{"id":426,"date":"2023-06-18T14:28:11","date_gmt":"2023-06-18T14:28:11","guid":{"rendered":"https:\/\/physigeek.com\/pt\/movimento-parabolico-ou-lancamento-parabolico\/"},"modified":"2023-06-18T14:28:11","modified_gmt":"2023-06-18T14:28:11","slug":"movimento-parabolico-ou-lancamento-parabolico","status":"publish","type":"post","link":"https:\/\/physigeek.com\/pt\/movimento-parabolico-ou-lancamento-parabolico\/","title":{"rendered":"Movimento parab\u00f3lico (ou tiro parab\u00f3lico)"},"content":{"rendered":"<p>Este artigo explica o que \u00e9 movimento parab\u00f3lico (ou tiro parab\u00f3lico) na f\u00edsica. Assim voc\u00ea encontrar\u00e1 as caracter\u00edsticas do movimento parab\u00f3lico, suas f\u00f3rmulas e, al\u00e9m disso, um exemplo passo a passo. <\/p>\n<h2 class=\"wp-block-heading\"><span class=\"ez-toc-section\" id=\"%C2%BFQue-es-el-movimiento-parabolico\"><\/span> O que \u00e9 movimento parab\u00f3lico?<span class=\"ez-toc-section-end\"><\/span><\/h2>\n<p> <strong>Movimento parab\u00f3lico<\/strong> , tamb\u00e9m chamado de <strong>tiro parab\u00f3lico<\/strong> ou <strong>tiro obl\u00edquo<\/strong> , \u00e9 esse movimento realizado por um corpo cuja trajet\u00f3ria descreve uma par\u00e1bola. Assim, um corpo que realiza um movimento parab\u00f3lico avan\u00e7a horizontalmente enquanto verticalmente primeiro sobe e depois desce.<\/p>\n<p> Por exemplo, lan\u00e7ar um proj\u00e9til \u00e9 um movimento parab\u00f3lico porque a trajet\u00f3ria de um proj\u00e9til \u00e9 uma par\u00e1bola. Assim, quando um proj\u00e9til \u00e9 lan\u00e7ado para cima, ele avan\u00e7a horizontalmente e eventualmente cai at\u00e9 atingir o solo sob a influ\u00eancia da gravidade. <\/p>\n<figure class=\"wp-block-image aligncenter size-full is-resized\"><img decoding=\"async\" loading=\"lazy\" src=\"https:\/\/physigeek.com\/wp-content\/uploads\/2023\/09\/mouvement-parabolique.png\" alt=\"movimento parab\u00f3lico, tiro parab\u00f3lico, tiro obl\u00edquo\" class=\"wp-image-8056\" width=\"545\" height=\"389\" srcset=\"https:\/\/physigeek.com\/wp-content\/uploads\/2023\/09\/mouvement-parabolique-300x214.png 300w, https:\/\/physigeek.com\/wp-content\/uploads\/2023\/09\/mouvement-parabolique.png 704w\" sizes=\"auto, (max-width: 300px) 100vw, 300px\"><\/figure>\n<h2 class=\"wp-block-heading\"><span class=\"ez-toc-section\" id=\"Caracteristicas-del-movimiento-parabolico\"><\/span> Caracter\u00edsticas do movimento parab\u00f3lico<span class=\"ez-toc-section-end\"><\/span><\/h2>\n<p> Agora que conhecemos a defini\u00e7\u00e3o de movimento parab\u00f3lico, vamos ver quais s\u00e3o as caracter\u00edsticas dos movimentos parab\u00f3licos.<\/p>\n<ul style=\"color:#4fd12f; font-weight: bold;\">\n<li style=\"margin-bottom:20px\"> <span style=\"color:#101010;font-weight: normal;\"><strong>A principal caracter\u00edstica do movimento parab\u00f3lico<\/strong> \u00e9 que a trajet\u00f3ria descrita pelo m\u00f3bile \u00e9 uma par\u00e1bola.<\/span><\/li>\n<li style=\"margin-bottom:20px\"> <span style=\"color:#101010;font-weight: normal;\">Outra caracter\u00edstica do movimento parab\u00f3lico \u00e9 que ele \u00e9 causado pela acelera\u00e7\u00e3o da gravidade. O corpo que descreve a trajet\u00f3ria parab\u00f3lica come\u00e7a com uma velocidade vertical positiva, ent\u00e3o a princ\u00edpio sobe, mas sob o efeito da gravidade a velocidade vertical diminui at\u00e9 se tornar negativa e ent\u00e3o o corpo desce.<\/span><\/li>\n<li style=\"margin-bottom:20px\"> <span style=\"color:#101010;font-weight: normal;\">Assim, a componente horizontal da velocidade de um movimento parab\u00f3lico \u00e9 constante, enquanto a componente vertical da velocidade diminui.<\/span><\/li>\n<li style=\"margin-bottom:20px\"> <span style=\"color:#101010;font-weight: normal;\">O movimento parab\u00f3lico \u00e9 portanto a uni\u00e3o de dois tipos de movimentos: o movimento horizontal \u00e9 um <a href=\"https:\/\/physigeek.com\/pt\/movimento-retilineo-uniforme-mru\/\">movimento retil\u00edneo uniforme<\/a> e, por outro lado, o movimento vertical \u00e9 um <a href=\"https:\/\/physigeek.com\/pt\/o-movimento-retilineo-acelera-uniformemente-mrua\/\">movimento retil\u00edneo uniformemente acelerado<\/a> .<\/span><\/li>\n<li style=\"margin-bottom:20px\"> <span style=\"color:#101010;font-weight: normal;\">A altura m\u00e1xima do movimento parab\u00f3lico \u00e9 alcan\u00e7ada quando a componente vertical da velocidade \u00e9 zero.