><strong>Explicare<\/strong> quomodo diversae phases motus (ascensus, punctum maximum, descensus) cum aequationibus cinematicis referantur.<\/li>\n<li><strong>Solvere<\/strong> problemata quae computationem altitudinis maximae, amplitudinis horizontis et temporis totius volatus proiectilis per aequationes motus parabolici requirunt.<\/li>\n<li><strong>Decomponere<\/strong> aequationes motus proiectilis ut intellegatur quomodo singulae partes (velocitas initialis, angulus iactus, acceleratio ex gravitate) trajectoriam totam afficiant.<\/li>\n<\/ol>\n<p><center><\/p>\n<p><strong>INDEX CONTENTORUM<\/strong><br \/>\n<a href=\"#1\"><strong>Introductio<\/strong><\/a><br \/>\n<a href=\"#2\"><strong>Evolutio iactus proiectilis<\/strong><\/a><br \/>\n<a href=\"#3\">Quomodo determinare altitudinem maximam a proiectili adeptam?<\/a><br \/>\n<a href=\"#4\">Quomodo determinare amplitudinem iactus proiectilium?<\/a><br \/>\n<a href=\"#5\">Quis angulus iactus amplitudinem proiectilis maximam efficit?<\/a><br \/>\n<a href=\"#6\"><strong>Exercitia Proposita<\/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><\/div>\n<p><a name=\"1\"><\/a><\/p>\n<h2>Introductio<\/h2>\n<p style=\"text-align: justify;\">In praelectionibus prioribus investigavimus <a href=\"https:\/\/www.youtube.com\/watch?v=P21bsNFF9Fw\" rel=\"noopener\" target=\"_blank\">motum rectilineum uniformiter acceleratum<\/a> et vidimus quid eveniret cum acceleratio constans in eadem directione motus applicaretur. Cum restrictionem de directione tollimus, obtinemus motum uniformiter acceleratum, sed iam non rectilineum. In hoc situ, motus in brachio parabolae explicatur, atque hic nascitur studium iactus proiectilis.<\/p>\n<p style=\"text-align: justify;\">In iactu proiectilis, velocitas initialis in quacumque directione datur, dum acceleratio directionem gravitatis propriam sequitur. Cum iactus proiectilis directe sursum fiat, obtinetur iactus verticalis, qui est casus MRUA.<\/p>\n<p><a name=\"2\"><\/a><\/p>\n<h2>Evolutio iactus proiectilis<\/h2>\n<p style=\"text-align: justify;\"><a href=\"https:\/\/www.youtube.com\/watch?v=_dXgQ_7u5GE&amp;t=172s\" target=\"_blank\" rel=\"noopener\"><strong><span style=\"color: #ff0000;\">Ponamus nos habere proiectilem<\/span><\/strong><\/a> in a\u00ebra e solo tormento cum velocitate initiali <span class=\"katex-eq\" data-katex-display=\"false\">v_0<\/span> et angulo inclinationis <span class=\"katex-eq\" data-katex-display=\"false\">\\theta.<\/span> Motus huius proiectilis sine difficultate fingi potest extrahendo aequationes itineris ex informatione modo tradita. Hae sic manent:<\/p>\n<p style=\"text-align: center;\"><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>\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=\"Iactus proiectilium\" 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=\"Iactus proiectilium\" class=\"alignnone size-full lazyload\" \/><\/noscript><\/center><\/p>\n<p style=\"text-align: justify;\">Ubi <span class=\"katex-eq\" data-katex-display=\"false\">\\vec{v}_{0} = (v_{0x},v_{0y})<\/span> est velocitas initialis, <span class=\"katex-eq\" data-katex-display=\"false\">\\vec{r}_0=(x_0,y_0)<\/span> est positio initialis et <span class=\"katex-eq\" data-katex-display=\"false\">g=9,81[m\/s^2]<\/span> est magnitudo accelerationis gravitatis. Nunc, si paragraphum priorem observamus, animadvertimus non directe nobis dari velocitatem proiectilis, sed eius celeritatem atque angulum iactus. Ex hac informatione et paulo trigonometriae determinari potest velocitas initialis quia:<\/p>\n<p style=\"text-align: center;\"><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>\n<p