Hi, I appreciate it if someone can reply and help me out with this problem. A ball is thrown straight up from the edge of the roof of a building. A second ball is dropped from the roof 1 second later. Air resistance may be ignored. (a) If the height of the building is H m, what must be the initial speed of the first ball if both are to hit the ground at the same time? So far I understand that there is: X(t)=H+Vo(T+1)+1/2(-9.8)(T+1)^2 -> first ball X(t)=H+1/2(-9.8)(T)^2 -> second ball Then I know that the two formulas equal each other. From there I found T=(Vo+4.9)/(Vo-9.8) This is where I get stuck.