Newton's 3rd Law and a bouncing ball just want to clear something up.
Take a ball that's been dropped to the ground. Gravity acts and this ball as it has mass and then the ball now moves to the ground with a constant force of say ($X$).
Now when the ball makes contact with the ground, Newtons 3'rd Law takes effect (no air resistance):
If object A (the ball) exerts a force on object B (the floor), then object B will exert an equal force on object A in the opposite direction. (Action has equal opposite reaction).
Now here's where I get confused.
If the ball (which has a constant force when it hits the ground ($X$)) experiences the same constant force in the opposite direction ($-X$, minus indicating opposite direction), then the total force acting on the ball should be net ZERO ( $X + (-X) = 0$).
So there are no net forces acting on the ball, so why does it BOUNCE BACK? What am I missing?
Shouldn't the ball just stay on the ground? Bouncing back means a force greater than (-X) was applied to the ball giving it upward motion. Where did it come from?
 A: 
Now here's where I get confused.
If the ball (which has a constant force when it hits the ground ($X$))
experiences the same constant force in the opposite direction ($-X$,
minus indicating opposite direction), then the total force acting on
the ball should be net ZERO ( $X + (-X) = 0$).

Yes the ball and the floor (Earth) exert equal and opposite forces on one another per Newton’s third law but they don't "cancel" each other. To determine the effect of the equal and opposite force on each object you need to apply Newton’s second law $F=ma$ to each object individually.
$$a_{ball}=\frac{F}{m_{ball}}$$
$$a_{Earth}=\frac{F}{m_{earth}}$$
The acceleration of the earth is so small due to its large mass that only the acceleration of the ball is observed.
The actual force experienced by the ball and earth on impact depends on the nature of the impact (elastic or inelastic} as well as the stopping distance/time of the ball. In this regard one can apply the work energy theorem which states that the net work done on an object equals its change in kinetic energy.
Hope this helps.
A: 
moves to the ground with a constant force of say (X). Now when the ball makes contact with the ground, Newtons 3'rd Law takes effect

First, $X=mg$, force of Earth's mass pulling on the ball. Second, Newton's 3rd Law (N3L) is always in effect; the ball is pulling on the Earth while the ball falls.

If the ball (which has a constant force when it hits the ground (X)) experiences the same constant force in the opposite direction (−X, minus indicating opposite direction), then the total force acting on the ball should be net ZERO ( X+(−X)=0).

Objects do not have or possess or carry force. A force results from an interaction of two things (ball/Earth or ball/floor) and acts on an object. While the interaction of the ball with Earth manifests in a constant force on the ball, that force has very little to do with the force magnitude between the floor and ball, beyond the velocity which results from the acceleration of $g$. I believe that is the big mistake you are making.
The force which the floor exerts upward  on the ball cannot be X. If it was X, the ball would keep moving at constant velocity. You don't observe that, so you can conclude it isn't true.  The force of the floor on the ball is electromagnetic in nature and results in elastic or plastic deformation, and must be larger than X for some time interval unless the ball crashes through the floor.
A: Just to clear up your statement, the net forces will NOT be zero. Newton's 3rd law states: If an object A exerts a force on a object B, then the object B will exert a force with the same intensity but opposite direction in object A. In your example, the ball is exertion a force on the floor, and the floor is exerting a force on the ball, so the forces won't cancel out because they're not acting on the same object.
ALTERNATIVE ADDITIONAL METHOD OF LOOKING AT THE SITUATION:
With that in mind, you can think of it as conversion of Potential energy ( the ball being dropped) to kinetic energy, according to 1st Law of Thermodynamics. The ball will bounce because it converted potential energy into kinetic, however because it hits the floor there will be energy dissipated into heat ( the ground will warm up a little). That's why the second bounce couldn't be higher than the first ( and also of course air resistance). Hope this helps
A: 
If the ball (which has a constant force when it hits the ground $(X)$.