<\/span><\/li>\n<li style=\"margin-bottom:20px\"> <span style=\"color:#101010;font-weight: normal;\">Num movimento parab\u00f3lico, o atrito do corpo com o ar ao longo da trajet\u00f3ria \u00e9 desprezado.<\/span> <\/li>\n<\/ul>\n<h2 class=\"wp-block-heading\"><span class=\"ez-toc-section\" id=\"Ejemplos-de-movimientos-parabolicos\"><\/span> Exemplos de movimentos parab\u00f3licos<span class=\"ez-toc-section-end\"><\/span><\/h2>\n<p> Abaixo est\u00e3o v\u00e1rios exemplos de movimentos parab\u00f3licos (ou lan\u00e7amentos parab\u00f3licos):<\/p>\n<ol style=\"color:#4fd12f; font-weight: bold;\">\n<li style=\"margin-bottom:12px\"> <span style=\"color:#101010;font-weight: normal;\">O tiro de uma tacada de basquete.<\/span><\/li>\n<li style=\"margin-bottom:12px\"> <span style=\"color:#101010;font-weight: normal;\">O disparo de um proj\u00e9til.<\/span><\/li>\n<li style=\"margin-bottom:12px\"> <span style=\"color:#101010;font-weight: normal;\">O jato de \u00e1gua de uma mangueira.<\/span><\/li>\n<li style=\"margin-bottom:12px\"> <span style=\"color:#101010;font-weight: normal;\">O lan\u00e7amento de uma pedra.<\/span><\/li>\n<li style=\"margin-bottom:12px\"> <span style=\"color:#101010;font-weight: normal;\">O chute de uma bola de futebol.<\/span> <\/li>\n<\/ol>\n<h2 class=\"wp-block-heading\"><span class=\"ez-toc-section\" id=\"Ecuaciones-del-movimiento-parabolico\"><\/span> Equa\u00e7\u00f5es de movimento parab\u00f3lico<span class=\"ez-toc-section-end\"><\/span><\/h2>\n<p> A seguir veremos quais s\u00e3o todas as equa\u00e7\u00f5es e f\u00f3rmulas para o movimento parab\u00f3lico, tamb\u00e9m conhecido como tiro parab\u00f3lico ou tiro obl\u00edquo. Portanto, essas f\u00f3rmulas permitir\u00e3o resolver problemas de movimento parab\u00f3lico.<\/p>\n<h3 class=\"wp-block-heading\"><span class=\"ez-toc-section\" id=\"Posicion\"><\/span> Posi\u00e7\u00e3o<span class=\"ez-toc-section-end\"><\/span><\/h3>\n<p> No movimento parab\u00f3lico, o componente horizontal da posi\u00e7\u00e3o \u00e9 definido pela f\u00f3rmula do movimento retil\u00edneo uniforme (MRU), enquanto a express\u00e3o para o componente vertical da posi\u00e7\u00e3o \u00e9 a f\u00f3rmula do movimento retil\u00edneo uniformemente acelerado (MRUA). Assim, as equa\u00e7\u00f5es que descrevem a trajet\u00f3ria de um movimento parab\u00f3lico s\u00e3o as seguintes:<\/p>\n<p class=\"has-text-align-center\">\n<p class=\"has-text-align-center\"><img decoding=\"async\" loading=\"lazy\" src=\"https:\/\/physigeek.com\/wp-content\/ql-cache\/quicklatex.com-6a1223e776bfd3723cb49642515d7a4e_l3.png\" class=\"ql-img-inline-formula quicklatex-auto-format\" alt=\"\\begin{cases}x=v_0\\cdot \\text{cos}(\\alpha)\\cdot t \\\\[2ex]y=h+v_0\\cdot \\text{sin}(\\alpha)\\cdot t - \\cfrac{1}{2}\\cdot g\\cdot t^2\\end{cases}\" title=\"Rendered by QuickLaTeX.com\" height=\"86\" width=\"253\" style=\"vertical-align: 0px;\"><\/p>\n<\/p>\n<p style=\"margin-bottom:5px\"> Ouro: <\/p>\n<ul style=\"color:#4fd12f; font-weight: bold;\">\n<li style=\"margin-bottom:5px\"><span style=\"color:#101010;font-weight: normal;\">\n<p class=\"has-text-align-center\"><img decoding=\"async\" loading=\"lazy\" src=\"https:\/\/physigeek.com\/wp-content\/ql-cache\/quicklatex.com-7e5fbfa0bbbd9f3051cd156a0f1b5e31_l3.png\" class=\"ql-img-inline-formula quicklatex-auto-format\" alt=\"x\" title=\"Rendered by QuickLaTeX.com\" height=\"8\" width=\"10\" style=\"vertical-align: 0px;\"><\/p>\n<p> \u00e9 a coordenada horizontal do corpo. <\/span><\/li>\n<li style=\"margin-bottom:5px\"><span style=\"color:#101010;font-weight: normal;\">\n<p class=\"has-text-align-center\"><img decoding=\"async\" loading=\"lazy\" src=\"https:\/\/physigeek.com\/wp-content\/ql-cache\/quicklatex.com-38461fc041e953482219abf5d4cce1cb_l3.png\" class=\"ql-img-inline-formula quicklatex-auto-format\" alt=\"y\" title=\"Rendered by QuickLaTeX.com\" height=\"12\" width=\"9\" style=\"vertical-align: -4px;\"><\/p>\n<p> \u00e9 a coordenada vertical do corpo. <\/span><\/li>\n<li style=\"margin-bottom:5px\"><span style=\"color:#101010;font-weight: normal;\">\n<p class=\"has-text-align-center\"><img decoding=\"async\" loading=\"lazy\" src=\"https:\/\/physigeek.com\/wp-content\/ql-cache\/quicklatex.com-033c1c43e83708a4a975dedd88f197a8_l3.png\" class=\"ql-img-inline-formula