{"id":425,"date":"2023-06-18T15:54:24","date_gmt":"2023-06-18T15:54:24","guid":{"rendered":"https:\/\/physigeek.com\/pt\/queda-livre\/"},"modified":"2023-06-18T15:54:24","modified_gmt":"2023-06-18T15:54:24","slug":"queda-livre","status":"publish","type":"post","link":"https:\/\/physigeek.com\/pt\/queda-livre\/","title":{"rendered":"Queda livre"},"content":{"rendered":"<p>Este artigo explica o que \u00e9 queda livre na f\u00edsica. Voc\u00ea tamb\u00e9m encontrar\u00e1 as f\u00f3rmulas para queda livre e, al\u00e9m disso, um exemplo resolvido de um corpo caindo em queda livre. <\/p>\n<h2 class=\"wp-block-heading\"><span class=\"ez-toc-section\" id=\"%C2%BFQue-es-la-caida-libre\"><\/span> O que \u00e9 queda livre?<span class=\"ez-toc-section-end\"><\/span><\/h2>\n<p> Na f\u00edsica, <strong>a queda livre<\/strong> \u00e9 um tipo de movimento retil\u00edneo no qual um corpo cai verticalmente de uma certa altura apenas sob a influ\u00eancia da gravidade. Em outras palavras, em queda livre, um corpo cai em linha reta e o atrito com o ar ou qualquer outro obst\u00e1culo \u00e9 desprezado.<\/p>\n<p> A principal caracter\u00edstica da queda livre \u00e9, portanto, que o atrito (ou atrito) com o ar n\u00e3o \u00e9 levado em considera\u00e7\u00e3o. Assim, a \u00fanica for\u00e7a envolvida no movimento de queda livre \u00e9 a for\u00e7a da gravidade.<\/p>\n<p> Por exemplo, se deixarmos cair uma bola do telhado de um edif\u00edcio e ignorarmos o atrito da bola com o ar, ela executa um movimento de queda livre porque a \u00fanica for\u00e7a que atua sobre ela \u00e9 a for\u00e7a 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\/chute-libre.png\" alt=\"queda livre\" class=\"wp-image-8151\" width=\"332\" height=\"291\" srcset=\"https:\/\/physigeek.com\/wp-content\/uploads\/2023\/09\/chute-libre-300x262.png 300w, https:\/\/physigeek.com\/wp-content\/uploads\/2023\/09\/chute-libre.png 493w\" sizes=\"auto, (max-width: 300px) 100vw, 300px\"><\/figure>\n<h2 class=\"wp-block-heading\"><span class=\"ez-toc-section\" id=\"Formulas-de-la-caida-libre\"><\/span> F\u00f3rmulas de queda livre<span class=\"ez-toc-section-end\"><\/span><\/h2>\n<p> Abaixo voc\u00ea pode ver quais s\u00e3o todas as f\u00f3rmulas para o movimento de queda livre. Essas f\u00f3rmulas permitir\u00e3o resolver problemas de queda livre de um corpo.<\/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 cai livremente sem qualquer atrito realiza movimento retil\u00edneo uniformemente acelerado (MRUA). Assim, a f\u00f3rmula que permitir\u00e1 calcular a posi\u00e7\u00e3o de um corpo em queda livre \u00e9 deduzida da f\u00f3rmula da posi\u00e7\u00e3o de um <a href=\"https:\/\/physigeek.com\/pt\/o-movimento-retilineo-acelera-uniformemente-mrua\/\">MRUA<\/a> :<\/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-89b3f9bb6bf789f11429d59533f96622_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 em queda livre. <\/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 para a 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> Observe que a origem das coordenadas deste sistema de refer\u00eancia \u00e9 o solo. Em outras palavras, o corpo colidir\u00e1 com o solo na posi\u00e7\u00e3o y=0.