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Yahoo!奇摩知識+ - 分類問答 - 科學常識 - 已解決
Yahoo!奇摩知識+ - 分類問答 - 科學常識 - 已解決 
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物理斜向拋射問題
Oct 1st 2013, 14:33

一石子自斜角A=45°的斜面底端相對於斜面以Q角發射,石子落於斜面的瞬間,速

度方向與斜面垂直,求cotQ=?

(1) 求速度

B=A+Q, g=9.8(mps^2), t=time

Vx=V*cosB

Vy=V*sinB-gt

落下時水平角=-45° => Vx+Vy=0

V*cosB+V*sinB-gt=0 => gt=V(cosB+sinB)


(2) 求座標點: 45°斜坡 => x=y

x=V*cosQ*t

y=V*sinB*t-0.5g*t^2=V*cosQ*t

=> gt=2V(sinB-cosB)


(3) 求cotQ=?

兩者相等: 2(sinB-cosB)=sinB+cosB => sinB=3cosB

3=tanB

=tan(Q+45)

=(tanQ+tan45)/(1-tanQ*tan45)

=(1+tanQ)/(1-tanQ)

1+tanQ=3(1-tanQ)=3-3tanQ

4tanQ=2 => tanQ=1/2

So cotQ=2


 


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