Okay, using a more accurate way to estimate
(Young's equations, see:
http://www.globalsecurity.org/military/library/report/1997/penetration_equations.pdf )
And the following:
D=2.625ft / 0.8m
Subsonic Cd =0.3, transonic 0.9, supersonic 0.5 (conservative)
Mass = 30,000lb / 13600kg
Terminal velocity should be supersonic ( sqrt(2Mg/(CdA) yields >1000m/s for a supersonic Cd of 0.5, however, falling from 50,000ft, it is likely the bomb would only reach (I'm estimating, here, I don't feel like doing the integrals) about 500m/s (8m/s2 average, ~60seconds).
So V = 500m/s (1640fps)
Looking at the bomb, it looks like a tangent Ogive with Ln/D of about 1.5-2, so lets assume 1.5.
Penetrating 5000psi 1% reinforced concrete (F=20), and assuming the roof is 10 penetrator calibers across (30ft) (W1=10), cure time tc >1yr, and thickness >>>Dbomb (Tc=50)
That gives the following parameters:
Ke=(F/W1)^0.3 = 1.23
S= 0.085 Ke(11-P)(tcTc)^-0.06 * (FC/5000)^0.3 = 0.827
Nose factor N =0.18(Ln/D)+0.56 = 0.83
Penetration is:
D = D = 0.000018 S N (m/A)^0.7 (V - 30.5)
Where m/A is 27850kg/m2
So D =7.5m , which doesn't make a lot of sense, we know it goes further than that.
Applying the hard target correction of Kh=0.46(m)^0.15 = ~2
Now we have D=15m, or about 50ft. Which is almost exactly what I would expect.
So more than Newtonian, but less than what I said before, but still, 50ft is a metric ass load of concrete.