<\/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> Um corpo que cai livremente sem qualquer atrito descreve um movimento retil\u00edneo uniformemente acelerado, 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 de tempo durante o qual 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 um corpo em queda livre est\u00e1 descendo, ent\u00e3o a velocidade tem sinal negativo. Portanto, quanto mais baixo estiver o corpo, mais negativa ser\u00e1 sua velocidade.<\/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> Durante uma queda livre, todo atrito \u00e9 desprezado, a \u00fanica for\u00e7a que interv\u00e9m \u00e9 a for\u00e7a gravitacional. Portanto, a acelera\u00e7\u00e3o \u00e9 constante ao longo da trajet\u00f3ria.<\/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=\"Resumen-de-las-formulas-de-la-caida-libre\"><\/span> Resumo das f\u00f3rmulas de queda livre<span class=\"ez-toc-section-end\"><\/span><\/h3>\n<p> Abaixo voc\u00ea pode ver uma tabela com todas as f\u00f3rmulas de queda livre como um resumo: <\/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-chute-libre.png\" alt=\"f\u00f3rmulas de queda livre\" class=\"wp-image-8293\" width=\"398\" height=\"242\" srcset=\"https:\/\/physigeek.com\/wp-content\/uploads\/2023\/09\/formules-de-chute-libre-300x182.png 300w, https:\/\/physigeek.com\/wp-content\/uploads\/2023\/09\/formules-de-chute-libre-768x467.png 768w, https:\/\/physigeek.com\/wp-content\/uploads\/2023\/09\/formules-de-chute-libre.png 814w\" sizes=\"auto, (max-width: 300px) 100vw, 300px\"><\/figure>\n<p> Note que a massa do corpo que desce em queda livre n\u00e3o aparece em nenhuma f\u00f3rmula, o que significa que a posi\u00e7\u00e3o, a velocidade e a acelera\u00e7\u00e3o de um corpo n\u00e3o dependem da sua massa se desprezarmos o atrito. <\/p>\n<h2 class=\"wp-block-heading\"><span class=\"ez-toc-section\" id=\"Ejercicio-resuelto-de-la-caida-libre\"><\/span> Exerc\u00edcio de queda livre resolvido<span class=\"ez-toc-section-end\"><\/span><\/h2>\n<p> Nesta se\u00e7\u00e3o resolveremos passo a passo um exerc\u00edcio de queda livre, para que voc\u00ea entenda melhor os conceitos desse tipo de movimento.<\/p>\n<ul>\n<li> Um corpo \u00e9 lan\u00e7ado em queda livre de uma altura de 80 metros e com velocidade inicial zero. Com que rapidez o corpo atinge o solo? Despreze o atrito com o ar e tome o valor da gravidade como g=10 m\/s <sup>2<\/sup> .<\/li>\n<\/ul>\n<p> Para determinar a velocidade do corpo em queda livre ao atingir o solo, devemos primeiro encontrar o tempo em queda livre. Quando o corpo estiver no solo, sua posi\u00e7\u00e3o ser\u00e1 y=0, ent\u00e3o igualamos a equa\u00e7\u00e3o de posi\u00e7\u00e3o 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-89b3f9bb6bf789f11429d59533f96622_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 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-a8b3abf7ff1bae725abd60c8d8b5370d_l3.png\" class=\"ql-img-inline-formula quicklatex-auto-format\" alt=\"0=80-0 \\cdot (t-0)-\\cfrac{1}{2}\\cdot 10 \\cdot (t-0)^2\" title=\"Rendered by QuickLaTeX.com\" height=\"38\" width=\"278\" 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-d302b4bf2089721a04f6dc49a12c6109_l3.png\" class=\"ql-img-inline-formula