style=\"text-align: justify;\">Ubi <span class=\"katex-eq\" data-katex-display=\"false\">v_0 = \\|\\vec{v}_0\\|<\/span> est magnitudo velocitatis initialis. Si insuper ponimus positionem initialem <span class=\"katex-eq\" data-katex-display=\"false\">(x_0,y_0)=(0,0)<\/span>, aequationes itineris sic exprimuntur:<\/p>\n<p style=\"text-align: center;\"><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>\n<p style=\"text-align: justify;\">His iam ad manum possumus respondere ad quaestiones de iactu proiectilium: Quam longe perveniet? Quam altitudinem attinget? Quantum temporis cadendo consumet? etc.<\/p>\n<p><a name=\"3\"><\/a><\/p>\n<h3>Quomodo determinare altitudinem maximam a proiectili adeptam?<\/h3>\n<p style=\"text-align: justify;\"><a href=\"https:\/\/www.youtube.com\/watch?v=_dXgQ_7u5GE&amp;t=398s\" target=\"_blank\" rel=\"noopener\"><strong><span style=\"color: #ff0000;\">Ad hanc quaestionem respondere<\/span><\/strong><\/a> debemus nosmet ipsos interrogare: Quid fit cum proiectile altitudinem maximam attingit? Fit ut componentis verticalis velocitatis eius nulla fiat et proinde:<\/p>\n<p style=\"text-align: center;\"><span class=\"katex-eq\" data-katex-display=\"false\"> -gt+v_{0}\\sin(\\theta) = 0 <\/span>\n<p style=\"text-align: justify;\">Hoc idem est ac dicere:<\/p>\n<p style=\"text-align: center;\"><span class=\"katex-eq\" data-katex-display=\"false\"> t = \\displaystyle \\frac{v_{0}\\sin(\\theta)}{g} <\/span>\n<p style=\"text-align: justify;\">Id est, proiectile altitudinem maximam attingit post tempus <span class=\"katex-eq\" data-katex-display=\"false\">t=v_0\\sin(\\theta)\/g<\/span> a momento iactus. Hoc \u00abtempus altitudinis maximae\u00bb appellamus et scribimus:<\/p>\n<p style=\"text-align: center;\"><span class=\"katex-eq\" data-katex-display=\"false\">\\color{blue}{t_{alt.max} = \\frac{v_{0}\\sin(\\theta)}{g}}<\/span>\n<p style=\"text-align: justify;\">Deinde, altitudo maxima quam proiectile consequi potest obtineri potest substituendo <span class=\"katex-eq\" data-katex-display=\"false\">t=t_{alt.max}<\/span> in componente verticali positionis proiectilis, unde provenit:<\/p>\n<p style=\"text-align: center;\"><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>\n<p><a name=\"4\"><\/a><\/p>\n<h3>Quomodo determinare amplitudinem iactus proiectilium?<\/h3>\n<p style=\"text-align: justify;\"><a href=\"https:\/\/www.youtube.com\/watch?v=_dXgQ_7u5GE&amp;t=653s\" target=\"_blank\" rel=\"noopener\"><strong><span style=\"color: #ff0000;\">Si cognoscere vis distantiam<\/span><\/strong><\/a> quam proiectile percurrit usque ad momentum quo solum tangit, id solum est quod aequationibus itineris ad iactum proiectilium associatis quaerendum est. Sed quomodo id facimus? Simplex est: Quid accidit cum proiectile solum tangit? Accidit ut coordinata positionis altitudini associata fiat nulla, id est:<\/p>\n<p style=\"text-align: center;\"><span class=\"katex-eq\" data-katex-display=\"false\">\\displaystyle -\\frac{1}{2}gt^2+v_{0}\\sin(\\theta)t = 0\n\n<\/span>\n<p style=\"text-align: justify;\">Hic possumus eruere tempus quo proiectile solum tangit, quod fit bis: in momento quo iacitur et cum cadit, quia solutiones huius aequationis possibilis sunt:<\/p>\n<p style=\"text-align: center;\"><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>\n<p style=\"text-align: justify;\">Eventum non nullum \u00abtempus casus\u00bb appellamus et scribimus:<\/p>\n<p style=\"text-align: center;\"><span class=\"katex-eq\" data-katex-display=\"false\">\\color{blue}{t_{caida} = \\displaystyle \\frac{2v_0 \\sin(\\theta)}{g}}<\/span>\n<p style=\"text-align: justify;\">Si supra inspicis animadvertes <span class=\"katex-eq\" data-katex-display=\"false\">t_{caida} = 2t_{alt.max}<\/span> quia tempus quod proiectile consumit ad altitudinem maximam attingendam idem est atque quod consumit ad cadendum ex puncto summo. Hoc est indicium quaedam symmetriae in motu proiectilis. Re vera haec symmetria iam apparet cum animadvertis coordinatam altitudini associatam formam parabolae habere.