Sorry the ball doesn't have constant force. Better to say that it is experiencing a constant force $mg$.
As per Newton's second law,
$$F=\frac{dP}{dt}$$
Now since the ball is dropped from a certain height $h$ it would have gained some velocity $\sqrt {2gh}$ just before hitting the ground and thus gained momentum $m\sqrt{2gh}$ in time $\sqrt \frac{2h}{g}$.
But when it strikes with the floor, it quickly comes to rest i.e change in momentum is still $m\sqrt{2gh}$ but this time the time in which this change has come is too small and thus the ball experiences a larger force from the ground in the upward direction than its weight and hence it jumps again.
Also remember that it is a contact force and the deeper you press something the more force you feel and thus the ball bounces once it's downward velocity becomes zero.
If the ground was not rigid (like sand) which can be displaced sideways due to collision then the Normal contact force doesn't increase much and thus it doesn't rise up again.
Summary: don't forget time.
Note : As commented below by TBissinger, I would like to add that I have assumed that the ball doesn't bounce back as per OP's question and this assumption clearly indicates that it should bounce back.
A: A lot of correct answers, maybe just one more sort of mathy one. As already mentioned, the force in question is not the force due to gravity. You could have the ball bounce off a room's wall (or a basketball hoop's backboard), so let's consider that situation instead and ignore gravity.
Imagine the ball being a point mass and the wall having being an elastic material that may be deformed by the ball. We will go to a case of infinite elasticity soon, which returns us to a solid wall, but for now we start from the soft wall. You can model the interaction between the ball and the wall by Hooke's law. That is, if the wall is at $x > 0$, the ball will experience a position-dependent force $F(x) = - k x$ for $x > 0$ and $F(x) = 0$ otherwise (because there is no wall and the ball moves freely).
That means it experiences a position-dependent acceleration of
$$a(x) = \frac{F(x)}{m} = - k \frac{x}{m},$$
so its equation of motion reads
$$\ddot{x} = - \frac{k}{m} x,$$
which is solved by
$$x(t) = A \cos\left(\frac{k}{m}t\right) + B \sin\left(\frac{k}{m}t\right),$$
and if you insert initial conditions $x(t=0) = 0$ and $\dot{x}(t=0) = v_0$, you will find the final result to be
$$x(t) = \frac{v_0}{\omega} \sin\left(\frac{k}{m}t\right)$$
for the position and
$$\dot{x}(t) = v(t) = v_0 \cos\left(\frac{k}{m}t\right). \tag{$\Delta$}$$
Note that $v_0 > 0$, since the ball flew into the wall from the left. This is true as long as the ball interacts with the wall, when it leaves the wall, the interaction is gone and the ball moves freely with constant velocity. We can actually calculate this velocity at the moment the ball leaves the wall. We just need to find the time where the ball leaves the wall, which is the time $t^* > 0$ for which $x(t^*) = 0$, and it should be the first time that this happens (because after that our model breaks down because the interaction is gone). We find this time by analyzing $\sin(kt^*/m) = 0$, and find that the time is $t^* = \pi m/k$. We can insert this time into ($\Delta$) and find that the velocity
$$v^* = v(t^*) = v_0 \cos\left(\frac{k}{m}t^* \right) = v_0 \cos(\pi) = - v_0.$$
That's the final result, the ball leaves the wall with the exactly opposite velocity to the one with which it originally touched the wall. In the entire process, we only relied on Newtonian mechanics, starting out from the assumption that the wall can be modelled by a simple law of ealsticity, i.e. Hooke's law. The result is even independent of the "wall stiffness" $k$. We can send $k/m \to \infty$ to arrive at the usual scenario where the interaction between ball and wall appears to be instantaneous. But on a very short time scale, this is a good first approximation to what happens (although $k$ should maybe rather be considered as the elasticity modulus of the ball, but I thought for exemplatory purposes, thinking about an elastic wall is easier).
This is what happens in the interaction between wall and ball. While they interact, the wall exerts a force $F_{wb}(x) = - kx$ on the ball and experiences itself the force $F_{bw}(x) = + kx$. As was already pointed out by Bob, this force would move the wall which is attached to the earth, but due to the immense differences between ball and wall or ball and (wall + earth), only the velocity of the ball is notably altered.
A: The statement of Newton's 3rd law can be clarified if we introduce the idea of "who/what feels" the force.
Consider a book resting on a table. Assuming it can respond, if I ask the book, "What force do you feel on your back?", it will say "my weight and the table pushing me up." If we ask the table, it will say "my weight and the book pushing me down."
The weight is extraneous, but it makes the example more familiar. The force of interaction is the force that the book and table "feel" against each other. What Newton's third law states is that interactions have a kind of symmetry. When things touch, they each push away from each other with the same magnitude.
It is important to note that neither of the parties feel both forces of interaction, they each feel one pushing away. So, a ball bouncing is described by the ball feeling it's downward weight and an upward force from the ground that is larger than the ball's weight, this causes the ball to slow and eventually rise back up.
Aside: the impact with the ground has some non-zero duration, and over that time the interaction force (the size of the force) will vary, but the ball will always feel that the ground pushes it up. The ball does not feel the force from the ball on the floor.
Other aside: not all interactions point away from each other, but the force of two bodies in contact always do because that is from matter is composed of atoms with a barrier of electrons, and electrons repel. Even for the weird forces that don't abide by Newton's 3rd law, there are more complex symmetries that arise.
A: In the most simplistic terms, your description is incorrect.  The forces as you have described are:

*

*The ball exerts a force on the earth.

