{"id":429,"date":"2023-06-18T12:09:07","date_gmt":"2023-06-18T12:09:07","guid":{"rendered":"https:\/\/physigeek.com\/pt\/plano-parabolico-horizontal\/"},"modified":"2023-06-18T12:09:07","modified_gmt":"2023-06-18T12:09:07","slug":"plano-parabolico-horizontal","status":"publish","type":"post","link":"https:\/\/physigeek.com\/pt\/plano-parabolico-horizontal\/","title":{"rendered":"Tiro parab\u00f3lico horizontal"},"content":{"rendered":"<p>Este artigo explica o que \u00e9 o lan\u00e7amento parab\u00f3lico horizontal, tamb\u00e9m chamado de lan\u00e7amento horizontal ou lan\u00e7amento horizontal, na f\u00edsica e quais s\u00e3o suas caracter\u00edsticas. Al\u00e9m disso, voc\u00ea encontrar\u00e1 as f\u00f3rmulas para o disparo parab\u00f3lico horizontal, bem como um exemplo concreto passo a passo. <\/p>\n<h2 class=\"wp-block-heading\"><span class=\"ez-toc-section\" id=\"%C2%BFQue-es-el-tiro-parabolico-horizontal\"><\/span> O que \u00e9 calado parab\u00f3lico horizontal?<span class=\"ez-toc-section-end\"><\/span><\/h2>\n<p> O <strong>lan\u00e7amento parab\u00f3lico horizontal<\/strong> , <strong>lan\u00e7amento horizontal<\/strong> ou <strong>lan\u00e7amento horizontal<\/strong> , \u00e9 um movimento em forma de par\u00e1bola que parte de uma altura e a velocidade inicial \u00e9 horizontal.<\/p>\n<p> O lan\u00e7amento parab\u00f3lico horizontal \u00e9 a uni\u00e3o de dois movimentos: o movimento vertical \u00e9 um <a href=\"https:\/\/physigeek.com\/pt\/movimento-retilineo-uniforme-mru\/\">MRU<\/a> e o movimento horizontal \u00e9 um <a href=\"https:\/\/physigeek.com\/pt\/o-movimento-retilineo-acelera-uniformemente-mrua\/\">MRUA<\/a> .<\/p>\n<p> Por exemplo, lan\u00e7ar uma bola horizontalmente do telhado de um edif\u00edcio \u00e9 um lan\u00e7amento parab\u00f3lico horizontal. A bola inicia o movimento desde uma altura, sua velocidade inicial \u00e9 totalmente horizontal e faz um movimento parab\u00f3lico devido \u00e0 gravidade, portanto \u00e9 um tiro parab\u00f3lico horizontal. <\/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\/lancement-de-tir-parabolique-horizontal.png\" alt=\"tiro parab\u00f3lico horizontal, tiro horizontal, tiro horizontal\" class=\"wp-image-8376\" width=\"523\" height=\"476\" srcset=\"https:\/\/physigeek.com\/wp-content\/uploads\/2023\/09\/lancement-de-tir-parabolique-horizontal-300x273.png 300w, https:\/\/physigeek.com\/wp-content\/uploads\/2023\/09\/lancement-de-tir-parabolique-horizontal.png 741w\" sizes=\"auto, (max-width: 300px) 100vw, 300px\"><\/figure>\n<h2 class=\"wp-block-heading\"><span class=\"ez-toc-section\" id=\"Caracteristicas-del-tiro-parabolico-horizontal\"><\/span> Caracter\u00edsticas do tiro parab\u00f3lico horizontal<span class=\"ez-toc-section-end\"><\/span><\/h2>\n<p> Depois de vermos a defini\u00e7\u00e3o de lan\u00e7amento parab\u00f3lico horizontal na f\u00edsica, vamos ver quais s\u00e3o as caracter\u00edsticas desse tipo de movimento.<\/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 disparo parab\u00f3lico horizontal<\/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;\">Da mesma forma, o disparo parab\u00f3lico horizontal \u00e9 caracterizado por uma velocidade inicial completamente horizontal.<\/span><\/li>\n<li style=\"margin-bottom:20px\"> <span style=\"color:#101010;font-weight: normal;\">A trajet\u00f3ria parab\u00f3lica do tiro parab\u00f3lico horizontal \u00e9 devida \u00e0 acelera\u00e7\u00e3o da gravidade. A princ\u00edpio, a componente vertical da velocidade \u00e9 zero, ent\u00e3o o corpo se move horizontalmente, mas sob a influ\u00eancia da gravidade, a velocidade vertical torna-se cada vez mais negativa e, como resultado, 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 disparo parab\u00f3lico horizontal \u00e9 constante, enquanto a componente vertical da velocidade diminui (torna-se cada vez mais negativa).