{"id":4087,"date":"2018-07-20T17:54:06","date_gmt":"2018-07-20T17:54:06","guid":{"rendered":"http:\/\/fisicaevestibular.com.br\/novo\/?page_id=4087"},"modified":"2024-08-21T17:20:02","modified_gmt":"2024-08-21T17:20:02","slug":"cinematica-11-lancamento-horizontal","status":"publish","type":"page","link":"https:\/\/fisicaevestibular.com.br\/novo\/formulario\/cinematica-11-lancamento-horizontal\/","title":{"rendered":"Cinem\u00e1tica 11 – Lan\u00e7amento horizontal"},"content":{"rendered":"

Formul\u00e1rio completo de F\u00edsica com informa\u00e7\u00f5es \u00fateis <\/b><\/span><\/span><\/span><\/p>\n

(dicas para vestibulares)<\/b><\/span><\/span><\/span><\/p>\n

Cinem\u00e1tica 11<\/b><\/span><\/span><\/span><\/p>\n

Lan\u00e7amento horizontal<\/b><\/span><\/span><\/span><\/p>\n

Colocando-se a\u00a0origem do sistema de refer\u00eancia<\/b><\/span><\/span><\/span>\u00a0no\u00a0<\/b><\/span><\/span><\/span>ponto de lan\u00e7amento, orienta-se<\/b><\/span><\/span><\/span>, por exemplo, o\u00a0<\/b><\/span><\/span><\/span>eixo X para a direita e o eixo Y para baixo.<\/b><\/span><\/span><\/span><\/p>\n

Decomp\u00f5e-se, em cada instante <\/b><\/span><\/span><\/span>o movimento em\u00a0<\/b><\/span><\/span><\/span>duas parcelas:<\/b><\/span><\/span><\/span><\/p>\n

Segundo o eixo X<\/b><\/span><\/span><\/span>\u00a0<\/b><\/span><\/span><\/span>\u00a0 <\/b><\/span><\/span><\/span>trata-se de um\u00a0movimento horizontal uniforme <\/b><\/span><\/span><\/span>com <\/b><\/span><\/span><\/span>velocidade constante <\/b><\/span><\/span><\/span>de<\/b><\/span><\/span><\/span>\u00a0<\/b><\/span><\/span><\/span>intensidade\u00a0V<\/b><\/span><\/span><\/span>o<\/b><\/span><\/span><\/sub><\/span>, <\/b><\/span><\/span><\/span>que \u00e9 a\u00a0<\/b><\/span><\/span><\/span>velocidade de lan\u00e7amento<\/b><\/span><\/span><\/span>\u00a0<\/b><\/span><\/span><\/span>\u00a0<\/b><\/span><\/span><\/span>S = S<\/b><\/span><\/span><\/span>o<\/b><\/span><\/span><\/sub><\/span>\u00a0+ V.t<\/b><\/span><\/span><\/span>\u00a0<\/b><\/span><\/span><\/span>\u00a0X= 0 + V<\/b><\/span><\/span><\/span>o<\/b><\/span><\/span><\/sub><\/span>.t\u00a0<\/b><\/span><\/span><\/span>\u00a0\u00a0<\/b><\/span><\/span><\/span>X = V<\/b><\/span><\/span><\/span>o<\/b><\/span><\/span><\/sub><\/span>.t<\/b><\/span><\/span><\/span><\/p>\n

<\/p>\n

Segundo o eixo Y<\/b><\/span><\/span><\/span>\u00a0<\/b><\/span><\/span><\/span>\u00a0 trata-se de um\u00a0<\/b><\/span><\/span><\/span>movimento uniformemente variado <\/b><\/span><\/span><\/span>com <\/b><\/span><\/span><\/span>velocidade vertical <\/b><\/span><\/span><\/span>inicia nula<\/b><\/span><\/span><\/span>\u00a0(<\/b><\/span><\/span><\/span>V<\/b><\/span><\/span><\/span>oy<\/b><\/span><\/span><\/sub><\/span> = 0,<\/b><\/span><\/span><\/span>\u00a0ou seja, \u00e9 uma\u00a0<\/b><\/span><\/span><\/span>queda livre com o corpo abandonado da origem<\/b><\/span><\/span><\/span>,\u00a0sujeito\u00a0<\/b><\/span><\/span><\/span>apenas \u00e0 acelera\u00e7\u00e3o da<\/b><\/span><\/span><\/span>\u00a0<\/b><\/span><\/span><\/span>gravidade, de intensidade\u00a0g, dire\u00e7\u00e3o vertical e sentido para baixo<\/b><\/span><\/span><\/span>.\u00a0<\/b><\/span><\/span><\/span><\/p>\n