*The earth exerts an equal and opposite force on the ball.

You then stated there are two forces on the ball, but there is actually only one.  Force #1 is acting on THE EARTH, not the ball.  So the ball has unbalanced force and accelerates upwards.
A: Forces are by one object, on a different object.  The "by" is whatever is causing the force to occur.  The "on" is whatever is receiving the force, and whose motion is being affected.
There are two forces on your bouncing ball:

*

*A force of gravity, by the Earth, on the ball.  Another name for this force is the weight of the ball.  Let's say it is 10 N.  Its direction is downward, because the Earth's gravity pulls downward.


*A surface force, by the Earth, on the ball.  Read my answer here explaining why this force is upward, and why it must be greater than the weight of the ball.  Let's say this force is 50 N upward.

Newton's Third Law states that if there is

*

*a force by object A on object B with a certain amount and direction

then there will also be

*

*a force by object B on object A with same amount and opposite direction.

Notice how the two bulleted clauses are nearly the same, except A and B have swapped roles, and the direction is reversed?  (Take a moment to think about this.)
The swapping of the object roles are particularly important; just because two forces happen to be "equal and opposite" does not necessarily make them a Third Law pair.
Okay, so the Third Law pair of the force #1 described above must be


*A force of gravity, by the ball, on the Earth.  It pulls upward on the Earth by 10 N.

and the Third Law pair of force #2 must be


*A surface force, by the ball, on the Earth.  It pushes downward on the Earth by 50 N.

So 1+3 are a Third Law pair, because the object roles are swapped, the amounts are the same, and the directions are reversed.  Similarly, 2+4 are a Third Law pair.  But 1+2 are not a pair: the object roles are not reversed, nor are the amounts the same.  Nor are 1+4, 2+3, or 3+4 Third Law pairs.

The net force is only meaningful to the forces on one given object.  We don't add in other forces on other objects, even if they were by the given object.
So if we want the net force on the ball,

*

*We include force #1, because it is on the ball.  10 N downward.


*We include force #2, because it is also on the ball.  50 N upward.


*We must not include force #3, because it is not on the ball.  It is on the Earth.  It does not matter that force #3 is by the ball.


*We also do not include force #4, because it is on the Earth, not on the ball.
The net force is 10 N downward + 50 N upward = 40 N upward.  The net force is not zero.  You now use the Second Law to calculate the acceleration of the ball, which (like the net force) will be upward.

You can also calculate the net force on the Earth.  In this case, only forces #3 and #4 act on the Earth.  We get a total of 10 N upward + 50 N downward = 40 N downward.
You can use this net force and the Second Law to calculate the acceleration of the Earth.  Because the Earth has an enormous amount of mass, the acceleration will be tiny, which is why we don't notice the Earth moving when the ball bounces.
A: 
If the ball (which has a constant force when it hits the ground ($X$)) experiences the same constant force in the opposite direction ($-X$, minus indicating opposite direction), then the total force acting on the ball should be net ZERO ($X + (-X) = 0$).

If the ball is lying on the ground, that's the case, and the ball is indeed not accelerating in that case.
But if the ball hits the ground at some speed, it will start to deform, and this deformation will create the opposing force that slows the ball down.
That force will be much larger than gravity. Several orders of magnitude larger, because the ball stops in several orders of magnitude shorter distance than from which it fell. This force is what stops the ball—at merely no force, the ball would continue to move at constant velocity, but there is the floor in the way.
Now whether it bounces depends on whether the deformation was elastic.
With elastic deformation, the deformed ball exerts force (roughly) proportional to the deformation. When it stops, it is squashed, and therefore still exerting this huge force from the deformation on the ground—and the ground on it. And this force accelerates it back up. It is as strong and acts as long as it did when stopping the ball, so it accelerates the ball to almost the same speed again, but in upward direction (except for the bit of speed it gained from gravity in the time and any losses in the elastic compression).
While if the deformation is not elastic—think a cloth sack stuffed with sand—the force is proportional to the rate of deformation. When the ball stops, the deformation stops increasing, the force from deformation stops, and the sack remains lying on the floor, somewhat squashed, exerting its weight on the floor and the floor exerting the same force in opposite direction back so the net force is zero and the sack is not moving.