<\/span><\/li>\n<li style=\"margin-bottom:20px\"> <span style=\"color:#101010;font-weight: normal;\">O lan\u00e7amento parab\u00f3lico horizontal \u00e9 portanto a uni\u00e3o de dois tipos de movimentos: o movimento horizontal \u00e9 um movimento retil\u00edneo uniforme (MRU) e, por outro lado, o movimento vertical \u00e9 um movimento retil\u00edneo uniformemente acelerado (MRUA).<\/span><\/li>\n<li style=\"margin-bottom:20px\"> <span style=\"color:#101010;font-weight: normal;\">Na f\u00edsica, no tiro parab\u00f3lico horizontal, o atrito do corpo com o ar \u00e9 desprezado ao longo do movimento.<\/span> <\/li>\n<\/ul>\n<h2 class=\"wp-block-heading\"><span class=\"ez-toc-section\" id=\"Formulas-del-tiro-parabolico-horizontal\"><\/span> F\u00f3rmulas de tiro parab\u00f3lico horizontal<span class=\"ez-toc-section-end\"><\/span><\/h2>\n<p> Abaixo est\u00e3o as f\u00f3rmulas (ou equa\u00e7\u00f5es) para o disparo parab\u00f3lico horizontal. Essas f\u00f3rmulas nos ajudar\u00e3o a resolver problemas de calado parab\u00f3lico horizontal.<\/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> Em um plano parab\u00f3lico horizontal, 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 disparo parab\u00f3lico horizontal 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-7c778eda7ac01cd4c754ec97d99f3b28_l3.png\" class=\"ql-img-inline-formula quicklatex-auto-format\" alt=\"\\begin{cases}x=v_0\\cdot t \\\\[2ex]y=h -\\cfrac{1}{2}\\cdot g\\cdot t^2\\end{cases}\" title=\"Rendered by QuickLaTeX.com\" height=\"86\" width=\"136\" 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-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 disparo parab\u00f3lico horizontal, a componente horizontal da velocidade \u00e9 constante ao longo da trajet\u00f3ria e equivale ao valor da velocidade inicial.<\/p>\n<p> Por outro lado, a componente vertical de um disparo parab\u00f3lico horizontal \u00e9 definida pela equa\u00e7\u00e3o do movimento retil\u00edneo uniformemente acelerado. Portanto, a componente vertical da velocidade \u00e9 igual a 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-923c94b2f351c8a83b26edb8bac9164b_l3.png\" class=\"ql-img-inline-formula quicklatex-auto-format\" alt=\"\\begin{cases}v_x=v_0 \\\\[2ex]v_y=-g\\cdot t\\end{cases}\" title=\"Rendered by QuickLaTeX.com\" height=\"65\" width=\"95\" 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-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 planos parab\u00f3licos horizontais, 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 (pois \u00e9 uma acelera\u00e7\u00e3o negativa).<\/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 que leva para o corpo fazer o disparo parab\u00f3lico horizontal tocar 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> Assim, a f\u00f3rmula para calcular o tempo de voo de um disparo parab\u00f3lico horizontal \u00e9 a seguinte:<\/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-319194c7a708eb2fa52ef3b34624a5ff_l3.png\" class=\"ql-img-inline-formula quicklatex-auto-format\" alt=\"\\displaystyle t_{vol}=\\sqrt{\\frac{2h}{g}}\" title=\"Rendered by QuickLaTeX.com\" height=\"54\" width=\"90\" style=\"vertical-align: -21px;\"><\/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-bd24a64e7d49c4ad9d16a92a468d51f2_l3.png\" class=\"ql-img-inline-formula quicklatex-auto-format\" alt=\"t_{flight}\" title=\"Rendered by QuickLaTeX.com\" height=\"18\" width=\"43\" style=\"vertical-align: -6px;\"><\/p>\n<p> \u00e9 o tempo de v\u00f4o. <\/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<div class=\"wp-block-otfm-box-spoiler-start otfm-sp__wrapper otfm-sp__box js-otfm-sp-box__closed otfm-sp__FFF8E1\" role=\"button\" tabindex=\"0\" aria-expanded=\"false\" data-otfm-spc=\"#FFF8E1\" style=\"text-align:center\">\n<div class=\"otfm-sp__title\"> <strong>Veja a demonstra\u00e7\u00e3o da f\u00f3rmula<\/strong><\/div>\n<\/div>\n<p class=\"has-text-align-left\"> 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 para a posi\u00e7\u00e3o vertical do tiro parab\u00f3lico horizontal igual a zero e, em seguida, 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-2ae71e3ad8efaa87d2cf99e31908530f_l3.png\" class=\"ql-img-inline-formula quicklatex-auto-format\" alt=\"y=h -\\cfrac{1}{2}\\cdot g\\cdot t^2\" title=\"Rendered by QuickLaTeX.com\" height=\"38\" width=\"123\" 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-7854e3c3d9a1c8173b868c63ce4bcb85_l3.png\" class=\"ql-img-inline-formula quicklatex-auto-format\" alt=\"0=h -\\cfrac{1}{2}\\cdot g\\cdot t_{vol}^2\" title=\"Rendered by QuickLaTeX.com\" height=\"38\" width=\"134\" 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-32dd191b224eb6a2340a95a1d2a57ded_l3.png\" class=\"ql-img-inline-formula quicklatex-auto-format\" alt=\"\\cfrac{1}{2}\\cdot g\\cdot t_{vol}^2=h\" title=\"Rendered by QuickLaTeX.com\" height=\"38\" width=\"102\" 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-adef29e16bcc19813ea6a7034698027c_l3.png\" class=\"ql-img-inline-formula quicklatex-auto-format\" alt=\"t_{vol}^2=\\cfrac{2h}{g}\" title=\"Rendered by QuickLaTeX.com\" height=\"42\" width=\"71\" style=\"vertical-align: -16px;\"><\/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-319194c7a708eb2fa52ef3b34624a5ff_l3.png\" class=\"ql-img-inline-formula quicklatex-auto-format\" alt=\"\\displaystyle t_{vol}=\\sqrt{\\frac{2h}{g}}\" title=\"Rendered by QuickLaTeX.com\" height=\"54\" width=\"90\" style=\"vertical-align: -21px;\"><\/p>\n<\/p>\n<div class=\"wp-block-otfm-box-spoiler-end otfm-sp_end\"><\/div>\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 posteriormente substituir o valor do tempo de voo na equa\u00e7\u00e3o da posi\u00e7\u00e3o horizontal do tiro parab\u00f3lico horizontal.<\/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-3daa5420bc4156b383762b58a0e8cc79_l3.png\" class=\"ql-img-inline-formula quicklatex-auto-format\" alt=\" x_{m\\'ax}=v_0\\cdot t_{vol}\" title=\"Rendered by QuickLaTeX.com\" height=\"15\" width=\"115\" style=\"vertical-align: -3px;\"><\/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-96befedb14fc2622547c54ae1f9fc906_l3.png\" class=\"ql-img-inline-formula quicklatex-auto-format\" alt=\"x_{m\\'ax}\" title=\"Rendered by QuickLaTeX.com\" height=\"11\" width=\"38\" style=\"vertical-align: -3px;\"><\/p>\n<p> \u00e9 o alcance horizontal m\u00e1ximo. <\/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-bd24a64e7d49c4ad9d16a92a468d51f2_l3.png\" class=\"ql-img-inline-formula quicklatex-auto-format\" alt=\"t_{flight}\" title=\"Rendered by QuickLaTeX.com\" height=\"18\" width=\"43\" style=\"vertical-align: -6px;\"><\/p>\n<p> \u00e9 o tempo de v\u00f4o. <\/span><\/li>\n<\/ul>\n<h3 class=\"wp-block-heading\"><span class=\"ez-toc-section\" id=\"Resumen-de-las-formulas-del-tiro-parabolico-horizontal\"><\/span> Resumo das f\u00f3rmulas de rascunho parab\u00f3lico horizontal<span class=\"ez-toc-section-end\"><\/span><\/h3>\n<p> Em resumo, deixamos uma tabela com todas as f\u00f3rmulas para o tiro parab\u00f3lico horizontal: <\/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-paraboliques-horizontales.png\" alt=\"f\u00f3rmulas de proje\u00e7\u00e3o parab\u00f3lica horizontal\" class=\"wp-image-8396\" width=\"494\" height=\"400\" srcset=\"https:\/\/physigeek.com\/wp-content\/uploads\/2023\/09\/formules-paraboliques-horizontales-300x244.png 300w, https:\/\/physigeek.com\/wp-content\/uploads\/2023\/09\/formules-paraboliques-horizontales-768x624.png 768w, https:\/\/physigeek.com\/wp-content\/uploads\/2023\/09\/formules-paraboliques-horizontales.