<\/p>\n<p style=\"text-align: justify;\">Cognito tempore casus, iam possibile est computare distantiam quam proiectile percurrit in momento quo solum tangit simpliciter substituendo eam in prima coordinata positionis:<\/p>\n<p style=\"text-align: center;\"><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>\n<p><a name=\"5\"><\/a><\/p>\n<h3>Quis angulus iactus amplitudinem proiectilis maximam efficit?<\/h3>\n<p style=\"text-align: justify;\"><a href=\"https:\/\/www.youtube.com\/watch?v=_dXgQ_7u5GE&amp;t=999s\" target=\"_blank\" rel=\"noopener\"><strong><span style=\"color: #ff0000;\">Si scire vis quis sit angulus<\/span><\/strong><\/a> iactus quo iactus proiectilis amplitudinem maximam habet, aut demonstrare vis quod id quod scis revera verum sit, solum est ut inter expressiones quas demonstravimus eligas eam quae sinat quaestionem modo mathematico formare. Iam computavimus distantiam casus in sectione superiore, et haec fit functio anguli iactus:<\/p>\n<p style=\"text-align: center;\"><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>\n<p style=\"text-align: justify;\">Functio sinus duos eventus extremos habet: +1 et -1, sed nobis primus interest. Ut <span class=\"katex-eq\" data-katex-display=\"false\">\\sin(2\\theta)=+1<\/span>, necesse est <span class=\"katex-eq\" data-katex-display=\"false\">2\\theta = 90^o<\/span> (+<span class=\"katex-eq\" data-katex-display=\"false\">2k\\pi,<\/span> sed illam partem omittimus quia non opus est) et proinde <span class=\"katex-eq\" data-katex-display=\"false\">\\theta=45^o<\/span> est angulus iactus qui amplitudinem maximam efficit. Hoc problema etiam solvi potest si proponitur ut problema optimizationis (utendo instrumentis <a href=\"https:\/\/www.youtube.com\/watch?v=OscTlX3raaE&#038;list=PL_C8rbeFjqAVaR_sgLJRBvMm5t6E1GxGI\" target=\"_blank\" rel=\"noopener\">huius generis calculi<\/a>) sed hoc iter elegi quod celerius est atque aeque illustrativum.<\/p>\n<p><a name=\"6\"><\/a><\/p>\n<h2>Exercitia Proposita<\/h2>\n<ol style=\"text-align: justify;\">\n<li>Proiectile e solo iactum est, cum angulo elevationis <span class=\"katex-eq\" data-katex-display=\"false\">\\theta=30^o<\/span> et celeritate initiali <span class=\"katex-eq\" data-katex-display=\"false\">v_0=70[km\/h].<\/span> a) Quae est altitudo maxima a proiectili consecuta? b) Quam distantiam percurrit proiectile usque ad momentum quo solum tangit? c) Quantum temporis consumit proiectile cadendo?<\/li>\n<li>Tormentum in solo positum globum celeritate 90[km\/h] emittit. Quo angulo elevationis tormentum aptandum est ut globus ad distantiam horizontalem 20[m] cadat?<\/li>\n<li>Idem tormentum ex exercitio priore nunc ad altitudinem 5[m] ponitur. Quo angulo elevationis aptandum est ut globus adhuc ad distantiam horizontalem 20[m] cadat?<\/li>\n<li>Bombardarium ad altitudinem 3 000[m] supra solum celeritate 1500[km\/h] volat. Si hoc proiectilem suo pondere dimittit, quam distantiam percurrit proiectile ab initio dimissionis usque ad momentum quo solum tangit?