HazelNuts39:
The loss of kinetic energy between 475 and 200 kTAS is equivalent to 8230 ft height. The airplane has lost the energy equivalent of 35000 + 8230 ft height in 5 minutes, i.e. at an average rate of 8650 ft/min.
If the forward velocity had reduced to 50 kts, the equivalent altitude loss would be ~9850 ft and overall equivalent potential energy loss ~45000 ft.
Equivalent speed would be 1010 kts.
regarding linear deceleration that would equal 1,73 m/s^2 linear deceleration. That requires significant drag.
More than what an aircraft in mormal configuration in straight and level flight would achieve.
So, regarding energy state the final horizontal velocity doesn't make that much of a difference, i.e. we have a good idea of the medium energy dissipation over that time span.
/Edit:
As a comparison, to give an idea of the level of deceleration: The average acceleration during takeoff and landing is roughly comparable: 150kts in 40-50s ~1,75 m/s^2 (45s). To achieve that by aerodynamic drag alone wouldn't be easy