Equa\u00e7\u00f5es:<\/b><\/span><\/span><\/span><\/span><\/p>\n

S = S<\/b><\/span><\/span><\/span><\/span>o<\/b><\/span><\/span><\/span><\/sub><\/span>\u00a0+ V<\/b><\/span><\/span><\/span><\/span>oy<\/b><\/span><\/span><\/span><\/sub><\/span>.t + at<\/b><\/span><\/span><\/span><\/span>2<\/b><\/span><\/span><\/span><\/sup><\/span>\/2\u00a0<\/b><\/span><\/span><\/span><\/span>\u00a0 <\/b><\/span><\/span><\/span><\/span>Y= 0 + 0.t + gt<\/b><\/span><\/span><\/span><\/span>2<\/b><\/span><\/span><\/span><\/sup><\/span>\/2\u00a0<\/b><\/span><\/span><\/span><\/span>\u00a0<\/b><\/span><\/span><\/span><\/span>Y=g.t<\/b><\/span><\/span><\/span><\/span>2<\/b><\/span><\/span><\/span><\/sup><\/span>\/2<\/b><\/span><\/span><\/span><\/span><\/p>\n

V<\/b><\/span><\/span><\/span><\/span>y<\/b><\/span><\/span><\/span><\/sub><\/span>\u00a0= V<\/b><\/span><\/span><\/span><\/span>oy<\/b><\/span><\/span><\/span><\/sub><\/span>\u00a0+ a.t\u00a0<\/b><\/span><\/span><\/span><\/span>\u00a0 <\/b><\/span><\/span><\/span><\/span>V<\/b><\/span><\/span><\/span><\/span>y<\/b><\/span><\/span><\/span><\/sub><\/span>= 0 + g.t\u00a0<\/b><\/span><\/span><\/span><\/span>\u00a0\u00a0<\/b><\/span><\/span><\/span><\/span>V<\/b><\/span><\/span><\/span><\/span>y<\/b><\/span><\/span><\/span><\/sub><\/span> = g.t<\/b><\/span><\/span><\/span><\/span><\/p>\n

V<\/b><\/span><\/span><\/span><\/span>2<\/b><\/span><\/span><\/span><\/sup><\/span> = V<\/b><\/span><\/span><\/span><\/span>o<\/b><\/span><\/span><\/span><\/sub><\/span>2<\/b><\/span><\/span><\/span><\/sup><\/span>\u00a0+ 2.a.\u0394S\u00a0<\/b><\/span><\/span><\/span><\/span>\u00a0 <\/b><\/span><\/span><\/span><\/span>V<\/b><\/span><\/span><\/span><\/span>y<\/b><\/span><\/span><\/span><\/sub><\/span>2<\/b><\/span><\/span><\/span><\/sup><\/span>\u00a0= V<\/b><\/span><\/span><\/span><\/span>oy<\/b><\/span><\/span><\/span><\/sub><\/span>2<\/b><\/span><\/span><\/span><\/sup><\/span>\u00a0+ 2.g.\u0394h<\/b><\/span><\/span><\/span><\/span>\u00a0<\/b><\/span><\/span><\/span><\/span>\u00a0<\/b><\/span><\/span><\/span><\/span>V<\/b><\/span><\/span><\/span><\/span>y<\/b><\/span><\/span><\/span><\/sub><\/span>2<\/b><\/span><\/span><\/span><\/sup><\/span>\u00a0= 0<\/b><\/span><\/span><\/span><\/span>2<\/b><\/span><\/span><\/span><\/sup><\/span>\u00a0+ 2.g.\u0394h<\/b><\/span><\/span><\/span><\/span>\u00a0 <\/b><\/span><\/span><\/span><\/span> \u00a0<\/b><\/span><\/span><\/span><\/span>V<\/b><\/span><\/span><\/span><\/span>y<\/b><\/span><\/span><\/span><\/sub><\/span>2<\/b><\/span><\/span><\/span><\/sup><\/span>\u00a0= 2.g.\u0394h<\/b><\/span><\/span><\/span><\/span>\u00a0\u00a0<\/b><\/span><\/span><\/span><\/span><\/p>\n

Informa\u00e7\u00f5es \u00fateis (Dicas)<\/b><\/span><\/span><\/span><\/p>\n

<\/p>\n

<\/p>\n

\u00a0<\/span>Observe na figura abaixo <\/b><\/span><\/span><\/span>que<\/b><\/span><\/span><\/span>,\u00a0<\/b><\/span><\/span><\/span>\u00e0 medida que o\u00a0<\/b><\/span><\/span><\/span>abacaxi<\/b><\/span><\/span><\/span>\u00a0vai caindo, a <\/b><\/span><\/span><\/span>flecha tamb\u00e9m cai<\/b><\/span><\/span><\/span>, na <\/b><\/span><\/span><\/span>mesma propor\u00e7\u00e3o, <\/b><\/span><\/span><\/span>pois ambos caem sujeitos\u00a0<\/b><\/span><\/span><\/span>apenas \u00e0 a\u00e7\u00e3o da acelera\u00e7\u00e3o da gravidade g.<\/b><\/span><\/span><\/span><\/p>\n