quicklatex-auto-format\" alt=\"v_0\" title=\"Rendered by QuickLaTeX.com\" height=\"11\" width=\"16\" style=\"vertical-align: -3px;\"><\/p>\n<p> \u00e9 a velocidade inicial. <\/span><\/li>\n<li style=\"margin-bottom:5px\"><span style=\"color:#101010;font-weight: normal;\">\n<p class=\"has-text-align-center\"><img decoding=\"async\" loading=\"lazy\" src=\"https:\/\/physigeek.com\/wp-content\/ql-cache\/quicklatex.com-5f44d9bbc8046069be4aa2989bff19aa_l3.png\" class=\"ql-img-inline-formula quicklatex-auto-format\" alt=\"\\alpha\" title=\"Rendered by QuickLaTeX.com\" height=\"8\" width=\"11\" style=\"vertical-align: 0px;\"><\/p>\n<p> \u00e9 o \u00e2ngulo inicial da trajet\u00f3ria. <\/span><\/li>\n<li style=\"margin-bottom:5px\"><span style=\"color:#101010;font-weight: normal;\">\n<p class=\"has-text-align-center\"><img decoding=\"async\" loading=\"lazy\" src=\"https:\/\/physigeek.com\/wp-content\/ql-cache\/quicklatex.com-fd9cb27edab3f0a8a249bc80cc9c6ee2_l3.png\" class=\"ql-img-inline-formula quicklatex-auto-format\" alt=\"t\" title=\"Rendered by QuickLaTeX.com\" height=\"12\" width=\"6\" style=\"vertical-align: 0px;\"><\/p>\n<p> \u00e9 o tempo decorrido. <\/span><\/li>\n<li style=\"margin-bottom:5px\"><span style=\"color:#101010;font-weight: normal;\">\n<p class=\"has-text-align-center\"><img decoding=\"async\" loading=\"lazy\" src=\"https:\/\/physigeek.com\/wp-content\/ql-cache\/quicklatex.com-2ce27f7d2d82e3b238176ec7e7ee9118_l3.png\" class=\"ql-img-inline-formula quicklatex-auto-format\" alt=\"h\" title=\"Rendered by QuickLaTeX.com\" height=\"12\" width=\"10\" style=\"vertical-align: 0px;\"><\/p>\n<p> \u00e9 a altura inicial do corpo. <\/span><\/li>\n<li style=\"margin-bottom:5px\"><span style=\"color:#101010;font-weight: normal;\">\n<p class=\"has-text-align-center\"><img decoding=\"async\" loading=\"lazy\" src=\"https:\/\/physigeek.com\/wp-content\/ql-cache\/quicklatex.com-e88010d25c51c0c42c505ee1004ed182_l3.png\" class=\"ql-img-inline-formula quicklatex-auto-format\" alt=\"g\" title=\"Rendered by QuickLaTeX.com\" height=\"12\" width=\"9\" style=\"vertical-align: -4px;\"><\/p>\n<p> \u00e9 a acelera\u00e7\u00e3o da gravidade, cujo valor \u00e9 9,81 m\/s <sup>2<\/sup> .<\/span><\/li>\n<\/ul>\n<h3 class=\"wp-block-heading\"><span class=\"ez-toc-section\" id=\"Velocidad\"><\/span> Velocidade<span class=\"ez-toc-section-end\"><\/span><\/h3>\n<p> No movimento parab\u00f3lico, a componente horizontal da velocidade \u00e9 constante ao longo de toda a trajet\u00f3ria, portanto para calcul\u00e1-la basta multiplicar a velocidade inicial pelo cosseno do \u00e2ngulo de inclina\u00e7\u00e3o.<\/p>\n<p> Por outro lado, a componente vertical de um disparo parab\u00f3lico \u00e9 definida pela equa\u00e7\u00e3o do movimento retil\u00edneo uniformemente acelerado. Portanto, a componente vertical da velocidade \u00e9 equivalente \u00e0 velocidade inicial vezes o seno do \u00e2ngulo de inclina\u00e7\u00e3o menos a acelera\u00e7\u00e3o da gravidade vezes o tempo decorrido.<\/p>\n<p class=\"has-text-align-center\">\n<p class=\"has-text-align-center\"><img decoding=\"async\" loading=\"lazy\" src=\"https:\/\/physigeek.com\/wp-content\/ql-cache\/quicklatex.com-9ad54222c1f929421b7d82314bd355db_l3.png\" class=\"ql-img-inline-formula quicklatex-auto-format\" alt=\"\\begin{cases}v_x=v_0\\cdot \\text{cos}(\\alpha) \\\\[2ex]v_y=v_0\\cdot \\text{sin}(\\alpha)-g\\cdot t\\end{cases }\" title=\"Rendered by QuickLaTeX.com\" height=\"65\" width=\"179\" style=\"vertical-align: 0px;\"><\/p>\n<\/p>\n<p style=\"margin-bottom:5px\"> Ouro: <\/p>\n<ul style=\"color:#4fd12f; font-weight: bold;\">\n<li style=\"margin-bottom:5px\"><span style=\"color:#101010;font-weight: normal;\">\n<p class=\"has-text-align-center\"><img decoding=\"async\" loading=\"lazy\" src=\"https:\/\/physigeek.com\/wp-content\/ql-cache\/quicklatex.com-49b889f48d662e0aaa8d7cb4fe10f013_l3.png\" class=\"ql-img-inline-formula quicklatex-auto-format\" alt=\"v_x\" title=\"Rendered by QuickLaTeX.com\" height=\"11\" width=\"17\" style=\"vertical-align: -3px;\"><\/p>\n<p> \u00e9 a componente horizontal da velocidade. <\/span><\/li>\n<li style=\"margin-bottom:5px\"><span style=\"color:#101010;font-weight: normal;\">\n<p class=\"has-text-align-center\"><img