Solve it relative to the Earth and it becomes quite simple.

Fine..I can do that.

I’ll use a 5 lb. sphere 10 in2 in area.

In both RE and FE, we can use all the same variables, except for “acceleration due to gravity”. In FE, we have to use “acceleration due to UA” and from the FoR of the earth, the sphere wouldn’t be accelerating toward the earth at 9.8m/s2. It would be accelerating at the relative acceleration between the sphere and the earth. I don't know exactly what that is because I have no way of knowing how much the atmowhateveryouwanttocallit is accelerating the sphere, beyond your description of "a little bit". However, I do know that it would be something less than 9.8 m/s2 (if it wasn’t the distance would never decrease). So, let’s say the relative acceleration of the sphere is half of 9.8 m/s2. in FE…4.9 m/s2.

A 5lb. sphere with a drag coefficient of .5, (

https://www.engineeringtoolbox.com/drag-coefficient-d_627.html) average air density of 1.275 kg/m3 (

https://www.theweatherprediction.com/habyhints2/444/), area of 10 in2 and accelerating at 9.8m/s2, has a TV of 103.96 m/s.

A 5lb. sphere with a drag coefficient of .5, average air density of 1.275 kg/m3, area of 10 in2 and accelerating at 4.9 m/s2, has a TV of 73.511 m/s.

Clearly, the same formula from the same FoR, gives different speeds for FE and RE. Of course, the FE number is meaningless anyway because it represents the speed at which the forces are balanced, and we know that in FE TV, the forces are not balanced.

The EP has nothing to do with it.