A blue ball is thrown upward with an initial speed of $v_{b0}$, from a height of $y_{b0}$ meters above the ground. $t_r$ seconds after the blue ball is thrown, a red ball is thrown down with an initial speed of $v_{r0}$ from a height of $y_{r0}$ meters above the ground. The force of gravity due to the earth results in the balls each having a constant downward acceleration of $g=9.8$ m/s2.
(e)How long after the blue ball is released do the two balls meet?
We can set out expressions for the position of the red and blue balls equal to each other and then solve for $t$.
$$y_r = y_{r0} - v_{r0}(t-t_r) - \frac{1}{2}g(t-t_r)^2$$ $$y_b = y_{b0} + v_{b0}t -\frac{1}{2}gt^2$$ (The solution manual had the wrong signs in front of $v_{b0}$ and $v_{r0}$ but their end result was correct.)
set $y_r = y_b$: $$ y_{r0} - v_{r0}(t-t_r) - \frac{1}{2}g(t-t_r)^2 = y_{b0} + v_{b0}t -\frac{1}{2}gt^2$$ Expand: $$ y_{r0} - v_{r0}(t-t_r) - \frac{1}{2}g(t^2 -2t_rt + t_r^2) = y_{b0} + v_{b0}t -\frac{1}{2}gt^2$$ $$ y_{r0} - v_{r0}(t-t_r) - {\color{red} \frac{1}{2}gt^2} +gt_rt - \frac{1}{2}gt_r^2 = y_{b0} + v_{b0}t - {\color{red} \frac{1}{2}gt^2}$$ $$ y_{r0} - v_{r0}(t-t_r) +gt_rt - \frac{1}{2}gt_r^2 = y_{b0} + v_{b0}t $$ $$ y_{r0} - {\color{blue} v_{r0}t}+v_{r0}t_r +{\color{blue}gt_rt } - \frac{1}{2}gt_r^2 = y_{b0} + {\color{blue}v_{b0}t} $$ $$ y_{r0}-y_{b0 } +v_{r0}t_r - \frac{1}{2}gt_r^2 = {\color{blue}( v_{b0}+ v_{r0}-gt_r)t }$$ $$t = \frac{ y_{r0}-y_{b0 } +v_{r0}t_r - \frac{1}{2}gt_r^2}{ v_{b0}+ v_{r0}-gt_r}$$