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-tiro-parabolico-horizontal\"><\/span> Exerc\u00edcio de tiro parab\u00f3lico horizontal resolvido<span class=\"ez-toc-section-end\"><\/span><\/h2>\n<p> Para assimilar melhor os conceitos explicados, voc\u00ea encontrar\u00e1 abaixo um passo a passo do exerc\u00edcio de tiro parab\u00f3lico horizontal.<\/p>\n<ul>\n<li> Uma bola \u00e9 lan\u00e7ada horizontalmente de uma altura de 8 metros com velocidade inicial de 6 m\/s. Calcule o seguinte desprezando o atrito do ar em todo o problema e aproximando o valor da gravidade em 10 m\/s <sup>2<\/sup> .\n<ol>\n<li> O momento em que a bola est\u00e1 no ar.<\/li>\n<li> A dist\u00e2ncia horizontal que a bola percorre at\u00e9 atingir o solo.<\/li>\n<li> A magnitude da velocidade com que a bola atinge o solo.<\/li>\n<\/ol>\n<\/li>\n<\/ul>\n<p> Para saber o hor\u00e1rio do voo, basta aplicar a f\u00f3rmula que vimos acima:<\/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-1ac5d065ded030390562ce814b72dd1e_l3.png\" class=\"ql-img-inline-formula quicklatex-auto-format\" alt=\"\\begin{aligned}\\displaystyle t_{vol}&amp;=\\sqrt{\\frac{2h}{g}}\\\\[2ex]t_{vol}&amp;=\\sqrt{\\frac{2\\cdot 8} {10}}\\\\[2ex]t_{vol}&amp;=1,26 \\ s\\end{aligned}\" title=\"Rendered by QuickLaTeX.com\" height=\"159\" width=\"102\" style=\"vertical-align: 0px;\"><\/p>\n<\/p>\n<p> Uma vez conhecido o tempo de voo, podemos determinar o alcance horizontal m\u00e1ximo substituindo o valor do tempo de voo na equa\u00e7\u00e3o pela componente horizontal da posi\u00e7\u00e3o.<\/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-866c53030ef8b4738e3fa18a4553bcdf_l3.png\" class=\"ql-img-inline-formula quicklatex-auto-format\" alt=\"\\begin{aligned}x_{m\\'ax}&amp;=v_0\\cdot t_{vol}\\\\[2ex]x_{m\\'ax}&amp;=6\\cdot 1.26\\\\[2ex]x_ {m \\'ax}&amp;=7.56 \\ m\\end{aligned}\" title=\"Rendered by QuickLaTeX.com\" height=\"99\" width=\"115\" style=\"vertical-align: 0px;\"><\/p>\n<\/p>\n<p> Para calcular a velocidade final, precisamos determinar a sua componente horizontal e a sua componente vertical no \u00faltimo instante. A componente horizontal \u00e9 constante ao longo da trajet\u00f3ria e constitui o valor da velocidade inicial.<\/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-80721b6f97cfa44a795ec3e11c336bf7_l3.png\" class=\"ql-img-inline-formula quicklatex-auto-format\" alt=\"v_x=v_0=6 \\ \\cfrac{m}{s}\" title=\"Rendered by QuickLaTeX.com\" height=\"34\" width=\"113\" style=\"vertical-align: -12px;\"><\/p>\n<\/p>\n<p> Por outro lado, para encontrar a componente vertical da velocidade, aplicamos a sua equa\u00e7\u00e3o 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-1a07841e4f02139417f0a340a02d5738_l3.png\" class=\"ql-img-inline-formula quicklatex-auto-format\" alt=\"\\begin{aligned}v_{y_f}&amp;=-g\\cdot t_{flight}\\\\[2ex]v_{y_f}&amp;=-10\\cdot 1.26\\\\[2ex]v_{y_f}&amp; =- 12.6 \\ \\cfrac{m}{s}\\end{aligned}\" title=\"Rendered by QuickLaTeX.com\" height=\"125\" width=\"127\" style=\"vertical-align: 0px;\"><\/p>\n<\/p>\n<p> Assim, o m\u00f3dulo da velocidade \u00e9 igual \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-36f3b2b6bf5f5864c1bb969f89eca196_l3.png\" class=\"ql-img-inline-formula quicklatex-auto-format\" alt=\"\\begin{aligned}|v_f|&amp;=\\sqrt{v_x^2+v_{y_f}^2}\\\\[2ex]|v_f|&amp;=\\sqrt{6^2+(-12,6) ^2}\\\\[2ex]|v_f|&amp;=13.96 \\ \\cfrac{m}{s}\\end{aligned}\" title=\"Rendered by QuickLaTeX.com\" height=\"141\" width=\"175\" style=\"vertical-align: 0px;\"><\/p>\n<\/p>\n<h2 class=\"wp-block-heading\"><span class=\"ez-toc-section\" id=\"Tiro-parabolico-horizontal-y-tiro-parabolico-oblicuo\"><\/span> Tiro parab\u00f3lico horizontal e tiro parab\u00f3lico obl\u00edquo<span class=\"ez-toc-section-end\"><\/span><\/h2>\n<p> Por fim, vamos ver qual a diferen\u00e7a entre o tiro parab\u00f3lico horizontal e o tiro parab\u00f3lico obl\u00edquo, pois s\u00e3o dois tipos de movimentos parab\u00f3licos que podem ser confundidos.<\/p>\n<p> O <strong>lan\u00e7amento parab\u00f3lico obl\u00edquo<\/strong> \u00e9 o movimento realizado por um corpo que primeiro sobe e depois desce enquanto avan\u00e7a horizontalmente. Em outras palavras, a trajet\u00f3ria de um plano parab\u00f3lico obl\u00edquo \u00e9 uma par\u00e1bola completa.<\/p>\n<p> A <strong>diferen\u00e7a entre o tiro parab\u00f3lico horizontal e o tiro parab\u00f3lico obl\u00edquo<\/strong> \u00e9 a velocidade inicial. Em um tiro parab\u00f3lico horizontal, a velocidade inicial \u00e9 horizontal, por\u00e9m, em um tiro parab\u00f3lico obl\u00edquo, a velocidade inicial forma um \u00e2ngulo positivo com o eixo horizontal.<\/p>\n<p> Assim, a trajet\u00f3ria de um disparo parab\u00f3lico horizontal come\u00e7a completamente horizontalmente, enquanto a trajet\u00f3ria de um disparo parab\u00f3lico obl\u00edquo come\u00e7a em um \u00e2ngulo com o eixo horizontal, uma vez que a velocidade inicial tem uma componente horizontal e uma vertical.<\/p>\n<p> Al\u00e9m disso, se o disparo parab\u00f3lico obl\u00edquo come\u00e7ar no solo, o disparo parab\u00f3lico horizontal come\u00e7ar\u00e1 no meio do disparo parab\u00f3lico obl\u00edquo. Portanto, o alcance m\u00e1ximo e o tempo de voo do disparo parab\u00f3lico horizontal s\u00e3o metade do alcance m\u00e1ximo e do tempo de voo do disparo parab\u00f3lico obl\u00edquo. <\/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\/tir-parabolique-horizontal-et-tir-parabolique-oblique.png\" alt=\"lan\u00e7amento parab\u00f3lico horizontal e lan\u00e7amento parab\u00f3lico obl\u00edquo\" class=\"wp-image-8404\" width=\"390\" height=\"404\" srcset=\"https:\/\/physigeek.com\/wp-content\/uploads\/2023\/09\/tir-parabolique-horizontal-et-tir-parabolique-oblique-289x300.png 289w, https:\/\/physigeek.com\/wp-content\/uploads\/2023\/09\/tir-parabolique-horizontal-et-tir-parabolique-oblique.png 622w\" sizes=\"auto, (max-width: 289px) 100vw, 289px\"><\/figure>\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\/movimento-parabolico-ou-lancamento-parabolico\/\">Tiro parab\u00f3lico obl\u00edquo<\/a><\/div>\n","protected":false},"excerpt":{"rendered":"<p>Este artigo explica o que \u00e9 o lan\u00e7amento parab\u00f3lico horizontal, tamb\u00e9m chamado de lan\u00e7amento horizontal ou lan\u00e7amento horizontal, na f\u00edsica e quais s\u00e3o suas caracter\u00edsticas. Al\u00e9m disso, voc\u00ea encontrar\u00e1 as f\u00f3rmulas para o disparo parab\u00f3lico horizontal, bem como um exemplo concreto passo a passo. O que \u00e9 calado parab\u00f3lico horizontal? O lan\u00e7amento parab\u00f3lico horizontal , &hellip;<\/p>\n<p class=\"read-more\"> <a class=\"\" href=\"https:\/\/physigeek.com\/pt\/plano-parabolico-horizontal\/\"> <span class=\"screen-reader-text\">Tiro parab\u00f3lico horizontal<\/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-429","post","type-post","status-publish","format-standard","hentry","category-cinematografico"],"yoast_head":"<!-- This site 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