Perhaps 200t of mass overwhelms the influence of fluid density, but theoretically, I should think Terminal Velocity (where drag=weight) would vary by altitude...
Hi GreatBear,
The Terminal velocity should be roughtly constant, if yuu consider it being IAS/CAS and not the absolute value.
At 35000 ft this would be roughly 1.8 times higher.
So let's assume for a very crude first guess a factor of 1.4 - 1.5 for the average speed.
That would give you (again very roughly) somewhere between 200 - 250 kts.
Which would mean ~20000 ft/min.
So it would take close to two minutes for that vertical free fall.
But this is indeed an extremely crude calculation, not considering any aerodynamic or even local Mach effects.
To achieve terminal velocity at altitude (~280 - 300kts) would only take about 15sec. amd a drop of ~3000ft.
Edit:
This is pure theoretical as it would only apply for a pure vertical drop.
In reality this will be a transition and the plane will not immediately start to fall like a rock at the point of stall.
If you look at a wing polar you will notice that at the stall point the lift drops quite a bit but directly at the point of stall there is still quite some Cl, then decreasing with increasing AoA
/Edit
It has to be considered though that the initial altitude may have been higher, as this is one of the more likely sceanrios to get quickly into a stall in the given circumstances.
The initiating altitude may have been as much as 38000ft considering the initial Energy potential of ~45000ft (see posts above) and a stall speed of 220 kts IAS (equivalent altitude of 220kts ~7000ft. Alternatively: Speed loss 270kts to 220kts gives an altitude difference of ~3000ft).
Again, all numbers very crude, just to give an idea about the orders of magnitude we are talking