{"id":423,"date":"2023-06-18T17:54:24","date_gmt":"2023-06-18T17:54:24","guid":{"rendered":"https:\/\/physigeek.com\/pt\/tiro-vertical\/"},"modified":"2023-06-18T17:54:24","modified_gmt":"2023-06-18T17:54:24","slug":"tiro-vertical","status":"publish","type":"post","link":"https:\/\/physigeek.com\/pt\/tiro-vertical\/","title":{"rendered":"Tiro vertical"},"content":{"rendered":"<p>Este artigo explica o que \u00e9 tiro vertical em f\u00edsica. Assim, voc\u00ea encontrar\u00e1 as caracter\u00edsticas do plano vertical, os tipos de planos verticais, suas equa\u00e7\u00f5es e um exemplo trabalhado passo a passo. <\/p>\n<h2 class=\"wp-block-heading\"><span class=\"ez-toc-section\" id=\"%C2%BFQue-es-el-tiro-vertical\"><\/span>O que \u00e9 tiro vertical?<span class=\"ez-toc-section-end\"><\/span><\/h2>\n<p> <strong>O lan\u00e7amento vertical<\/strong> \u00e9 um movimento causado pelo lan\u00e7amento de um corpo verticalmente. Ou seja, em f\u00edsica, o arremesso vertical \u00e9 um movimento em que o corpo se move apenas verticalmente, seja jogando o corpo para cima (lan\u00e7amento vertical para cima) ou para baixo (lan\u00e7amento vertical para baixo).<\/p>\n<p> Por exemplo, lan\u00e7ar uma bola verticalmente para cima \u00e9 um lan\u00e7amento vertical. A bola ir\u00e1 primeiro subir verticalmente devido \u00e0 for\u00e7a aplicada a ela e depois ir\u00e1 descer verticalmente at\u00e9 atingir o solo sob a influ\u00eancia da gravidade. <\/p>\n<h2 class=\"wp-block-heading\"><span class=\"ez-toc-section\" id=\"Caracteristicas-del-tiro-vertical\"><\/span> Recursos de tiro vertical<span class=\"ez-toc-section-end\"><\/span><\/h2>\n<p> Agora que conhecemos a defini\u00e7\u00e3o de tiro vertical na f\u00edsica, vamos ver quais s\u00e3o as caracter\u00edsticas desse tipo de movimento para entender melhor o conceito.<\/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 tiro vertical<\/strong> \u00e9 que a trajet\u00f3ria do corpo \u00e9 totalmente vertical. Assim, o corpo que d\u00e1 o tiro vertical se move ao longo de uma linha reta vertical.<\/span><\/li>\n<li style=\"margin-bottom:20px\"> <span style=\"color:#101010;font-weight: normal;\">Outra caracter\u00edstica do disparo vertical \u00e9 que o atrito com o ar \u00e9 desprezado. Assim, na f\u00edsica, em um plano vertical, qualquer tipo de atrito ou obst\u00e1culo que possa existir no caminho do movimento \u00e9 ignorado.<\/span><\/li>\n<li style=\"margin-bottom:20px\"> <span style=\"color:#101010;font-weight: normal;\">Assim, num plano vertical, apenas a gravidade afeta o movimento do corpo em movimento. Em outras palavras, a \u00fanica for\u00e7a que atua sobre o corpo \u00e9 a for\u00e7a da gravidade.<\/span><\/li>\n<li style=\"margin-bottom:20px\"> <span style=\"color:#101010;font-weight: normal;\">Portanto, o calado vertical \u00e9 o movimento retil\u00edneo uniformemente acelerado (MRUA), tamb\u00e9m chamado <a href=\"https:\/\/physigeek.com\/pt\/o-movimento-retilineo-acelera-uniformemente-mrua\/\">de movimento retil\u00edneo uniformemente variado (MRUV)<\/a> .<\/span> <\/li>\n<\/ul>\n<h2 class=\"wp-block-heading\"><span class=\"ez-toc-section\" id=\"Tipos-de-tiro-vertical\"><\/span> Tipos de disparo vertical<span class=\"ez-toc-section-end\"><\/span><\/h2>\n<p> Existem dois tipos de disparo vertical:<\/p>\n<ul style=\"color:#4fd12f; font-weight: bold;\">\n<li style=\"margin-bottom:20px\"> <span style=\"color:#101010;font-weight: normal;\"><strong>Tiro vertical para cima:<\/strong> Um tiro vertical em que o corpo \u00e9 jogado para cima e depois abaixado. Por exemplo: lan\u00e7ar uma bola verticalmente para cima.