decoding=\"async\" loading=\"lazy\" src=\"https:\/\/physigeek.com\/wp-content\/ql-cache\/quicklatex.com-5f9cca35a3ff6c43b956a6f9fe4e3921_l3.png\" class=\"ql-img-inline-formula quicklatex-auto-format\" alt=\"v_y\" title=\"Rendered by QuickLaTeX.com\" height=\"14\" width=\"16\" style=\"vertical-align: -6px;\"><\/p>\n<p> \u00e9 a componente vertical da velocidade. <\/span><\/li>\n<li style=\"margin-bottom:5px\"><span style=\"color:#101010;font-weight: normal;\">\n<p class=\"has-text-align-center\"><img decoding=\"async\" loading=\"lazy\" src=\"https:\/\/physigeek.com\/wp-content\/ql-cache\/quicklatex.com-033c1c43e83708a4a975dedd88f197a8_l3.png\" class=\"ql-img-inline-formula quicklatex-auto-format\" alt=\"v_0\" title=\"Rendered by QuickLaTeX.com\" height=\"11\" width=\"16\" style=\"vertical-align: -3px;\"><\/p>\n<p> \u00e9 a velocidade inicial. <\/span><\/li>\n<li style=\"margin-bottom:5px\"><span style=\"color:#101010;font-weight: normal;\">\n<p class=\"has-text-align-center\"><img decoding=\"async\" loading=\"lazy\" src=\"https:\/\/physigeek.com\/wp-content\/ql-cache\/quicklatex.com-5f44d9bbc8046069be4aa2989bff19aa_l3.png\" class=\"ql-img-inline-formula quicklatex-auto-format\" alt=\"\\alpha\" title=\"Rendered by QuickLaTeX.com\" height=\"8\" width=\"11\" style=\"vertical-align: 0px;\"><\/p>\n<p> \u00e9 o \u00e2ngulo inicial da trajet\u00f3ria. <\/span><\/li>\n<li style=\"margin-bottom:5px\"><span style=\"color:#101010;font-weight: normal;\">\n<p class=\"has-text-align-center\"><img decoding=\"async\" loading=\"lazy\" src=\"https:\/\/physigeek.com\/wp-content\/ql-cache\/quicklatex.com-fd9cb27edab3f0a8a249bc80cc9c6ee2_l3.png\" class=\"ql-img-inline-formula quicklatex-auto-format\" alt=\"t\" title=\"Rendered by QuickLaTeX.com\" height=\"12\" width=\"6\" style=\"vertical-align: 0px;\"><\/p>\n<p> \u00e9 o tempo decorrido. <\/span><\/li>\n<li style=\"margin-bottom:5px\"><span style=\"color:#101010;font-weight: normal;\">\n<p class=\"has-text-align-center\"><img decoding=\"async\" loading=\"lazy\" src=\"https:\/\/physigeek.com\/wp-content\/ql-cache\/quicklatex.com-e88010d25c51c0c42c505ee1004ed182_l3.png\" class=\"ql-img-inline-formula quicklatex-auto-format\" alt=\"g\" title=\"Rendered by QuickLaTeX.com\" height=\"12\" width=\"9\" style=\"vertical-align: -4px;\"><\/p>\n<p> \u00e9 a acelera\u00e7\u00e3o da gravidade, cujo valor \u00e9 9,81 m\/s <sup>2<\/sup> .<\/span><\/li>\n<\/ul>\n<h3 class=\"wp-block-heading\"><span class=\"ez-toc-section\" id=\"Aceleracion\"><\/span> Acelera\u00e7\u00e3o<span class=\"ez-toc-section-end\"><\/span><\/h3>\n<p> Em todos os movimentos parab\u00f3licos a acelera\u00e7\u00e3o do corpo tem sempre o mesmo valor. A componente horizontal da acelera\u00e7\u00e3o \u00e9 zero, enquanto a componente vertical da acelera\u00e7\u00e3o \u00e9 o valor da gravidade com sinal negativo.<\/p>\n<p class=\"has-text-align-center\">\n<p class=\"has-text-align-center\"><img decoding=\"async\" loading=\"lazy\" src=\"https:\/\/physigeek.com\/wp-content\/ql-cache\/quicklatex.com-b317db933cced3fd619deeff201818c8_l3.png\" class=\"ql-img-inline-formula quicklatex-auto-format\" alt=\"\\begin{cases}a_x=0 \\\\[2ex]a_y=-g\\end{cases}\" title=\"Rendered by QuickLaTeX.com\" height=\"65\" width=\"77\" style=\"vertical-align: 0px;\"><\/p>\n<\/p>\n<p style=\"margin-bottom:5px\"> Ouro: <\/p>\n<ul style=\"color:#4fd12f; font-weight: bold;\">\n<li style=\"margin-bottom:5px\"><span style=\"color:#101010;font-weight: normal;\">\n<p class=\"has-text-align-center\"><img decoding=\"async\" loading=\"lazy\" src=\"https:\/\/physigeek.com\/wp-content\/ql-cache\/quicklatex.com-9eccaa1194e3cd8a77d62f4a2fd89a51_l3.png\" class=\"ql-img-inline-formula quicklatex-auto-format\" alt=\"a_x\" title=\"Rendered by QuickLaTeX.com\" height=\"11\" width=\"17\" style=\"vertical-align: -3px;\"><\/p>\n<p> \u00e9 a componente horizontal da acelera\u00e7\u00e3o. <\/span><\/li>\n<li style=\"margin-bottom:5px\"><span style=\"color:#101010;font-weight: normal;\">\n<p class=\"has-text-align-center\"><img decoding=\"async\" loading=\"lazy\" src=\"https:\/\/physigeek.com\/wp-content\/ql-cache\/quicklatex.com-774db4a74587d8bf0c9a33ba6b3db0d3_l3.png\" class=\"ql-img-inline-formula quicklatex-auto-format\" alt=\"a_y\" title=\"Rendered by QuickLaTeX.com\" height=\"14\" width=\"16\" style=\"vertical-align: -6px;\"><\/p>\n<p> \u00e9 a componente vertical da acelera\u00e7\u00e3o. <\/span><\/li>\n<li style=\"margin-bottom:5px\"><span style=\"color:#101010;font-weight: normal;\">\n<p class=\"has-text-align-center\"><img decoding=\"async\" loading=\"lazy\" src=\"https:\/\/physigeek.com\/wp-content\/ql-cache\/quicklatex.com-e88010d25c51c0c42c505ee1004ed182_l3.png\" class=\"ql-img-inline-formula quicklatex-auto-format\" alt=\"g\" title=\"Rendered by QuickLaTeX.com\" height=\"12\" width=\"9\" style=\"vertical-align: -4px;\"><\/p>\n<p> \u00e9 a acelera\u00e7\u00e3o da gravidade, cujo valor \u00e9 9,81 m\/s <sup>2<\/sup> .