Related rates - ladder sliding down a wall physics question

[Circularity]

Hopefully I posted this in the right place.

The setup is standard - A ladder 10 ft long rests against a vertical wall. The bottom of the ladder slides away from the wall at a rate of 1 ft/sec. If we let y be the vertical distance and x the horizontal distance, we use the Pythagorean theorem to get

x^2+y^2 = 100,

so

dy/dt = -x/y dx/dt,

i.e.,

dy/dt = -x/y (since dx/dt=1)

My question is this. What is the speed of the head of the ladder (the part sliding down the wall) as it strikes the ground? Calculus texts never seem to look at this situation - they always ask something benign like the speed of the head of the ladder when x=4 or something.

The equation is staring us in the face, and it seems to indicate that the speed goes to...infinity?! (as the head of the ladder moves towards the ground, x goes to 10 and y goes to 0.)

And since momentum is mass times velocity, the head of the ladder (constant mass) has arbitrarily large momentum in the vertical direction in the moments just before it strikes the ground?

Is this right? What is going on here?

 

[HallsofIvy]

Your model does not apply in that case: it is impossible for a ladder to slide down the wall with the foot of the ladder moving at a constant speed.

 

[Circularity]

n/m, got it. Thanks.