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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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