1. The problem statement, all variables and given/known data A red ball is thrown down with an initial speed of 1.0 m/s from a height of 27.0 meters above the ground. Then, 0.4 seconds after the red ball is thrown, a blue ball is thrown upward with an initial speed of 24.8 m/s, from a height of 0.8 meters above the ground. The force of gravity due to the earth results in the balls each having a constant downward acceleration of 9.81 m/s2. How long after the red ball is thrown, will the two balls be in the air at the same height. 2. Relevant equations X=X(initial)+v(initial)t+1/2at^2 3. The attempt at a solution I set the equations equal to eachother for each ball (making x the same) so I got.. 27+1t+4.9t^2=24.8(t-.4)+4.9(t-.4)^2 After factoring and so on, I got to.. 35.96=19.88t t=1.8 This is not correct, where am I going wrong?