<\/span><\/li>\n<li style=\"margin-bottom:20px\"> <span style=\"color:#101010;font-weight: normal;\"><strong>Tiro vertical para baixo:<\/strong> Tiro vertical em que o corpo \u00e9 lan\u00e7ado para baixo, de forma que o corpo n\u00e3o suba em nenhum ponto, mas des\u00e7a at\u00e9 atingir o solo. Por exemplo: lan\u00e7ar verticalmente um objeto contra o ch\u00e3o.<\/span> <\/li>\n<\/ul>\n<div class=\"wp-block-columns is-layout-flex wp-container-3 wp-block-columns-is-layout-flex\">\n<div class=\"wp-block-column is-vertically-aligned-bottom is-layout-flow wp-block-column-is-layout-flow\">\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-vertical-vers-le-haut.png\" alt=\"tiro vertical para cima\" class=\"wp-image-8281\" width=\"250\" height=\"384\" srcset=\"https:\/\/physigeek.com\/wp-content\/uploads\/2023\/09\/tir-vertical-vers-le-haut-195x300.png 195w, https:\/\/physigeek.com\/wp-content\/uploads\/2023\/09\/tir-vertical-vers-le-haut.png 416w\" sizes=\"auto, (max-width: 195px) 100vw, 195px\"><\/figure>\n<\/div>\n<div class=\"wp-block-column is-vertically-aligned-bottom is-layout-flow wp-block-column-is-layout-flow\">\n<figure class=\"wp-block-image size-full is-resized\"><img decoding=\"async\" loading=\"lazy\" src=\"https:\/\/physigeek.com\/wp-content\/uploads\/2023\/09\/evacuation-verticale.png\" alt=\"vertical abatido\" class=\"wp-image-8254\" width=\"240\" height=\"301\" srcset=\"https:\/\/physigeek.com\/wp-content\/uploads\/2023\/09\/evacuation-verticale-239x300.png 239w, https:\/\/physigeek.com\/wp-content\/uploads\/2023\/09\/evacuation-verticale.png 420w\" sizes=\"auto, (max-width: 239px) 100vw, 239px\"><\/figure>\n<\/div>\n<\/div>\n<p> Observe que o tiro vertical pode come\u00e7ar no solo ou, como nos exemplos anteriores, em uma altura diferente do solo. <\/p>\n<h2 class=\"wp-block-heading\"><span class=\"ez-toc-section\" id=\"Formulas-del-tiro-vertical\"><\/span> F\u00f3rmulas de tiro vertical<span class=\"ez-toc-section-end\"><\/span><\/h2>\n<p> Abaixo est\u00e3o as f\u00f3rmulas (ou equa\u00e7\u00f5es) para disparo vertical. Estas f\u00f3rmulas ser\u00e3o, portanto, \u00fateis para resolver problemas de disparo vertical.<\/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> Um corpo que executa um tiro vertical descreve um movimento retil\u00edneo uniformemente acelerado (MRUA). Assim, a f\u00f3rmula que permitir\u00e1 calcular a posi\u00e7\u00e3o vertical de um corpo durante um lan\u00e7amento vertical \u00e9 deduzida da f\u00f3rmula da posi\u00e7\u00e3o de um MRUA:<\/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-f092aa5505874d9399136ba407bc6248_l3.png\" class=\"ql-img-inline-formula quicklatex-auto-format\" alt=\"y=H+v_0\\cdot (t-t_0)-\\cfrac{1}{2}\\cdot g \\cdot (t-t_0)^2\" title=\"Rendered by QuickLaTeX.com\" height=\"38\" width=\"286\" style=\"vertical-align: -12px;\"><\/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-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 altura do corpo fazendo o tiro vertical. <\/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-a7cedbc00aa5531f310166df85e3a9bb_l3.png\" class=\"ql-img-inline-formula