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This problem involves a related rates scenario. Let's solve it step by step.

1. **Define the variables**:
   - Let \( x \) be the distance from the base of the ladder to the wall (house).
   - Let \( y \) be the height of the ladder on the wall.
   - The length of the ladder, \( L \), is 10 feet (constant).

2. **Apply the Pythagorean Theorem**:
   Since the ladder, wall, and ground form a right triangle, we have:
   \[
   x^2 + y^2 = 10^2
   \]
   \[
   x^2 + y^2 = 100
   \]

3. **Differentiate with respect to time \( t \)**:
   Differentiate both sides of the equation with respect to \( t \):
   \[
   2x \frac{dx}{dt} + 2y \frac{dy}{dt} = 0
   \]
   Simplifying, we get:
   \[
   x \frac{dx}{dt} + y \frac{dy}{dt} = 0
   \]

4. **Substitute the given values**:
   - When the base is 8 feet from the house, \( x = 8 \).
   - The rate at which the base is moving away from the wall is \( \frac{dx}{dt} = 2 \) feet per second.
   - We need to find \( \frac{dy}{dt} \), the rate at which the top of the ladder is sliding down the wall.

5. **Find \( y \) when \( x = 8 \)**:
   Substitute \( x = 8 \) into the Pythagorean equation:
   \[
   8^2 + y^2 = 100
   \]
   \[
   64 + y^2 = 100
   \]
   \[
   y^2 = 36
   \]
   \[
   y = 6
   \]

6. **Substitute \( x = 8 \), \( y = 6 \), and \( \frac{dx}{dt} = 2 \) into the differentiated equation**:
   \[
   8(2) + 6 \frac{dy}{dt} = 0
   \]
   \[
   16 + 6 \frac{dy}{dt} = 0
   \]
   \[
   6 \frac{dy}{dt} = -16
   \]
   \[
   \frac{dy}{dt} = -\frac{16}{6} = -\frac{8}{3}
   \]

The ladder's top is sliding down the wall at a rate of \( \frac{8}{3} \) feet per second when the base is 8 feet from the house.

Would you like further details on any step? Here are some related questions:

1. What happens to the rate of descent if the base is moving at a different speed?
2. How would the solution change if the ladder length were different?
3. What is the significance of the negative sign in the rate of change of \( y \)?
4. Could we solve this problem using another method, such as energy conservation?
5. How does this problem relate to real-world applications, like sliding objects?

**Tip:** Always differentiate with respect to time when solving related rates problems.

A 10-foot ladder is leaning against a house on flat ground. The house is to the left of the ladder. The base of the ladder starts to slide away from the house. When the base has slid to 8 feet from the house, it is moving horizontally at the rate of 2 feet per second. How fast is the ladder’s top sliding down the wall when the base is 8 feet from the house?

This detailed solution covers a related rates problem involving a 10-foot ladder sliding down a wall. Using the Pythagorean Theorem and differentiation, we calculate how fast the top of the ladder is sliding when the base is 8 feet away from the wall and moving at 2 feet per second. Ideal for students in Grades 10-12.

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