Atennis ball is tossed up off a building with a velocity of 22m/s. it takes 6.4s to reach the ground. how high is the building and what is the maximum height of the tennis ball?

Right, so this isn't exactly straight up or straight down; the ball goes up (taller than the building) and then goes down.

Our knowns: 
Vi (initial velocity) = 22m/s
Vf (final velocity) = 0m/s (when the ball reaches its apex)
a (acceleration, rounded to make life easier) = -10m/s²
t (total time) = 6.4s

Our unknown:
Δx (displacement)

So now with all of that, we can get started. 

First, we'll calculate the height of the ball at its apex:
Vf² = Vi² + 2aΔx
0 = 484 + 2(-10)Δx
0 = 484 - 20Δx
-484 = -20Δx
24.2 = Δx

This means that, when the final velocity is zero (max height), the height of the ball is 24.2 m. Now that you know this, you can find the remaining height the ball has to fall to reach the ground (now we need the time). 

Tip: Remember that this is symmetrical movement, so the final velocity will be equal to the initial velocity, but only negative because the now the ball is falling. Similarly, the initial velocity will now be the final velocity the ball had at its apex.

You can use a few equations, but I chose this one:
Δx = (Vi + Vf)t
Δx = (0 - 22)6.4
Δx = -70.4 m (it's negative because the ball is falling) 

NOT done yet. That does not mean that the building is 70.4 m tall; it is the displacement from which the ball fell from its apex. However, knowing this and the height of the ball at its apex, we can find what is in between, AKA the height of the building. 

Height of building = 70.4 m - 24.2m = 46.2m 
So, the height of the building is 46.2m. And the maximum height of the ball is 70.4m. For the sake of significant figures, it would be 46m and 70.m. 

I've included a graphic to help visualize it. Hopefully it's helpful.