A ladder 13 meters long rests on horizontal ground and leans against a vertical wall. The foot of the ladder is pulled away from the wall at the rate of 0.4 m/sec. How fast is the top sliding down the wall when the foot of the ladder is 5 m from the wall?

Answer 1

If the base of the ladder is x feet from the wall, and the ladder reaches up y feet, then we have

x^2 + y^2 = 13^2

taking derivatives wrt t,

x dx/dt + y dy/dt = 0

Now just plug in your numbers and solve for dy/dt

(note that at the moment in question, you conveniently have a 5-12-13 right triangle.)