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[obm-l] Re:

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Subject: [obm-l] Re:
From: "saulo nilson" <saulo.nilson@xxxxxxxxx>
Date: Mon, 1 May 2006 16:37:39 -0300
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se teta e o angulo de lançamento em relaçao a horizontal, a formula do alcance maximo e funçao de sen(2*teta), que e maximo quando sen2teta=1

ou seja

2teta=90

teta=45

On 5/1/06, Natan Padoin <natanpadoin@yahoo.com.br> wrote:

Alguém poderia me ajudar a provar que o ângulo que proprociona um alcance máximo no movimento oblíquo é 45°? Ficaria muito grato. Abraço.

Yahoo! doce lar. Faça do Yahoo! sua homepage.

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