<\/span><\/li>\n<\/ul>\n<h3 class=\"wp-block-heading\"><span class=\"ez-toc-section\" id=\"Tiempo-de-vuelo\"><\/span> Hora do voo<span class=\"ez-toc-section-end\"><\/span><\/h3>\n<p> O tempo de v\u00f4o \u00e9 o tempo necess\u00e1rio para que o corpo que realiza o movimento parab\u00f3lico toque o solo. Portanto, o tempo de voo \u00e9 o tempo desde o momento em que o corpo inicia a par\u00e1bola at\u00e9 atingir o solo.<\/p>\n<p> Quando o corpo atingir o solo, a coordenada vertical de sua posi\u00e7\u00e3o ser\u00e1 zero. Portanto, para calcular o tempo de v\u00f4o, voc\u00ea precisa definir a equa\u00e7\u00e3o da posi\u00e7\u00e3o vertical do movimento parab\u00f3lico igual a zero e ent\u00e3o resolver a equa\u00e7\u00e3o do tempo. <\/p>\n<p class=\"has-text-align-center\">\n<p class=\"has-text-align-center\"><img decoding=\"async\" loading=\"lazy\" src=\"https:\/\/physigeek.com\/wp-content\/ql-cache\/quicklatex.com-01fcd09349d7c25e11fc42ae3744f423_l3.png\" class=\"ql-img-inline-formula quicklatex-auto-format\" alt=\"y=0 \\quad \\color{orange}\\bm{\\longrightarrow}\\color{black}\\quad t_{vol}\" title=\"Rendered by QuickLaTeX.com\" height=\"16\" width=\"225\" style=\"vertical-align: -4px;\"><\/p>\n<\/p>\n<h3 class=\"wp-block-heading\"><span class=\"ez-toc-section\" id=\"Alcance-horizontal\"><\/span> Escopo horizontal<span class=\"ez-toc-section-end\"><\/span><\/h3>\n<p> O alcance horizontal m\u00e1ximo ser\u00e1 alcan\u00e7ado quando o corpo tocar o solo, instante que equivale ao tempo de voo. Portanto, para calcular o alcance horizontal, primeiro deve-se tomar o tempo de voo e depois substituir o valor do tempo de voo na equa\u00e7\u00e3o da posi\u00e7\u00e3o horizontal do movimento parab\u00f3lico. <\/p>\n<p class=\"has-text-align-center\">\n<p class=\"has-text-align-center\"><img decoding=\"async\" loading=\"lazy\" src=\"https:\/\/physigeek.com\/wp-content\/ql-cache\/quicklatex.com-09a2bd6165dae0d8de909888cc1483ab_l3.png\" class=\"ql-img-inline-formula quicklatex-auto-format\" alt=\" t_{vol}\\quad \\color{orange}\\bm{\\longrightarrow}\\color{black}\\quad x(t_{vol})\" title=\"Rendered by QuickLaTeX.com\" height=\"19\" width=\"232\" style=\"vertical-align: -5px;\"><\/p>\n<\/p>\n<h3 class=\"wp-block-heading\"><span class=\"ez-toc-section\" id=\"Altura-maxima\"><\/span> Altura m\u00e1xima<span class=\"ez-toc-section-end\"><\/span><\/h3>\n<p> Num movimento parab\u00f3lico, a altura m\u00e1xima \u00e9 atingida quando a componente vertical da velocidade do corpo \u00e9 zero. Assim, para determinar a altura m\u00e1xima, a componente vertical da velocidade deve ser igual a zero, a partir da\u00ed encontraremos o instante em que a altura m\u00e1xima \u00e9 atingida e, por fim, devemos substituir o instante de tempo calculado no calculado momento. equa\u00e7\u00e3o.posi\u00e7\u00e3o vertical. <\/p>\n<p class=\"has-text-align-center\">\n<p class=\"has-text-align-center\"><img decoding=\"async\" loading=\"lazy\" src=\"https:\/\/physigeek.com\/wp-content\/ql-cache\/quicklatex.com-a822a7a4e291ee420791e704d53820f3_l3.png\" class=\"ql-img-inline-formula quicklatex-auto-format\" alt=\"v_y=0 \\quad \\color{orange}\\bm{\\longrightarrow}\\color{black}\\quad t_{y_{m\\'ax}}\\quad \\color{orange}\\bm{\\longrightarrow}\\ couleur{noir}\\quad y_{m\\'ax}\" title=\"Rendered by QuickLaTeX.com\" height=\"18\" width=\"500\" style=\"vertical-align: -6px;\"><\/p>\n<\/p>\n<h3 class=\"wp-block-heading\"><span class=\"ez-toc-section\" id=\"Angulo-de-la-trayectoria\"><\/span> \u00e2ngulo de trajet\u00f3ria<span class=\"ez-toc-section-end\"><\/span><\/h3>\n<p> O \u00e2ngulo da trajet\u00f3ria em um determinado ponto \u00e9 equivalente ao \u00e2ngulo formado pelas duas componentes da velocidade. Assim, a tangente do \u00e2ngulo da trajet\u00f3ria \u00e9 igual ao quociente entre a componente vertical e a componente horizontal da velocidade.