Question

A ladder 5m long is standing vertically, flat against a vertical wall, while its lower end in on the horizontal floor. ... Find the speed at which the upper end is moving down the wall 4 seconds after the lower end has left the wall ?

A ladder 5m long is standing vertically, flat against a vertical wall, while its lower end in on the horizontal floor. The lower end moves horizontally away from the wall at a constant speed of 1m/s while the upper end stays in contact with the wall. Find the speed at which the upper end is moving down the wall 4 seconds after the lower end has left the wall.

Answer 1

Speed is `4/3 " m/s"`

Explanation:

Using `x` and `y` for horizontal and vertical displacements, with:

• `{(x_o = 0),(dot x = 1),(y_0 = 5):}`

The relationship between `x` and `y` is Pythagorean:

• `x^2 + y^2 = 25`

Differentiate wrt `t`:

`2 x dot x + 2 y dot y = 0`

`dot y = - (x dot x)/y= - (x dot x)/sqrt(25 - x^2)`

`= - ((4*1) * 1)/sqrt(25 - (4*1)^2) = - 4/3 [ " = velocity "]`

So speed is `4/3" m/s"`

EDIT :

I found this image on YouTube which might help you understand what is going on. NB the numbers are different, but the basic idea is the same: