Fictitious force on a free falling body? So I have read the first answer to this question and I have trouble making sense of the "Newton's explanation" part.
If a person was inside a car moving on a straight road with acceleration of $9.8ms^{-2}$ and let go of a ball in his hand, the ball will go backward because of a fictitious force (considering the car as our frame of reference which is non-inertial). Ok this I understand.
But now let's say this car with the person inside fell off a cliff. If he let go of a ball in his hand, the ball will not go to the back of the car. Where did the fictitious force go? Aren't the two cases equivalent in Newton's mechanics? the only difference I see is the direction of acceleration ($x$ in the first case and $-z$ in the second)
 A: With the car falling off the cliff there is a downward gravitational attractive force on the ball due to the Earth and yet the ball (the system) does not move relative to the car.
So there is a fictitious force on the ball which equal in magnitude and opposite in direction (upwards) to the downward gravitational attractive force on the ball due to the Earth.
This means that there is no net force on the ball and the ball is not moving relative to the car.
A: 
But now let's say this car with the person inside fell off a cliff. If he let go of a ball in his hand, the ball will not go to the back of the car.

First of all, there is not equivalence between a car accelerating at $9.8m/s^2$ and a car falling down the cliff. 
The equivalence is for a car accelerating and a car is at rest on the surface of the Earth (of course the latter should be oriented upwards in order to get the same directions), due to the reaction force of the car itself to the person or the surface, respectively.
A freely falling car does not feel a net force, it would be like it is at rest in the space far far away from any kind of gravitational potential.
If the car was moving at a constant speed, only then it would be just like a free falling car. And the ball would not go backward or something inside the car in both cases here.
On the other hand, inside a car at rest on the surface of the Earth (assuming it is oriented upwards somehow) the ball will simply fall towards the center of the Earth (which is the back of the car in that weird orientation) just like the ball inside the car that is accelerating at 9.8m/$s^2$.
EDIT:
The equivalence principle is a part of the Newtonian gravity as well as Einsteinian. Because it is first postulated by Galileo. Newtonian mechanics sees the principle in terms of force as expected. 
Indeed, in Newtonian picture, you only feel the presence of the gravity because you are exposed to the reaction force by the ground you stand or the chair you sit. Otherwise you would freely fall just like being at rest on interstellar space. 
So does the person in the accelerated car, that is, the car is accelerate the seat, the seat resist due to its inertia so it encounters a reaction force by the car, and the driver does the same because s/he is resisting to the seat's acceleration and encounters a reaction force and so on. Just like an object on the surface of the Earth tries to fall but is stopped by the ground as a reaction.
A: 
So I have read the first answer to this question and I have trouble making sense of the "Newton's explanation" part.

It's not that great of an explanation. I'll redo it. The fact that you feel no force acting on you when you're in free fall can be explained like this:
What you "feel" are stresses and strains in your body. In a uniform gravitational field, every particle in your body experiences the same acceleration, resulting in no stresses or strains. The Earth is large compared to a human or a car. The Earth's gravitational field is very close to uniform across the extent of a human body or the length of a car. When you drive off a cliff in a car and release a ball, the accelerations of the car, you, and the ball toward the Earth are nearly equal. You feel a pit in your stomach because you are in free fall, and the ball appears to be moving with you.
If you instead were in a spacecraft passing very close to a neutron star, the gradient across the spacecraft would be rather significant. Even Newtonian gravity would say that the gradient could tear your body apart. Relativistic effects make this spaghettification even stronger in strong gravitational fields.
Excluding neutron stars and black holes, deviations between Newtonian gravity and general relativity are tiny. This is true even for Mercury, which exhibits the greatest non-Newtonian acceleration. Mercury's orbit precesses primarily because of the Newtonian influence of the other planets. Those Newtonian influences could not explain all of the precession that had been observed over the centuries by the latter half of the nineteenth century. One of the key successes of general relativity is that it fully explains this very tiny 43 arc seconds of precession per century.  
