# 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.

Jul 6, 2018

Speed is $\frac{4}{3} \text{ m/s}$

#### Explanation:

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

• $\left\{\begin{matrix}{x}_{o} = 0 \\ \dot{x} = 1 \\ {y}_{0} = 5\end{matrix}\right.$

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

• ${x}^{2} + {y}^{2} = 25$

Differentiate wrt $t$:

$2 x \dot{x} + 2 y \dot{y} = 0$

$\dot{y} = - \frac{x \dot{x}}{y} = - \frac{x \dot{x}}{\sqrt{25 - {x}^{2}}}$

$= - \frac{\left(4 \cdot 1\right) \cdot 1}{\sqrt{25 - {\left(4 \cdot 1\right)}^{2}}} = - \frac{4}{3} \left[\text{ = velocity }\right]$

So speed is $\frac{4}{3} \text{ m/s}$

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: