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This problem involves related rates, where the ladder forms a right triangle with the wall and the ground. The ladder's length is constant at 10 feet, and we are given the rate at which the top of the ladder slides down the wall. We need to find how fast the bottom of the ladder moves away from the wall when the bottom is 5 feet from the wall.

### Step-by-Step Solution:

Let:
- \( x(t) \) be the distance between the bottom of the ladder and the wall (the horizontal distance).
- \( y(t) \) be the height of the top of the ladder from the ground (the vertical distance).
- \( L \) be the length of the ladder, which is 10 feet. This length stays constant.

From the Pythagorean theorem, we know:
\[
x^2 + y^2 = L^2
\]
Substituting \( L = 10 \) feet:
\[
x^2 + y^2 = 100
\]

### Differentiating with Respect to Time:
Now, we differentiate both sides of the equation with respect to time \( t \):
\[
2x \frac{dx}{dt} + 2y \frac{dy}{dt} = 0
\]
This simplifies to:
\[
x \frac{dx}{dt} + y \frac{dy}{dt} = 0
\]
We are given:
- \( \frac{dy}{dt} = -2 \) ft/sec (since the top of the ladder is sliding down, the rate is negative).
- \( x = 5 \) ft when we need to find \( \frac{dx}{dt} \).

### Finding \( y \) when \( x = 5 \):
Using the Pythagorean theorem:
\[
5^2 + y^2 = 100
\]
\[
25 + y^2 = 100
\]
\[
y^2 = 75 \quad \Rightarrow \quad y = \sqrt{75} = 5\sqrt{3} \quad \text{ft}
\]

### Substituting Values into the Rate Equation:
Now we substitute \( x = 5 \), \( y = 5\sqrt{3} \), and \( \frac{dy}{dt} = -2 \) into the differentiated equation:
\[
5 \frac{dx}{dt} + 5\sqrt{3}(-2) = 0
\]
\[
5 \frac{dx}{dt} - 10\sqrt{3} = 0
\]
\[
5 \frac{dx}{dt} = 10\sqrt{3}
\]
\[
\frac{dx}{dt} = 2\sqrt{3} \quad \text{ft/sec}
\]

Thus, the bottom of the ladder is moving away from the wall at a rate of \( 2\sqrt{3} \) ft/sec when the bottom is 5 feet from the wall.

### Do you need more details or have any questions?

Here are some related questions:
1. What happens to the rate of change of the bottom's position as the ladder gets closer to lying flat?
2. How does the speed of the top of the ladder change as the ladder slides further down the wall?
3. How would the solution change if the ladder was shorter or longer?
4. Can you find the rate of change of the area of the triangle formed by the ladder, the wall, and the ground?
5. What would happen if the top of the ladder slid at a different rate?

**Tip:** When dealing with related rates problems, always start by identifying the relationships between the variables and differentiate with respect to time to link the rates of change.

A 10-ft ladder is leaning against a wall. If the top of the ladder slides down the wall at a rate of 2 ft/sec, how fast is the bottom moving along the ground when the bottom of the ladder is 5 ft from the wall?

Solve a related rates problem involving a 10-ft ladder leaning against a wall. As the top of the ladder slides down the wall at 2 ft/sec, calculate how fast the bottom moves along the ground when it is 5 ft from the wall. Using the Pythagorean Theorem and implicit differentiation, this problem covers key calculus concepts.

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