Assuming the bullet was fired from the ground, we need to solve for t in the equation
-4.9t2 + vt = 2500 m
But we first need to find the initial speed v, as Sean pointed out. We can find v from kinematics.
vf2 - v2 = 2ahmax
vf = 0 [instantaneous rest at hmax]
v = initial speed = ? [what we want to find]
a = -g = -9.8 m/s2
hmax = 2500 m
v = √(2ghmax) = √[2(9.8)(2500)] m/s ≅ √(49,000) m/s
h(t) = -4.9t2 + [√(49,000)]t
with h in meters, t in seconds.
Now set h = 2500 m and solve for t.
2500 = -4.9t2 + [√(49,000)]t
-4.9t2 + [√(49,000)]t - 2500 = 0
You now have a quadratic equation in t, which you should know how to solve, and get
t ≅ 22.59 seconds.