>>430
>(√(2k/y0))tanθ/2y0sinθcosθ=(√(k/2y0^3))/cos^3θ
=(√(k/2y0^3))/cos^2θ
dθcos^2θ=dx√(k/2y0^3)
(θ+sinθcosθ)/2=x√(k/2y0^3) (θ=0 for x=0)
(x, y)=((θ+sinθcosθ)√(y0^3/2k), y0cos^2θ) (θ:0→π/2)
y=0 for x=π√(y0^3/8k)