Question: I was once asked the following question by a student I was tutoring; and I was stumped by it:

When one throws a stone why does it take the same amount of time for a stone to rise to its peak and then down to the ground?

* One could say that this is an experimental observation; after one could envisage, hypothetically, where this is not the case.
* One could say that the curve that the stone describes is a parabola; and the two halves are symmetric around the perpendicular line through its apex. But surely the description of the motion of a projectile as a parabola was the outcome of observation; and even if it moves along a parabola, it may (putting observation aside) move along it with its descent speed different from its ascent; or varying; and this, in part, leads to the observation or is justified by the Newtons description of time - it flows equably everywhere.
* It's because of the nature of the force. It's independent of the motion of the stone.

I prefer the last explanation - but is it true? And is this the best explanation?

Answer: 
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Your arm is capable of propelling an object at up to around 150km/h (and that's with some practice). At that speed the many factors like air resistance are negligible.

Let’s load a 16-inch shell into a gun (you will find several on the USS Iowa), aim it 45 degrees up and press the button. The shell is going to go up at an initial rate of Mach 1.7 (vertical vector), decreasing to zero at apogee, and it will come down accelerating at *g* until it reaches its terminal velocity. Things tend not to fall at supersonic speeds so it should be clear that the up time will not equal the down time.
**Assuming that the only forces acting on the stone are the initial force exerted by the hand on the stone AND the force of gravity**, we have the following situation:

Immediately after throwing, the stone has a kinetic energy in the x-direction that never changes (**conservation of energy**) and the stone has a kinetic energy in the y-direction that is constantly being depleted by gravity as it rises to its apex. Once the stone is at its apex (vertical kinetic energy = 0), gravity **works** to restore all of the stone's original vertical kinetic energy as it falls back to its original height.

So, the x-direction **kinetic energy** remained constant throughout the flight and the y-direction **total energy (kinetic + potential)** remained constant throughout the flight, independent of each other. This combination causes the symmetry of the situation.

If air resistance and terminal velocity are introduced into the problem, the character of the problem is completely changed. Even a very simple situation as the one you described can become very complex and confusing to most people. Most people would be stymied by the complexity of the math involved even in this simple situation, let alone the complexity of putting a probe into orbit about Earth.