How long after the blue ball is thrown are the two balls in the air at the same height?

Question

A blue ball is thrown upward with an initial speed of 21.4 m/s, from a height of 0.7 meters above the ground. 2.6 seconds after the blue ball is thrown, a red ball is thrown down with an initial speed of 7.4 m/s from a height of 25.7 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.

Answer 1

Using the kinematic equations for height we find:

h1=.7+21.4T-1/2gT^2 where T is the Time of equal heights

h2=25.5-7.4(T-2.6)-1/2g(T-2.6)^2

Set h1=h2 to find .7+21.4T-1/2gT^2=25.5-7.4(T-2.6)-1/2g(T-2.6)^2

Solve to find T=0.69 sec