quicklatex-auto-format\" alt=\"H\" title=\"Rendered by QuickLaTeX.com\" height=\"12\" width=\"16\" style=\"vertical-align: 0px;\"><\/p>\n<p> \u00e9 a altura a partir da qual o corpo \u00e9 projetado. <\/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 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-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 instante em que a posi\u00e7\u00e3o do corpo \u00e9 calculada. <\/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-c1a3f4a217f20d31e6f72a2f42a2e7dd_l3.png\" class=\"ql-img-inline-formula quicklatex-auto-format\" alt=\"t_0\" title=\"Rendered by QuickLaTeX.com\" height=\"15\" width=\"13\" style=\"vertical-align: -3px;\"><\/p>\n<p> \u00e9 o momento 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-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 na Terra \u00e9 9,81 m\/s <sup>2<\/sup> .<\/span><\/li>\n<\/ul>\n<p> <strong>Nota:<\/strong> Tenha em mente que a origem coordenada deste referencial \u00e9 o solo. Ent\u00e3o o corpo ir\u00e1 colidir com o solo na posi\u00e7\u00e3o y=0, da mesma forma, se for um tiro vertical para cima, a velocidade inicial ser\u00e1 positiva (o corpo est\u00e1 subindo), mas se for um tiro vertical para baixo, o a velocidade inicial ser\u00e1 negativa. (o corpo desce).<\/p>\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> Num lan\u00e7amento vertical n\u00e3o h\u00e1 atrito, portanto o m\u00f3bile descreve um movimento retil\u00edneo uniformemente acelerado e, portanto, a f\u00f3rmula que nos permitir\u00e1 encontrar a velocidade em qualquer momento \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-2b86a879f38e8f8da7c676a34ef04aba_l3.png\" class=\"ql-img-inline-formula quicklatex-auto-format\" alt=\"v=v_0-g\\cdot (t-t_0)\" title=\"Rendered by QuickLaTeX.com\" height=\"19\" width=\"148\" style=\"vertical-align: -5px;\"><\/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-796872219106704832bd95ce08640b7b_l3.png\" class=\"ql-img-inline-formula quicklatex-auto-format\" alt=\"v\" title=\"Rendered by QuickLaTeX.com\" height=\"8\" width=\"9\" style=\"vertical-align: 0px;\"><\/p>\n<p> \u00e9 a velocidade do corpo em um determinado momento. <\/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 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 na Terra \u00e9 9,81 m\/s <sup>2<\/sup> . <\/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 instante em que a velocidade do corpo \u00e9 calculada. <\/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-c1a3f4a217f20d31e6f72a2f42a2e7dd_l3.png\" class=\"ql-img-inline-formula quicklatex-auto-format\" alt=\"t_0\" title=\"Rendered by QuickLaTeX.com\" height=\"15\" width=\"13\" style=\"vertical-align: -3px;\"><\/p>\n<p> \u00e9 o momento inicial.<\/span><\/li>\n<\/ul>\n<p> <strong>Nota:<\/strong> Tenha em mente que quando o corpo acelera tem sinal positivo, enquanto quando o corpo desacelera tem sinal negativo. Portanto, em um tiro vertical para cima a velocidade inicial ser\u00e1 positiva, por\u00e9m, em um tiro vertical descendente a velocidade inicial ser\u00e1 para baixo.<\/p>\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> Num tiro vertical, qualquer tipo de atrito \u00e9 desprezado, a \u00fanica for\u00e7a que interv\u00e9m \u00e9 a for\u00e7a gravitacional. Portanto, a acelera\u00e7\u00e3o do corpo \u00e9 constante ao longo da trajet\u00f3ria e seu valor \u00e9 o da gravidade.