1. The problem statement, all variables and given/known data A blue ball is thrown upward with an initial speed of 20.4 m/s, from a height of 0.5 meters above the ground. 2.5 seconds after the blue ball is thrown, a red ball is thrown down with an initial speed of 9.8 m/s from a height of 23.9 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 blue ball is thrown are the two balls in the air at the same height 2. Relevant equations Xf=Xi+Vit+0.5a^2 3. The attempt at a solution Blue ball: Vi=20.4 m/s a=-9.81 Xi=0 t=t-2.5 Red ball: Vi=-9.8 a=-9.81 xi=23.4 t=t The height is different because i chose the blue ball as reference point. I plugged the values into the equation and equated them giving me: 20.4(t-2.5)-4.9(t+2.5)^2=23.4-9.8t-4.9t^2 further simplifying i get.... -4.1t-4.9t^2-81.6=23.4-9.8t-4.9t^2 then... 5.7t=105 t=18.4s which is wrong. can someone tell me where i went wrong?