So the earth is accelerated upwards but if you jump then you are not affected by the force of UA - if you were then you wouldn't fall.

That’s not what the wiki says. It says that when you jump…you don’t fall…the earth catches up. Things don’t” fall” in UA, that is the point. I don't see why an object being released would be affected any differently by UA than someone who is jumping. Both should just remain suspended, with no force acting upon it until the earth "catches up".

When you jump, your upward velocity is for a moment, greater than the Earth's so you rise above it. But after a few moments, the Earth's increasing velocity due to its acceleration eventually catches up

In addition, nobody has really addressed the initial question I asked is how you would calculate the rate of acceleration. As I pointed out in a previous post

The calculations should be the same except the direction of the UA force and normal force would be opposite of gravity and normal force.

So let’s compare the calculations for both, using negative to indicate the gravity/normal direction and positive to indicate the UA/normal direction.

Using negative for gravity to calculate the net force on a 10k object at a 20 ⁰ angle

Fnorm = 10.000 * -9.810 * cos(20) = -92.184N

F⊥= 10.000 * -9.810 * cos(20) =-92.184N

F// = 0.000 -.810 * sin(20) = -33.552N

So in gravity the net force on the object would be -33.552N…IOW, down, in the negative direction

Using the same process using positive for UA to calculate the net force on a 10k object at a 20 ⁰ angle

Net UA force 10kg @ 20 ⁰

Fnorm = 10.000 * +9.810 * cos(20) = +92.184N

F⊥= 10.000 * +9.810 * cos(20) =+92.184N

F// = 0.000 -.810 * sin(20) = +33.552N

So with UA, the net force on the object would be +33.552N…IOW, up in the positive direction.

So using the same calculation formula, with UA the object is actually being pushed up. So if UA is actually responsible for accelerating it down, how would you calculate it?