1. The problem statement, all variables and given/known data Suppose you throw a stone straight up with an initial velocity of 20.5 m/s and, 4.0 s later you throw a second stone straight up with the same initial velocity. The first stone going down will meet the second stone going up. At what height above the point of release do the two stones meet? 2. Relevant equations xf = xi + vit + (at2)/(2) 3. The attempt at a solution So far I have setup an equation for each stone involving 't' and set them equal to find the time the stone meets in the air, and then attempted to solve for the height of either stone at time 't'. My answer in consistently 1.84 meters. I keep getting told this is the wrong answer. So far I have this vt-gt^2 = v(t-4) - g(t-4)^2 simplified to: t = (v/g) + 2 Plug in numbers to get: t = 4.0918; plug 't' into the position function for stone 1: height = 20.5(4.0918)-9.8(4.0918^2) = 1.84 meters. I'm not sure whats going on, I have also tried using (t+4) for tone #2, but the problem still turns out to be wrong.