Ok so the time to hit the ground is 6,1809 s and it hits the ground 72.248 meters from the building.Still not sure how to figure out with what speed. I really truly appreciate your help.

If you can, try and remember from memory the following equations. They are often known as the "SUVAT" equations. They basically describe the motion of an object under a constant acceleration:

\(\displaystyle v = u + at\)

\(\displaystyle v^2 = u^2 + 2as\)

\(\displaystyle s = ut + \frac{1}{2} at^2\)

\(\displaystyle s = \frac{1}{2}(u + v)t\)

In these equations:

s = displacement from origin (in m)

u = initial speed (in m/s)

v = final speed (in m/s)

t = duration (for object initially at speed u to reach speed v; in s)

a = acceleration (constant; in m/s2)

These are extremely useful equations since most terrestrial motion problems can be approximated as constant acceleration problems (e.g. by neglecting air resistance).

So... for question 1 you used \(\displaystyle s = ut+\frac{1}{2} at^2\). For question 2, which formula will help you get the speed when the object hits the ground?