<\/p>\n<p class=\"has-text-align-center\">\n<p class=\"has-text-align-center\"><img decoding=\"async\" loading=\"lazy\" src=\"https:\/\/physigeek.com\/wp-content\/ql-cache\/quicklatex.com-83a1b49fe71033c804ad8193164bddd6_l3.png\" class=\"ql-img-inline-formula quicklatex-auto-format\" alt=\"\\text{tan}(\\alpha)=\\cfrac{v_y}{v_x}\" title=\"Rendered by QuickLaTeX.com\" height=\"37\" width=\"94\" style=\"vertical-align: -15px;\"><\/p>\n<\/p>\n<p style=\"margin-bottom:5px\"> Ouro: <\/p>\n<ul style=\"color:#4fd12f; font-weight: bold;\">\n<li style=\"margin-bottom:5px\"><span style=\"color:#101010;font-weight: normal;\">\n<p class=\"has-text-align-center\"><img decoding=\"async\" loading=\"lazy\" src=\"https:\/\/physigeek.com\/wp-content\/ql-cache\/quicklatex.com-5f9cca35a3ff6c43b956a6f9fe4e3921_l3.png\" class=\"ql-img-inline-formula quicklatex-auto-format\" alt=\"v_y\" title=\"Rendered by QuickLaTeX.com\" height=\"14\" width=\"16\" style=\"vertical-align: -6px;\"><\/p>\n<p> \u00e9 a componente vertical da velocidade. <\/span><\/li>\n<li style=\"margin-bottom:5px\"><span style=\"color:#101010;font-weight: normal;\">\n<p class=\"has-text-align-center\"><img decoding=\"async\" loading=\"lazy\" src=\"https:\/\/physigeek.com\/wp-content\/ql-cache\/quicklatex.com-49b889f48d662e0aaa8d7cb4fe10f013_l3.png\" class=\"ql-img-inline-formula quicklatex-auto-format\" alt=\"v_x\" title=\"Rendered by QuickLaTeX.com\" height=\"11\" width=\"17\" style=\"vertical-align: -3px;\"><\/p>\n<p> \u00e9 a componente horizontal da velocidade. <\/span><\/li>\n<li style=\"margin-bottom:5px\"><span style=\"color:#101010;font-weight: normal;\">\n<p class=\"has-text-align-center\"><img decoding=\"async\" loading=\"lazy\" src=\"https:\/\/physigeek.com\/wp-content\/ql-cache\/quicklatex.com-5f44d9bbc8046069be4aa2989bff19aa_l3.png\" class=\"ql-img-inline-formula quicklatex-auto-format\" alt=\"\\alpha\" title=\"Rendered by QuickLaTeX.com\" height=\"8\" width=\"11\" style=\"vertical-align: 0px;\"><\/p>\n<p> \u00e9 o \u00e2ngulo do caminho. <\/span><\/li>\n<\/ul>\n<h3 class=\"wp-block-heading\"><span class=\"ez-toc-section\" id=\"Resumen-de-las-formulas-del-movimiento-parabolico\"><\/span> Resumo das f\u00f3rmulas de movimento parab\u00f3lico<span class=\"ez-toc-section-end\"><\/span><\/h3>\n<p> Em resumo, deixamos uma tabela com as f\u00f3rmulas do movimento parab\u00f3lico. <\/p>\n<figure class=\"wp-block-image aligncenter size-full is-resized\"><img decoding=\"async\" loading=\"lazy\" src=\"https:\/\/physigeek.com\/wp-content\/uploads\/2023\/09\/formules-de-tir-parabolique-a-mouvement-parabolique.png\" alt=\"f\u00f3rmulas de movimento parab\u00f3lico\" class=\"wp-image-8135\" width=\"508\" height=\"493\" srcset=\"https:\/\/physigeek.com\/wp-content\/uploads\/2023\/09\/formules-de-tir-parabolique-a-mouvement-parabolique-300x291.png 300w, https:\/\/physigeek.com\/wp-content\/uploads\/2023\/09\/formules-de-tir-parabolique-a-mouvement-parabolique-768x746.png 768w, https:\/\/physigeek.com\/wp-content\/uploads\/2023\/09\/formules-de-tir-parabolique-a-mouvement-parabolique.png 1014w\" sizes=\"auto, (max-width: 300px) 100vw, 300px\"><\/figure>\n<h2 class=\"wp-block-heading\"><span class=\"ez-toc-section\" id=\"Ejercicio-resuelto-del-movimiento-parabolico\"><\/span> Exerc\u00edcio resolvido de movimento parab\u00f3lico<span class=\"ez-toc-section-end\"><\/span><\/h2>\n<ul>\n<li> Um objeto \u00e9 lan\u00e7ado do solo com velocidade inicial de 15 m\/s e \u00e2ngulo de inclina\u00e7\u00e3o de 30\u00ba. Calcule o alcance horizontal m\u00e1ximo e o m\u00f3dulo da velocidade com que o corpo atinge o solo. Despreze o atrito com o ar em todo o problema e considere o valor da gravidade como 10 m\/s <sup>2<\/sup> .