1. The problem statement, all variables and given/known data A ball is dropped from the top of a 55.0 m -high cliff. At the same time, a carefully aimed stone is thrown straight up from the bottom of the cliff with a speed of 21.0 m/s . The stone and ball collide part way up. How far above the base of the cliff does this happen? 2. Relevant equations d = vt+1/2at^2 d1 + d2 = 55 3. The attempt at a solution Ball v = 0 d = 1/2*9.8t^2 Stone v = 21 d = 21t-1/2*9.8t^2 (I made it -9.8 because the stone is going up so it's fighting gravity, I hope I'm right) Ball + Stone = 55 1/2*9.8t^2 + 21t-1/2*9.8t^2 = 55 21t = 55 t = 2.619 Stone: 21(2.619)-(1/2)(9.8)(2.619)^2 = 21.389 Ball: (1/2)(9.8)(2.619)^2 = 33.6109 I don't understand what I'm supposed to do next.