quicklatex-auto-format\" alt=\"0=80-5t^2\" title=\"Rendered by QuickLaTeX.com\" height=\"15\" width=\"94\" style=\"vertical-align: 0px;\"><\/p>\n<\/p>\n<p> Resolvemos a equa\u00e7\u00e3o quadr\u00e1tica resultante: <\/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-acf2f09abd8f75c1ee0e570a6c0a93bc_l3.png\" class=\"ql-img-inline-formula quicklatex-auto-format\" alt=\"\\begin{aligned}5t^2&amp;=80\\\\[2ex]t^2&amp;=\\cfrac{80}{5}\\\\[2ex]t^2&amp;=16\\\\[2ex]t&amp;=\\sqrt{ 16}=\\begin{cases}4\\\\[2ex]-4 \\ \\color{red}\\bm{\\times}\\end{cases}\\\\[2ex]t&amp;=4\\ s\\end{aligned}[\/ latex] Et maintenant que nous connaissons le temps n\u00e9cessaire au corps pour atteindre le sol, nous utilisons l'\u00e9quation de la vitesse pour trouver la vitesse finale du corps : [latex]\\begin{aligned}v&amp;=v_0-g\\cdot (t-t_0)\\\\[2ex]v&amp;=0-10\\cdot (4-0)\\\\[2ex]v&amp;=-40 \\ \\cfrac{ m}{s}\\end{align\u00e9}\" title=\"Rendered by QuickLaTeX.com\" height=\"409\" width=\"584\" style=\"vertical-align: 0px;\"><\/p>\n<\/p>\n<h2 class=\"wp-block-heading\"><span class=\"ez-toc-section\" id=\"Caida-libre-y-tiro-vertical\"><\/span> Queda livre e tiro vertical<span class=\"ez-toc-section-end\"><\/span><\/h2>\n<p> Por fim, veremos qual a diferen\u00e7a entre a queda livre e o tiro vertical, pois s\u00e3o dois tipos de movimentos muito semelhantes, mas ligeiramente diferentes.<\/p>\n<p> Um lan\u00e7amento vertical envolve lan\u00e7ar um corpo verticalmente para cima, de modo que o corpo primeiro suba e depois caia at\u00e9 atingir o solo. Da mesma forma, num tiro vertical o corpo tamb\u00e9m pode ser lan\u00e7ado para baixo, o importante \u00e9 que ele se mova em linha reta e seja aplicada uma velocidade inicial.<\/p>\n<p> A <strong>diferen\u00e7a entre queda livre e lan\u00e7amento vertical<\/strong> \u00e9 que em geral, durante a queda livre, o corpo n\u00e3o tem velocidade inicial, pois simplesmente cai. Em contrapartida, durante um lan\u00e7amento vertical, o corpo tem uma velocidade inicial, uma vez que uma for\u00e7a \u00e9 aplicada ao lan\u00e7ar o corpo.<\/p>\n<p> Portanto, a velocidade final do corpo e o tempo para atingir o solo s\u00e3o diferentes na queda livre e no lan\u00e7amento vertical. Por\u00e9m, a acelera\u00e7\u00e3o \u00e9 a mesma nos dois tipos de movimento, pois \u00e9 a acelera\u00e7\u00e3o da gravidade. <\/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\/tiro-vertical\/\">Caracter\u00edsticas do plano vertical<\/a><\/div>\n","protected":false},"excerpt":{"rendered":"<p>Este artigo explica o que \u00e9 queda livre na f\u00edsica. Voc\u00ea tamb\u00e9m encontrar\u00e1 as f\u00f3rmulas para queda livre e, al\u00e9m disso, um exemplo resolvido de um corpo caindo em queda livre. O que \u00e9 queda livre? Na f\u00edsica, a queda livre \u00e9 um tipo de movimento retil\u00edneo no qual um corpo cai verticalmente de uma &hellip;<\/p>\n<p class=\"read-more\"> <a class=\"\" href=\"https:\/\/physigeek.com\/pt\/queda-livre\/\"> <span class=\"screen-reader-text\">Queda livre<\/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-425","post","type-post","status-publish","format-standard","hentry","category-cinematografico"],"yoast_head":"<!-- This site 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