<\/li>\n<\/ul>\n<p> Para encontrar a amplitude horizontal do movimento parab\u00f3lico, devemos primeiro determinar o tempo de voo. E, para isso, devemos igualar a zero a equa\u00e7\u00e3o da componente vertical da posi\u00e7\u00e3o, pois quando o corpo tocar o solo, a posi\u00e7\u00e3o vertical ser\u00e1 y=0. <\/p>\n<p class=\"has-text-align-center\">\n<p class=\"has-text-align-center\"><img decoding=\"async\" loading=\"lazy\" src=\"https:\/\/physigeek.com\/wp-content\/ql-cache\/quicklatex.com-9ab5a4669f2bed735498c28d335e04e9_l3.png\" class=\"ql-img-inline-formula quicklatex-auto-format\" alt=\"y=h+v_0\\cdot \\text{sin}(\\alpha)\\cdot t -\\cfrac{1}{2}\\cdot g\\cdot t^2\" title=\"Rendered by QuickLaTeX.com\" height=\"38\" width=\"240\" style=\"vertical-align: -12px;\"><\/p>\n<\/p>\n<p class=\"has-text-align-center\">\n<p class=\"has-text-align-center\"><img decoding=\"async\" loading=\"lazy\" src=\"https:\/\/physigeek.com\/wp-content\/ql-cache\/quicklatex.com-b302515222791249c838a2238182524f_l3.png\" class=\"ql-img-inline-formula quicklatex-auto-format\" alt=\"0=0+15\\cdot \\text{sin}(30^o)\\cdot t -\\cfrac{1}{2}\\cdot 10\\cdot t^2\" title=\"Rendered by QuickLaTeX.com\" height=\"38\" width=\"262\" style=\"vertical-align: -12px;\"><\/p>\n<\/p>\n<p class=\"has-text-align-center\">\n<p class=\"has-text-align-center\"><img decoding=\"async\" loading=\"lazy\" src=\"https:\/\/physigeek.com\/wp-content\/ql-cache\/quicklatex.com-96c994022b665bc74d1e0bd37437bdaa_l3.png\" class=\"ql-img-inline-formula quicklatex-auto-format\" alt=\"0=7,5\\cdot t -5\\cdot t^2\" title=\"Rendered by QuickLaTeX.com\" height=\"19\" width=\"134\" style=\"vertical-align: -4px;\"><\/p>\n<\/p>\n<p> Resolvemos a equa\u00e7\u00e3o quadr\u00e1tica que obtivemos removendo o fator comum: <\/p>\n<p class=\"has-text-align-center\">\n<p class=\"has-text-align-center\"><img decoding=\"async\" loading=\"lazy\" src=\"https:\/\/physigeek.com\/wp-content\/ql-cache\/quicklatex.com-9ea3476be25efa1f2917a1985dd8d9d9_l3.png\" class=\"ql-img-inline-formula quicklatex-auto-format\" alt=\"0=t(7,5-5t)\" title=\"Rendered by QuickLaTeX.com\" height=\"19\" width=\"114\" style=\"vertical-align: -5px;\"><\/p>\n<\/p>\n<p class=\"has-text-align-center\">\n<p class=\"has-text-align-center\"><img decoding=\"async\" loading=\"lazy\" src=\"https:\/\/physigeek.com\/wp-content\/ql-cache\/quicklatex.com-16852ce664cedc209b4f0c88311cf62c_l3.png\" class=\"ql-img-inline-formula quicklatex-auto-format\" alt=\"\\displaystyle t=\\begin{cases}t=0 \\ \\color{red}\\bm{\\times}\\color{black}\\\\[2ex]7.5 -5t=0 \\ \\longrightarrow \\ t= \\cfrac {7,5}{5}=1,5 \\ s\\end{cases}\" title=\"Rendered by QuickLaTeX.com\" height=\"86\" width=\"309\" style=\"vertical-align: 0px;\"><\/p>\n<\/p>\n<p> Portanto, o corpo atingir\u00e1 o alcance horizontal m\u00e1ximo no tempo t=1,5 s, ent\u00e3o substitu\u00edmos esse valor na equa\u00e7\u00e3o da posi\u00e7\u00e3o horizontal para calcular o alcance horizontal m\u00e1ximo:<\/p>\n<p class=\"has-text-align-center\">\n<p class=\"has-text-align-center\"><img decoding=\"async\" loading=\"lazy\" src=\"https:\/\/physigeek.com\/wp-content\/ql-cache\/quicklatex.com-0ee7aeb8984a471119007f3344290974_l3.png\" class=\"ql-img-inline-formula quicklatex-auto-format\" alt=\"\\begin{aligned}x&amp;=v_0\\cdot \\text{cos}(\\alpha)\\cdot t\\\\[2ex]x&amp;=15\\cdot \\text{cos}(30^o)\\cdot 1.5 \\\\ [2ex]x&amp;=19.49 \\ m \\end{aligned}\" title=\"Rendered by QuickLaTeX.com\" height=\"88\" width=\"197\" style=\"vertical-align: 0px;\"><\/p>\n<\/p>\n<p> Por outro lado, para calcular o m\u00f3dulo da velocidade final, \u00e9 necess\u00e1rio primeiro determinar as duas componentes da velocidade neste instante. Assim, calculamos a componente horizontal da velocidade:<\/p>\n<p class=\"has-text-align-center\">\n<p class=\"has-text-align-center\"><img decoding=\"async\" loading=\"lazy\" src=\"https:\/\/physigeek.com\/wp-content\/ql-cache\/quicklatex.com-73ca2c7f828a69c0ad9f6e1ffaf4302c_l3.png\" class=\"ql-img-inline-formula quicklatex-auto-format\" alt=\"\\begin{aligned}v_x&amp;=v_0\\cdot \\text{cos}(\\alpha) \\\\[2ex]v_x&amp;=15\\cdot \\text{cos}(30^o)\\\\[2ex]v_x&amp;=12 .99 \\ \\cfrac{m}{s}\\end{aligned}\" title=\"Rendered by QuickLaTeX.com\" height=\"125\" width=\"133\" style=\"vertical-align: 0px;\"><\/p>\n<\/p>\n<p> A seguir, calculamos a componente vertical da velocidade com a f\u00f3rmula correspondente:<\/p>\n<p class=\"has-text-align-center\">\n<p class=\"has-text-align-center\"><img decoding=\"async\" loading=\"lazy\" src=\"https:\/\/physigeek.com\/wp-content\/ql-cache\/quicklatex.com-ad9644c10727b8269a43c256214d0153_l3.png\" class=\"ql-img-inline-formula quicklatex-auto-format\" alt=\"\\begin{aligned}v_y&amp;=v_0\\cdot \\text{sin}(\\alpha)-g\\cdot t\\\\[2ex]v_y&amp;=15\\cdot \\text{sin}(30^o) -10\\ cdot 1.5\\\\[2ex]v_y&amp;=-7.5 \\ \\cfrac{m}{s}\\end{aligned}\" title=\"Rendered by QuickLaTeX.com\" height=\"125\" width=\"231\" style=\"vertical-align: 0px;\"><\/p>\n<\/p>\n<p> Finalmente, o m\u00f3dulo de velocidade \u00e9 equivalente \u00e0 raiz quadrada da soma dos quadrados de suas componentes vetoriais:<\/p>\n<p class=\"has-text-align-center\">\n<p class=\"has-text-align-center\"><img decoding=\"async\" loading=\"lazy\" src=\"https:\/\/physigeek.com\/wp-content\/ql-cache\/quicklatex.com-7b1424eebf6e1cecfa0e555aec2da0d5_l3.png\" class=\"ql-img-inline-formula quicklatex-auto-format\" alt=\"\\begin{aligned}|\\vv{v}|&amp;=\\sqrt{v_x^2+v_y^2}\\\\[2ex]|\\vv{v}|&amp;=\\sqrt{12.99^2 +( -7,5)^2}\\\\[2ex]|\\vv{v}|&amp;=15 \\ \\cfrac{m}{s}\\end{aligned}\" title=\"Rendered by QuickLaTeX.com\" height=\"141\" width=\"189\" style=\"vertical-align: 0px;\"><\/p>\n<\/p>\n<p> Concluindo este problema, podemos concluir que quando o movimento parab\u00f3lico parte do solo, a magnitude da velocidade final coincide com a magnitude da velocidade inicial. <\/p>\n<h2 class=\"wp-block-heading\"><span class=\"ez-toc-section\" id=\"Movimiento-parabolico-y-tiro-parabolico-horizontal\"><\/span> Movimento parab\u00f3lico e lan\u00e7amento parab\u00f3lico horizontal<span class=\"ez-toc-section-end\"><\/span><\/h2>\n<p> Por fim, veremos qual a diferen\u00e7a entre movimento parab\u00f3lico e lan\u00e7amento parab\u00f3lico horizontal, pois s\u00e3o dois tipos de movimentos comumente utilizados em f\u00edsica.<\/p>\n<p> O <strong>lan\u00e7amento parab\u00f3lico horizontal<\/strong> \u00e9 um tipo de movimento parab\u00f3lico em que o corpo inicialmente apresenta uma trajet\u00f3ria totalmente horizontal. Assim, em um lan\u00e7amento parab\u00f3lico horizontal, o corpo \u00e9 lan\u00e7ado de uma determinada altura e sua velocidade inicial \u00e9 horizontal.<\/p>\n<p> Portanto, a <strong>diferen\u00e7a entre o balan\u00e7o parab\u00f3lico e o lan\u00e7amento parab\u00f3lico horizontal<\/strong> \u00e9 a velocidade inicial. A velocidade inicial do disparo parab\u00f3lico horizontal \u00e9 completamente horizontal, por\u00e9m a velocidade inicial do movimento parab\u00f3lico forma um \u00e2ngulo positivo com o eixo horizontal. <\/p>\n<div style=\"background-color:#FFFDE7; padding-top: 10px; padding-bottom: 10px; padding-right: 10px; padding-left: 20px; border: 2.5px dashed #FFB74D; border-radius:20px;\"> <span style=\"color:#ff951b\">\u27a4<\/span> <strong>Veja:<\/strong> <a href=\"https:\/\/physigeek.com\/pt\/plano-parabolico-horizontal\/\">Tiro parab\u00f3lico horizontal<\/a><\/div>\n","protected":false},"excerpt":{"rendered":"<p>Este artigo explica o que \u00e9 movimento parab\u00f3lico (ou tiro parab\u00f3lico) na f\u00edsica. Assim voc\u00ea encontrar\u00e1 as caracter\u00edsticas do movimento parab\u00f3lico, suas f\u00f3rmulas e, al\u00e9m disso, um exemplo passo a passo. O que \u00e9 movimento parab\u00f3lico? Movimento parab\u00f3lico , tamb\u00e9m chamado de tiro parab\u00f3lico ou tiro obl\u00edquo , \u00e9 esse movimento realizado por um corpo &hellip;<\/p>\n<p class=\"read-more\"> <a class=\"\" href=\"https:\/\/physigeek.com\/pt\/movimento-parabolico-ou-lancamento-parabolico\/\"> <span class=\"screen-reader-text\">Movimento parab\u00f3lico (ou tiro parab\u00f3lico)<\/span> Leia mais &raquo;<\/a><\/p>\n","protected":false},"author":1,"featured_media":0,"comment_status":"open","ping_status":"open","sticky":false,"template":"","format":"standard","meta":{"site-sidebar-layout":"default","site-content-layout":"","ast-site-content-layout":"","site-content-style":"default","site-sidebar-style":"default","ast-global-header-display":"","ast-banner-title-visibility":"","ast-main-header-display":"","ast-hfb-above-header-display":"","ast-hfb-below-header-display":"","ast-hfb-mobile-header-display":"","site-post-title":"","ast-breadcrumbs-content":"","ast-featured-img":"","footer-sml-layout":"","theme-transparent-header-meta":"","adv-header-id-meta":"","stick-header-meta":"","header-above-stick-meta":"","header-main-stick-meta":"","header-below-stick-meta":"","astra-migrate-meta-layouts":"","footnotes":""},"categories":[3],"tags":[],"class_list":["post-426","post","type-post","status-publish","format-standard","hentry","category-cinematografico"],"yoast_head":"<!-- This site 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