Oh wow, I like these kind of questions, it will require some calculations though, but I'll explain it as best as I can
Well on average bullets travel at a speed of about 762m/s and if shot perpendicularly it will be under uniform acceleration or in this case, uniform retardation. The bullet will slowly deceleration untill it gets to its maximum height where it's velocity will be 0 and it will begin to drop. For me to answer this question, I'll ignore all external forces acting on the bullet on its way up.
Using one of the formulas for uniform projectile acceleration
H = u^2sin^2i/2g
v = final velocity at maximum height = 0
u = initial velocity = 760m/s
g = acceleration/deceleration due to gravity = -10m/s^2
H = distance traveled/maximum height reached
i = angle the bullet is shot at = 90
So we're basically looking for the distance traveled, putting all the values into the equation and calculating the value of H which is the total distance traveled or total height attained.
H = 28,880m
Thus the maximum height a bullet moving at 760m/s shot perpendicularly from the earth, neglecting air pressure and all outside disturbances is 28,880m