Let take vertical axis y and origin o on the grounnd. Then for blue: y_{1} = 0.7 + 23.5t - 4.905t^{2}, v_{1} = 23.5 - 9.81t

For red ball: y_{2} = 31.2 - 12(t - 2.9) - 4.905(t - 2.9))^{2}, v_{2} = - 12 - 9.81(t - 2.9)

1) At maximum height v_{1} = 0

2) v_{1} = 23.5 - 9.81t = 0, t = 23.5/9.81 = 2.396 s

3) To get maximum height we calculate y_{1} at t = 2.396 s; y_{1max}= 0.7 + 23.5·2.396 - 4.901·2.396^{2} = 28.847 m

4) y_{2} at = 3.77 s, y_{2 }= 31.2 - 12(3.77 - 2.9) - 4.905(3.77 - 2.9)^{2} = 17.047 m

5) y_{1} = y_{2}; 0.7 + 23.5t - 4.905t^{2} = 31.2 - 12(t - 2.9) - 4.905(t - 2.9))^{2},

0.7 - 31.2 - 12·2.9 + 4.905·2.9^{2} = - 12t - 23.5t + 9.81·2.9t, t = 3.411 s at height 23.79 m