1. The problem statement, all variables and given/known data You throw one ball vertically (V0= 15m/s) on top of a 100m tower. Two seconds later you release another ball from rest from the same tower. Create a motion equation for both balls At what moment do they pass each other? At what height? 2. Relevant equations s=vt+1/2at^2 3. The attempt at a solution First I thought I could figure out how long it took the first stone to come back down to the height of the top of the tower (where the other ball was released) and came up with about 3.06s via V0/a or 15/9.8 then multiply by two. Because this would reach 0 one second after ball two, I then made s=s with t-1 for ball one and t for ball two. I then came out to a formula: -4.9t^2=15t-15-4.9t^2+9.8t-4.9 After solving I got t=0.407 but if I insert this into the formulas (1/2*-9.8*t^2) and 15*(t-1)+1/2*-9.8*(t-1)^2 to find the distance that they have traveled it never works out. I have tried this in many different ways, including starting from the beginning with t-2 for ball two and still I come up with numbers that don't work. I think I am having trouble visualizing the time that the balls are actually in the air. Am i wrong in first trying to figure out when the first ball reaches s=0 (top of tower?) Is there a way that I can do this including the entire journey of the first ball in the equation? If I use t-1 what exactly does this mean? That the ball is in the air for one second less? If anyone could help I would be very appreciative. I am self studying physics for an exam, and I have a lot of trouble understanding it, so I really need to see clear steps to follow. Thank you!