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詢問 高一 基礎物理 問題 ?
Aug 4th 2013, 22:25

物體以初速v0被鉛直上拋，重力加速度g，則自拋出到最大高度的一半處，所需

時間為何？

Ans:

能量方程式: m*V0^2/2=m*g*h

獲得最大高度: h=V0^2/2g

最大高度一半: h/2=V0^2/4g

能量方程式: m*V0^2/2=m*g*(h/2)+m*V^2/2

獲得速度為: V^2=V0^2-V0^2/2=V0^2/2 => V=V0/√2

所以所需時間為:

t=(V0-V)/g

=(V0-V0/√2)/g

=V0(√2-1)/g√2.........ans


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