Is the top of my ladder really reaching infinite velocity? Here is a classic "related rate" maths problems:

A $10$ ft long ladder rests against a vertical wall. If the bottom of the ladder slides away from the wall at a rate of $1$ ft/s, how fast is the top of the ladder sliding down?

Let's represent the vertical wall as our $y$-axis oriented from bottom to top, and the ground being the $x$-axis oriented to the right. We call $x$ the distance from the bottom of the ladder to the vertical wall, and $y$ the distance from the ground to the top of the ladder. We are given that $\frac{dx}{dt}=1$ ft/s and we are looking for $\frac{dy}{dt}$.
Since $x^2+2y^2=100$, we have $2x\frac{dx}{dt}+2y\frac{dy}{dt}=0$, so $\frac{dy}{dt}=-\frac{x}{y}\frac{dx}{dt}=-\frac{x}{\sqrt{100-x^2}}$. Fine.
But, when the ladder finishes to slide, that is when $x$ is approaching $10$ from the left, $\frac{dy}{dt}$ is decreasing with no bound: $\displaystyle \lim_{x\to 10^-}\frac{dy}{dt}=-\infty$. 
This obviously does not match with everyday observation. 
What happens when the ladder finishes to slide?
 A: This is a good example of using kinematic equations and then being puzzled by the physical interpretation.  
Your problem is not for a ladder, a wall and the ground, it is just a line of length 100 feet constrained to move between two lines at right angles to one another at a speed of 1 foot per second along one of the lines.
The mathematics is correct.
Once you move into the physical world you have other constraints to worry about - Newton's laws of motion, conservation of energy etc.

It is obvious from the result of your calculation that ignoring Newton's laws of motion produces a result which cannot be realised in practice.  
A: You are making the false assumption that the ladder remains in contact with the wall. In reality, contact will be lost and the velocity remains finite.
To solve this properly you would compute the rate of rotation of the ladder, and you would notice that there isn't enough torque to keep the ladder turning as fast as you need to keep it touching the wall.
A: As has already been said, this problem is quite unphysical. The “ladder” and “wall” vocabulary is better thought of as a guide to visualization of the motions than as meaningfully physical objects.
Notice also that as $\frac{dy}{dt}$ hits infinity, we also find $x$ cannot continue to increase at 1 ft/s because the “ladder” is too short — so there is a sense in which both ends are problematic.
What I want to show you in this answer is that there is a type of mechanical system which works more like this than a falling ladder — the linkage.

If you imagine that there are rails mounted along the x and y axes, and sliders of some description on those rails, and a bar hinged at each end, then this is more like your problem than the ladder, because the bar cannot “pull away from the wall” as is the physical solution to the movement of an actual ladder. Instead, what will happen is the mechanism will jam, as the sideways force on the $y$-axis slider exceeds the limits of the bearings used to allow it to slide freely. You may find that you cannot even push it back to the freely-sliding region.
This directly corresponds to your deriving a velocity of infinity — useful linkages must be designed to avoid jamming. (They must also avoid the case of being temporarily under-constrained — imagine a rectangle with hinges for corners. If the rectangle is folded over flat, it can then be folded again in the middle — this is typically an undesired degree of freedom in a linkage which can also lead to actual jamming indirectly.)
Sometimes whether a linkage will jam depends entirely on which elements of the linkage are driven. For example, if instead of pulling the $x$ slider away from the wall at a supposedly constant rate, you apply a torque to the bar, then the linkage will move easily until it hits the limits of its travel, with entirely finite velocities and forces.
(Disclaimer: I'm not sure if “jam” is the actual term used in mathematical analysis of linkages. In fact, I suspect it isn't.)

There is a linkage known as the Trammel of Archimedes, recently popular as a “useless machine” design, which is the same system except that the rails are long enough so that continuous four-quadrant movement is possible. However, it will still jam up if you attempt to drive it by pushing on the sliders (yellow in the image below) rather than the bar (pink).

by Wikimedia Commons user Zephyris

A more common example of the problem of linkages jamming is the ordinary crank, as used in engines. There are two positions of a crank where the crank arm aligns with the connecting rod (called top dead center and bottom dead center), in which the system cannot be moved by pushing on the connecting rod. This was a historical problem in one-cylinder steam engines — they had to be manually rotated to a good starting position, but then could pass over those positions through inertia of the rotating parts once up to speed. Once you add multiple cylinders whose cranks are at different angles on the common crankshaft (minimally, two cylinders 90° apart), this is no longer a problem. (Internal combustion engines are a more complicated case, and need to be actually spinning to start, hence starter motors.)
