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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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