lsinθ= 1 2 gt 2 ,lcosθ=v 0 t,解得:t= 2 v 0 tanθ g , 由图知tan(α+θ)= v y v 0 = gt v 0 =2tanθ, 所以α与抛出速度v 0 无关,故α 1 =α 2 ,选项A正确,B错误; 小球做平抛运动,水平位移x=v 0 t 竖直位移, h= 1 2 g t 2 ∵ h x =tanθ ∴ t= 2 v 0 tanθ g 则水平位移, x= v 0 t= 2tanθ g v 20 ∵v 2 =3v 1 ,∴s 2 =9s 1 故C错误,D正确; 故选BD