<\/p>\n

Ent\u00e3o, se o\u00a0<\/b><\/span><\/span><\/span>garoto fizesse mira acima do abacaxi <\/b><\/span><\/span><\/span>e <\/b><\/span><\/span><\/span>ele n\u00e3o ca\u00edsse, o garoto\u00a0erraria o alvo,<\/b><\/span><\/span><\/span> pois a <\/b><\/span><\/span><\/span>flecha, \u00e0 medida que avan\u00e7a na horizontal com\u00a0velocidade constante V<\/b><\/span><\/span><\/span>o<\/b><\/span><\/span><\/sub><\/span>,\u00a0tamb\u00e9m est\u00e1 <\/b><\/span><\/span><\/span>caindo com velocidade\u00a0V<\/b><\/span><\/span><\/span>y<\/b><\/span><\/span><\/sub><\/span>.<\/b><\/span><\/span><\/span><\/p>\n

Se o\u00a0<\/b><\/span><\/span><\/span>instante\u00a0da queda do abacaxi e do lan\u00e7amento da flecha <\/b><\/span><\/span><\/span>for <\/b><\/span><\/span><\/span>simult\u00e2neo<\/b><\/span><\/span><\/span> e a <\/b><\/span><\/span><\/span>flecha mirada nele,<\/b><\/span><\/span><\/span> os dois <\/b><\/span><\/span><\/span>possuir\u00e3o em cada instante, <\/b><\/span><\/span><\/span>as\u00a0<\/b><\/span><\/span><\/span>mesmas velocidades horizontais e verticais atingindo sempre o abacaxi.\u00a0<\/b><\/span><\/span><\/span><\/p>\n

Para se <\/b><\/span><\/span><\/span>determinar o instante e a altura do encontro<\/b><\/span><\/span><\/span>, voc\u00ea deve\u00a0<\/b><\/span><\/span><\/span>isolar o tempo na equa\u00e7\u00e3o X = V<\/b><\/span><\/span><\/span>o<\/b><\/span><\/span><\/sub><\/span>t \u00a0<\/b><\/span><\/span><\/span>\u00a0 t =X\/V<\/b><\/span><\/span><\/span>o<\/b><\/span><\/span><\/sub><\/span>\u00a0e\u00a0substitu\u00ed-lo <\/b><\/span><\/span><\/span>na <\/b><\/span><\/span><\/span>equa\u00e7\u00e3o da altura Y = gt<\/b><\/span><\/span><\/span>2<\/b><\/span><\/span><\/sup><\/span>\/2.<\/b><\/span><\/span><\/span><\/p>\n","protected":false},"excerpt":{"rendered":"

Formul\u00e1rio completo de F\u00edsica com informa\u00e7\u00f5es \u00fateis (dicas para vestibulares) Cinem\u00e1tica 11 Lan\u00e7amento horizontal Colocando-se a\u00a0origem do sistema de refer\u00eancia\u00a0no\u00a0ponto de lan\u00e7amento, orienta-se, por exemplo, o\u00a0eixo X para a direita e o eixo Y para baixo. Decomp\u00f5e-se, em cada instante o movimento em\u00a0duas parcelas: Segundo o eixo X\u00a0\u00a0 trata-se de um\u00a0movimento horizontal uniforme com velocidade constante de\u00a0intensidade\u00a0Vo, que \u00e9 a\u00a0velocidade<\/p>\n","protected":false},"author":1,"featured_media":0,"parent":3987,"menu_order":0,"comment_status":"closed","ping_status":"closed","template":"","meta":{"om_disable_all_campaigns":false,"_monsterinsights_skip_tracking":false,"_monsterinsights_sitenote_active":false,"_monsterinsights_sitenote_note":"","_monsterinsights_sitenote_category":0,"footnotes":""},"class_list":["post-4087","page","type-page","status-publish","hentry"],"aioseo_notices":[],"_links":{"self":[{"href":"https:\/\/fisicaevestibular.com.br\/novo\/wp-json\/wp\/v2\/pages\/4087","targetHints":{"allow":["GET"]}}],"collection":[{"href":"https:\/\/fisicaevestibular.com.br\/novo\/wp-json\/wp\/v2\/pages"}],"about":[{"href":"https:\/\/fisicaevestibular.com.br\/novo\/wp-json\/wp\/v2\/types\/page"}],"author":[{"embeddable":true,"href":"https:\/\/fisicaevestibular.com.br\/novo\/wp-json\/wp\/v2\/users\/1"}],"replies":[{"embeddable":true,"href":"https:\/\/fisicaevestibular.com.br\/novo\/wp-json\/wp\/v2\/comments?post=4087"}],"version-history":[{"count":1,"href":"https:\/\/fisicaevestibular.com.br\/novo\/wp-json\/wp\/v2\/pages\/4087\/revisions"}],"predecessor-version":[{"id":4088,"href":"https:\/\/fisicaevestibular.com.br\/novo\/wp-json\/wp\/v2\/pages\/4087\/revisions\/4088"}],"up":[{"embeddable":true,"href":"https:\/\/fisicaevestibular.com.br\/novo\/wp-json\/wp\/v2\/pages\/3987"}],"wp:attachment":[{"href":"https:\/\/fisicaevestibular.com.br\/novo\/wp-json\/wp\/v2\/media?parent=4087"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}