When a ball is tossed straight up, does it experience momentary equilibrium at top of its path?
The forces are never balanced, as there is only ever one force - gravity.
The key is to remember Newton's second law: $F = ma$. Force and acceleration are paired, not force and velocity. Knowing just an object's current velocity tells you nothing about what forces are acting on it.
There are two ways to see how the velocity goes to zero. Either the initial impulse (instantaneous transfer of momentum) is depleted by a force applied over time, $$ m \Delta v = \int F \ \mathrm{d}t, $$ or the initial kinetic energy is depleted by a force applied over a distance, $$ \frac{1}{2} m \Delta (v^2) = \int F \ \mathrm{d}x. $$ In both cases, the force of gravity is acting continuously to slowly cancel the initial velocity, and there is nothing that turns off this force at the apex of the trajectory.
No. You are wrong. Even if we exclude drag, the force of gravity does not suddenly drop to zero.
The force acts to continuously accelerate the ball downwards. Because it starts off with a velocity upwards this will be reduced because of this force. At some point it will be zero as the velocity reverses sign, but there is no equilibrium as you can prove by the fact it continues to accelerate downwards.
The force that accelerated it upwards, for example the pitcher's arm, its removed as son as the ball leaves his hand. From that point the only forces on the ball are gravity, which is effectively a constant, and drag, which varies with speed.
Your argument does not work in Newtonian mechanics as fully explained by Chris White and Rory Alsop.
However, there is another perspective to all this and that is the one of general relativity. In GR, the spacetime manifold is "curved" by the Earth's presence and your ball simply follows a geodesic that begins at the spacetime point defined by where / when it was when you first threw it and whose initial direction (tangent) is defined by the ball's initial velocity four-vector. It is therefore in freefall and at equilibrium at all times in its flight (assuming no air resistance). The only time it is not in equilibrium is when it crashes into the ground and the interaction between the ground and ball matter accelerates the ball upwards: at impact there is a huge, transient impulse to the ball (and much mechanical deformation and rebound) but immediately thereafter the ball is undergoing a constant acceleration $g$ metres per second squared relative to the inertial frames that are momentarily co-moving with at any given time time after it hits the ground (the co-moving inertial frames keep on their merry way, heading through the ground towards the Earth's centre).
If you did your experiment again with a vectorial accelerometer embedded in the ball and with a telemetry system so that the ball could radio its acceleration back to recording equipment, that accelerometer data would give you the same description as the general relativity one just given. That is, you would see a big initial upwards acceleration as you threw the ball, followed by a long time interval where the accelerometer would read precisely nought (as long as the ball is thrown without spinning) whilst the ball were in flight. Then the accelerometer would register a huge impulse spike, but thereafter it would keep reading $g$ metres per second squared upwards as the ball sits stationary relative to the ground.
Notice how in this description the only real forces are the force you fling the ball with, and the reaction force from the ground when it hits. Gravity is an inertial force in the D'Alembertian sense, owing to our standing on the ground and thus being accelerated upwards at $g$ metres per second squared relative to our momentarily co-moving inertial reference frames, so we are indeed observing things from an accelerated reference frame in GR.