<\/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-e3282a7b306fa53bfdfa002aec864752_l3.png\" class=\"ql-img-inline-formula quicklatex-auto-format\" alt=\"a=-g\" title=\"Rendered by QuickLaTeX.com\" height=\"12\" width=\"56\" style=\"vertical-align: -4px;\"><\/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-0e55b0b3943237ccfc96979505679274_l3.png\" class=\"ql-img-inline-formula quicklatex-auto-format\" alt=\"a\" title=\"Rendered by QuickLaTeX.com\" height=\"8\" width=\"9\" style=\"vertical-align: 0px;\"><\/p>\n<p> \u00e9 a acelera\u00e7\u00e3o 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 na Terra \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 o corpo leva para o corpo fazer o disparo vertical tocar o solo. Simplificando, o tempo de voo \u00e9 o tempo entre o momento em que o corpo inicia o lan\u00e7amento vertical e o momento em que atinge o solo.<\/p>\n<p> Quando o corpo atingir o solo, sua posi\u00e7\u00e3o vertical 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 igual a zero e ent\u00e3o resolver o tempo a partir da equa\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-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=\"Altura-maxima\"><\/span> Altura m\u00e1xima<span class=\"ez-toc-section-end\"><\/span><\/h3>\n<p> Se for um tiro vertical para baixo, logicamente a altura m\u00e1xima ser\u00e1 a altura inicial. Por\u00e9m, em um disparo vertical para cima, a altura m\u00e1xima \u00e9 atingida quando a velocidade do corpo \u00e9 zero.<\/p>\n<p> Assim, para determinar a altura m\u00e1xima em um tiro vertical ascendente, a velocidade deve ser igual a zero, a partir da\u00ed encontraremos o instante em que a altura m\u00e1xima \u00e9 atingida e, a seguir, substituiremos o instante de tempo calculado na equa\u00e7\u00e3o de 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-52816ebc6396cf669207130cd1baef10_l3.png\" class=\"ql-img-inline-formula quicklatex-auto-format\" alt=\"v=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=\"492\" style=\"vertical-align: -6px;\"><\/p>\n<\/p>\n<h3 class=\"wp-block-heading\"><span class=\"ez-toc-section\" id=\"Resumen-de-las-formulas-del-tiro-vertical\"><\/span> Resumo das f\u00f3rmulas de tiro vertical<span class=\"ez-toc-section-end\"><\/span><\/h3>\n<p> Deixamos ent\u00e3o uma tabela com todas as f\u00f3rmulas de tiro vertical como resumo: <\/p>\n<figure class=\"wp-block-image aligncenter size-large is-resized\"><img decoding=\"async\" loading=\"lazy\" src=\"https:\/\/physigeek.com\/wp-content\/uploads\/2023\/09\/formules-de-tir-vertical.png\" alt=\"f\u00f3rmulas de rascunho vertical, equa\u00e7\u00f5es de rascunho vertical\" class=\"wp-image-8268\" width=\"496\" height=\"323\" srcset=\"https:\/\/physigeek.com\/wp-content\/uploads\/2023\/09\/formules-de-tir-vertical-300x196.png 300w, https:\/\/physigeek.com\/wp-content\/uploads\/2023\/09\/formules-de-tir-vertical-1024x670.png 1024w, https:\/\/physigeek.com\/wp-content\/uploads\/2023\/09\/formules-de-tir-vertical-768x503.png 768w, https:\/\/physigeek.com\/wp-content\/uploads\/2023\/09\/formules-de-tir-vertical.png 1027w\" sizes=\"auto, (max-width: 300px) 100vw, 300px\"><\/figure>\n<h2 class=\"wp-block-heading\"><span class=\"ez-toc-section\" id=\"Ejercicio-resuelto-del-tiro-vertical\"><\/span> Exerc\u00edcio de tiro vertical resolvido<span class=\"ez-toc-section-end\"><\/span><\/h2>\n<ul>\n<li> De uma altura de 7 m, um objeto \u00e9 lan\u00e7ado verticalmente para cima com uma velocidade inicial de 12 m\/s, de modo que o corpo descreve um lan\u00e7amento vertical para cima. Qual \u00e9 a altura m\u00e1xima atingida pelo corpo? Aproxime o valor da gravidade para 10 m\/s <sup>2<\/sup> .