Question

A ladder 10 feet long rests against a vertical wall. Initially, the top of the ladder is 8 feet above the ground. If the top of the ladder is sliding down the wall at a rate of 2 feet per second, how fast is the bottom of the ladder sliding away from the wall 1 second after the ladder starts sliding? Answer the question in a complete sentence and include appropriate units

Answer 1

Explanation:

Let vertical height of ladder from ground be y and

horizontal distance of the base of the ladder from the wall be x respectively.

Length of the ladder = l (constant) = 10 ft

Using Pythagoras theorem:

`{l}^(2) = {y}^(2) + {x}^(2)`

Differentiate both sides w.r.t time

`0 = 2y (dy)/(dt) + 2x (dx)/(dt)`

`y (dy)/(dt) + x (dx)/(dt) = 0`

We know that (After 1 sec, y = 6 ft and x = 8 ft ; dy/dt = 2 ft/sec)

`6* 2 + 8(dx)/(dt) = 0`

`(dx)/(dt) = - 1.5 ft \: per \: sec`

( Ignore - ive sign)

Therefore, bottom of the ladder is sliding away from the wall at a speed of 1.5 ft/sec one second after the ladder starts sliding.