Quoted: Just another dumb question, if I fired straight up, how long will it take to get back to the ground leaving the muzzel at 3100fps? And how high would that bullet go?

Not a dumb question, but a classic example of a one dimensional kinematics problem:
For the sake of simplicity, the muzzle is at ground level:
initial velocity (vo)=3100ft/sec=944.88m/s (I like SI units)
acceleration(a) due to gravity=9.8m/s/s
So, to derive the formula that we'll use:
a=final velocity(vf)vo/time
So, at=vfvo
So, vf=vo+at
Because the definition of average velocity is: average velocity(va)=(final position of bullet(xf)initial position of bullet(xo))/time, we can rewrite this equation as:
xf=xova*t
velocity changes at a uniform rate, so:
va=(vo+vf)/2
We can now combine these last two equations with our first derived equation to find that:
xf=xo+va*t=xo+((vo=vf)/2)*t=xo+((vo+vo+at)/t)
OR:
xf=xo+vot+(a*t*t)/2
Plugging in the numbers:
the position of the bullet, in our simple example, is the same at the beginning and at the end, so:
xf=0
xo=0
vo=944.88m/s
(t unknown)
a=9.8m/s/s
0=(944.88m/s)*t+((9.8m/s/s)*t*t)/2
factor out a t:
((944.88m/s4.9m/s/s)*t)*t=0
t=944.88m/s/4.9m/s/s=192s
Maybe someone could check that for me, and correct all the typos?