两物体平跑水平射程比为s1:s2=2:3,两物体抛出点A、B在同一铅垂线上,A点比B点高,B离地H,若AB=h不变,平抛初速不变,要使两水平射程相等,则AB应上移还是下移?移动多少?
S1=V1*t=V1*[√(2(H+h)/g)]
S2=V2*t=V2*[√(2H/g)]
S1/S2=2/3
由这三个式子得到：V1/V2=(2/3)*[√(H/(H+h))]-------(1)
同理可得到：V1/V2=(S1'/S2')*[√(H'/(H'+h))]-------(2)
令S1'/S2'=1-------据题意“两水平射程相等”.
则：[√(H'/(H'+h))]=(2/3)*[√(H/(H+h))]
两边平方：H'/(H'+h)=(4/9)*[H/(H+h)]
解出：H'=4Hh/(5H+9h)
不难看出,H'