Why does the vertical velocity of a projectile launched horizontally, close to the surface of the Earth, not change? I cannot find an answer to this question and every time I come close, I am lost again. This is a fairly amateur question.
I am assuming if a projectile is launched horizontally, from say Newton's mountain, at a velocity of $8000\ \mathrm{m/s}$, how come its vertical velocity stays constant even though it is accelerating towards the earth at $9.8 \ \mathrm{m/s^2}$? Its vertical velocity at the end of the first second should be $9.8 \ \mathrm{m/s^2}$ downwards, and therefore at the end of the following second it should be $19.6 \ \mathrm{m/s}$. If we assume this is how the motion proceeds, then its vertical velocity would increase and it will crash down. This is of course not how it seems though. What am I missing?
 A: If you are referring to the situation where the projectile has entered a circular orbit, then its distance to the earth's center is not changing. But velocity is a vector, with magnitude (speed) and direction. In a circular orbit the speed can be constant but the effect of the force of gravity is to continually change the direction of the velocity. That is acceleration. So your statement that the vertical velocity stays constant is not correct as it is changing direction.
It's a different situation from what you may be thinking of i.e. launching a projectile horizontally over a flat surface. If the speed is large enough, the surface will fall away as the projectile proceeds due to the curvature of the earth and the projectile enters an orbit. This is a valid solution to the equation of motion due to the force of gravity.
A: OP's question is understandable and here's an admittedly handwaving attempt at providing some intuition.
Consider the vector diagram below. The horizontal vector is the velocity of the projectile in the horizontal direction soon after it is launched. This is largely unaffected by gravity as it is perpendicular to the line connecting the center of the earth with the projectile. Due to the downward force of gravity, the vertical velocity increases from its initial zero value to the small vertical vector shown and the resultant velocity is shown in blue. But now that blue vector is largely in the new horizontal direction because the earth is curved. Gravity now causes some vertical (i.e. towards the center of the earth) velocity in this new position but that again leads to a new resultant direction in a new "horizontal" direction. Etc. etc.
The force of gravity does not provide a continuously increasing "vertical" velocity as the definition of "vertical" keeps changing.

A: It seems you a referring to this thought experiment. I think you may solve your uncertainty by noting that earth should not be approximated as flat at such great speeds. As the object falls toward the earth, the surface of the earth itself is curving away from the object. The animated diagram in the article may help visualize this.

Consider the trajectory image. Recall that velocities are relative (a velocity must be defined with respect to a reference frame) and velocities are based on distances. A naive approach would be to define the velocity with respect to the bottom of this image, i.e. the vertical component of velocity would be rate of change of the distance of the object from some absolute reference plane below the earth (yellow arrow). If you do this, you will find that an object shot horizontally will have a downward vertical velocity that increases with each passing second, as you expect... At least until it nears the bottom of the earth and the earth's gravity behind to pull it back up again. You see the earth's gravity is a spherical potential, meaning it draws objects inward toward it's center, and in this problem we cannot approximate it as a downwards potential toward a flat surface like we so often do. If we take into account the shape of the earth, the physically relevant distance is from the object to the center of the earth, and the donward "vertical" (more accurately inward "radial") velocity is the time derivative of that distance (orange arrow in diagram). If an object is shot horizontally (tangentially) along the surface of the earth with just the right velocity (the exact orbital speed), then the distance from the object to the surface of the earth will remain constant and it's velocity in that (radial) direction will be zero. Gravity is still acting on the object, and it is still falling. But due to to the earth's curvature, the surface of the earth is also "falling away" as the object advances (the "fallings" are by exactly the same rate, if the object it shot with that magical orbital speed).
A: Its vertical speed does not increase because its centrifugal force acts against gravity. If its horizontal speed were greater it would rise vertically. If its horizontal speed were less it would begin increasing vertical speed and eventually crash to the ground. But if horizontal speed makes its centrifugal force equal to its centripetal force (gravity) then it will have a stable orbit, discounting air resistance.
