Δh=1/2gt1^2+v0t1-(1/2gt2^2+v0t2)=1/2g(t1+t2)(t1-t2)-V0(t1-t2)t2>t1 设t2=t1+n,n>0 则原式=1/2(2t1+n)(-n)-V0(-n)=-1/2^2-nt1+nv0=-1/2n^2-nt1+nv0 可知Δh关于t1减小..Δh=o时 t1=1/2n+v0 ,Δv=v0-gt1-(v0-gt2)=g(t2-t1)=-ng=定值<0.所以歼宽答闷轮案选蚂改信B...