<\/li>\n<\/ol>\n","protected":false},"excerpt":{"rendered":"<p>Missile Proiectilis Summarium: In hac lectione omnes aspectus cinematicos iactus proiectilis recognoscemus, thema grave in physica quod studium nostrum prius de motu uniformiter accelerato extendit. Exponemus quomodo, sublata restrictione de directione motus, occurramus trajectoriis parabolicis propriis proiectilium. Investigabimus quomodo velocitates initiales in qualibet directione, cum acceleratione gravitatis coniunctae, hos motus effingant. . PROPOSITA DISCENDI: Ad [&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":174,"footnotes":""},"categories":[1262,1250],"tags":[],"class_list":["post-34461","post","type-post","status-publish","format-standard","has-post-thumbnail","hentry","category-fundamenta-mechanicae","category-physica"],"yoast_head":"<!-- This site is optimized with the Yoast SEO plugin v27.4 - https:\/\/yoast.com\/product\/yoast-seo-wordpress\/ -->\n<title>Missile Proiectilis - toposuranos.com\/material<\/title>\n<meta name=\"description\" content=\"Disce omnia de trajectoriis parabolicis et earum aequationibus, quomodo gravitas in iactum proiectilium influit.\" \/>\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\/la\/missile-proiectilis\/\" \/>\n<meta property=\"og:locale\" content=\"es_ES\" \/>\n<meta property=\"og:type\" content=\"article\" \/>\n<meta property=\"og:title\" content=\"Missile Proiectilis\" \/>\n<meta property=\"og:description\" content=\"Disce omnia de trajectoriis parabolicis et earum aequationibus, quomodo gravitas in iactum proiectilium influit.\" \/>\n<meta property=\"og:url\" content=\"https:\/\/toposuranos.com\/material\/la\/missile-proiectilis\/\" \/>\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=\"2021-09-25T00:00:00+00:00\" \/>\n<meta property=\"article:modified_time\" content=\"2025-09-08T03:36:29+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=\"Missile Proiectilis\" \/>\n<meta name=\"twitter:description\" content=\"Disce omnia de trajectoriis parabolicis et earum aequationibus, quomodo gravitas in iactum proiectilium influit.\" \/>\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=\"1 minuto\" \/>\n<script type=\"application\/ld+json\" class=\"yoast-schema-graph\">{\"@context\":\"https:\\\/\\\/schema.org\",\"@graph\":[{\"@type\":\"Article\",\"@id\":\"https:\\\/\\\/toposuranos.com\\\/material\\\/la\\\/missile-proiectilis\\\/#article\",\"isPartOf\":{\"@id\":\"https:\\\/\\\/toposuranos.com\\\/material\\\/la\\\/missile-proiectilis\\\/\"},\"author\":{\"name\":\"giorgio.reveco\",\"@id\":\"https:\\\/\\\/toposuranos.com\\\/material\\\/#\\\/schema\\\/person\\\/e15164361c3f9a2a02cf6c234cf7fdc1\"},\"headline\":\"Missile Proiectilis\",\"datePublished\":\"2021-09-25T00:00:00+00:00\",\"dateModified\":\"2025-09-08T03:36:29+00:00\",\"mainEntityOfPage\":{\"@id\":\"https:\\\/\\\/toposuranos.com\\\/material\\\/la\\\/missile-proiectilis\\\/\"},\"wordCount\":1148,\"commentCount\":0,\"publisher\":{\"@id\":\"https:\\\/\\\/toposuranos.com\\\/material\\\/#organization\"},\"image\":{\"@id\":\"https:\\\/\\\/toposuranos.com\\\/material\\\/la\\\/missile-proiectilis\\\/#primaryimage\"},\"thumbnailUrl\":\"https:\\\/\\\/toposuranos.com\\\/material\\\/wp-content\\\/uploads\\\/2021\\\/09\\\/proyectil.jpg\",\"articleSection\":[\"Fundamenta Mechanicae\",\"Physica\"],\"inLanguage\":\"es\",\"potentialAction\":[{\"@type\":\"CommentAction\",\"name\":\"Comment\",\"target\":[\"https:\\\/\\\/toposuranos.com\\\/material\\\/la\\\/missile-proiectilis\\\/#respond\"]}]},{\"@type\":\"WebPage\",\"@id\":\"https:\\\/\\\/toposuranos.com\\\/material\\\/la\\\/missile-proiectilis\\\/\",\"url\":\"https:\\\/\\\/toposuranos.com\\\/material\\\/la\\\/missile-proiectilis\\\/\",\"name\":\"Missile Proiectilis - 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