A red ball is dropped from rest at a height of 8.80 m. A blue ball at a height of 10.5 m is thrown down at the same instant at 6.00 m/s.?

Take g = 9.8 m/s²
Take all downward quantity as positive.

s = ut + (1/2)at²
Distance that the red traveled, s₁ = (0)t + (1/2)(9.8)t² = 4.9t²
Distance that the ball ball traveled, s₂ = (6.00)t + (1/2)(9.8)t² = 6.00t + 4.9t²

When the ball ball catches up with the red ball :
8.80 - s₁ = 10.5 - s₂
8.80 - 4.9t² = 10.5 - (6.00t + 4.9t²)
8.80 - 4.9t² = 10.5 - 6.00t - 4.9t²
6.00t = 1.7
Time taken, t = 0.28 s

Use displacement equation s = u*t + 0.5at^2, with down meaning positive.
Distance traveled by red ball after time t: R = 0.5gt^2 = 4.905t^2.
This is measured relative to its starting point.

The blue ball's distance from its own starting place is 6.00t + 4.09t^2, at the same moment in time as the red ball because they start going at the same time.
However, we need to measure their distances from a common point, otherwise we won't know when they're at the same place. Let this common point be red ball's starting position, 1.70 m below the blue's
Then blue ball's position relative to that point = 6.00t + 4.09t^2 - 1.70.
Now we need to know when they'll be equal.

Solve
4.905t^2 = 6.00t + 4.09t^2 - 1.70
t = 0.283
To check if this is correct, plug it back into each of their equations to see if the distances match.

Impossible to tell because there is no information on the surface area or size of each ball. An eight foot ball will present much more air resistance than one the size of a tennis ball.