A red ball is thrown down with an initial speed of 1.2 m/s from a height of 25 meters above the ground. Then, 0.6 seconds after the red ball is thrown, a blue ball is thrown upward with an initial speed of 23.8 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 are the two balls in the air at the same height?
[tex] x = x_o + v_ot + (1/2)at^2 [/tex]
The Attempt at a Solution
I tried setting two different versions of this equation equal to each other; one with information for the blue ball and one with information for the red ball, and then solve for time. I believe that I'm on the right track but that I'm somehow not properly figuring out the time offset for the blue ball. Any suggestions?