设任一时刻的速度为v,经t=1s,位移x1=v0t+1/2*gt^2=v0+4.9,经t'=2s,位移x2=v0t'+1/2*gt'^2=2v0+19.6,可见,前1s内位移为x1=v0+4.9,后1s内位移为x2-x1=v0+14.7,所以,后1s内位移比前1s内位移多9.8m.注意到“任一时刻的速...