I am having a problem solving this: A ball is thrown straight up from the edge of a 20m tall building (ignore air resistance). One second later a second ball is thrown. What must be the initial velocity of the second ball in order for both of them to hit the ground at the same time. Knowns: Ball 1 x=20m v0=0m/s a=g (I am using 9.8m/s^2) t-down=2.02s (x=vt+0.5at^2, 20=0t+0.5(9.8)t^2) Ball 2 x=20m v0 is the unknown, t is also unkown a=g I don't know how to calculate t-up (ball was thrown up originally) thus I don't know the total time. The answer in the book is 8.18m/s and t is 1.55s. Any help would be appreciated as to how to obtain those answers. Using the displacement formula: x=x0+v0+0.5at^2 I can calculate T-Down for ball #1 which is 2.02s (20=0+0+4.9t^2) I thought maybe to calculate time down for ball #2 I can subtract 1 second from 2.02s(ball 1 t down) which is 1.02s. Then substitute that into 20=v0(1.02)+4.9(1.02^2) = 14.6m which is not the correct answer. I assume the error is due to me missing t-up completely. I also tried solving the equations simoltaneously since they are equal but it seems to make little sense when I do (and I also have 2 unknown on 1 side): 20=v0t+4.9at^2, v0t+4.9at^2-20=0 20=4.9at^2, 4.9at^2-20=0 V0t+4.9at^2-20=4.9at^2-20 (the -20 just cancels and the result is v0t equals 0.