<\/li>\n<\/ul>\n<p> Por se tratar de um tiro vertical ascendente, a altura m\u00e1xima ser\u00e1 atingida quando a velocidade for zero. Assim, podemos encontrar o tempo durante o qual a altura m\u00e1xima \u00e9 produzida igualando a equa\u00e7\u00e3o da velocidade a zero: <\/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-4bf371d451315ad9d990fa089f1ec27c_l3.png\" class=\"ql-img-inline-formula quicklatex-auto-format\" alt=\"\\begin{array}{c} v=v_0-g\\cdot (t-t_0)\\\\[2ex]0=12-10\\cdot (t-0)\\\\[2ex]0=12-10t \\\\[2ex]10t=12\\\\[2ex]t=\\cfrac{12}{10}\\\\[2ex]t=1,2 \\ \\cfrac{m}{s}\\end{array}[\/latex ] Et une fois que l'on conna\u00eet le temps pendant lequel la hauteur maximale est atteinte, il suffit de substituer cet instant dans l'\u00e9quation de la position du tir vertical : [latex]\\begin{aligned}y&amp;=H+v_0\\cdot (t-t_0)-\\cfrac{1}{2}\\cdot g \\cdot (t-t_0)^2\\\\[2ex]y&amp;=7+ 12\\cdot (1,2-0)-\\cfrac{1}{2}\\cdot 10 \\cdot (1,2-0)^2 \\\\[2ex]y&amp;=14,2 \\ m\\end{aligned} \" title=\"Rendered by QuickLaTeX.com\" height=\"404\" width=\"709\" style=\"vertical-align: 0px;\"><\/p>\n<\/p>\n<h2 class=\"wp-block-heading\"><span class=\"ez-toc-section\" id=\"Tiro-vertical-y-caida-libre\"><\/span> Tiro vertical e queda livre<span class=\"ez-toc-section-end\"><\/span><\/h2>\n<p> Por fim, vamos ver qual \u00e9 a diferen\u00e7a entre o tiro vertical e a queda livre, pois s\u00e3o dois tipos de movimentos muito semelhantes.<\/p>\n<p> Na f\u00edsica, queda livre \u00e9 um movimento que envolve deixar cair um corpo de uma altura sem aplicar qualquer for\u00e7a. Para que o corpo des\u00e7a verticalmente em linha reta sob o efeito da gravidade, desprezando assim o atrito com o ar.<\/p>\n<p> A <strong>diferen\u00e7a entre o lan\u00e7amento vertical e a queda livre<\/strong> \u00e9 que no lan\u00e7amento vertical o corpo tem uma velocidade inicial, enquanto na queda livre geralmente n\u00e3o h\u00e1 velocidade inicial.<\/p>\n<p> Al\u00e9m disso, ao arremessar verticalmente o corpo pode subir, enquanto na queda livre o corpo sempre desce. <\/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\/queda-livre\/\">Caracter\u00edsticas da queda livre<\/a><\/div>\n","protected":false},"excerpt":{"rendered":"<p>Este artigo explica o que \u00e9 tiro vertical em f\u00edsica. Assim, voc\u00ea encontrar\u00e1 as caracter\u00edsticas do plano vertical, os tipos de planos verticais, suas equa\u00e7\u00f5es e um exemplo trabalhado passo a passo. O que \u00e9 tiro vertical? O lan\u00e7amento vertical \u00e9 um movimento causado pelo lan\u00e7amento de um corpo verticalmente. Ou seja, em f\u00edsica, o &hellip;<\/p>\n<p class=\"read-more\"> <a class=\"\" href=\"https:\/\/physigeek.com\/pt\/tiro-vertical\/\"> <span class=\"screen-reader-text\">Tiro vertical<\/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-423","post","type-post","status-publish","format-standard","hentry","category-cinematografico"],"yoast_head":"<!-- This site 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