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Let's solve this related rates problem by breaking down the information given.

### Problem Setup
- The ladder is 25 feet long, so it forms a right triangle with the wall and the ground, where the ladder is the hypotenuse.
- Let:
  - \( x \) be the horizontal distance of the foot of the ladder from the wall.
  - \( y \) be the vertical distance of the top of the ladder from the ground.
  - \( \frac{dx}{dt} = 0.6 \) ft/s, which is the rate at which the foot of the ladder is moving away from the wall.

### Goal
We need to find how fast the top of the ladder is sliding down the wall, or \( \frac{dy}{dt} \), when \( x = 24 \) feet.

### Step-by-Step Solution
1. **Set Up the Relationship Using Pythagoras' Theorem**:
   Since the ladder forms a right triangle, we have:
   \[
   x^2 + y^2 = 25^2
   \]
   Simplifying, we get:
   \[
   x^2 + y^2 = 625
   \]

2. **Differentiate Both Sides with Respect to \( t \)**:
   Differentiating implicitly with respect to time \( t \), we get:
   \[
   2x \frac{dx}{dt} + 2y \frac{dy}{dt} = 0
   \]
   Dividing by 2 throughout simplifies this to:
   \[
   x \frac{dx}{dt} + y \frac{dy}{dt} = 0
   \]

3. **Substitute Known Values**:
   - We are given \( x = 24 \) feet.
   - To find \( y \) when \( x = 24 \), substitute \( x = 24 \) into the Pythagorean equation:
     \[
     24^2 + y^2 = 625
     \]
     \[
     576 + y^2 = 625
     \]
     \[
     y^2 = 49
     \]
     \[
     y = 7
     \]
   - We are also given \( \frac{dx}{dt} = 0.6 \) ft/s.

4. **Solve for \( \frac{dy}{dt} \)**:
   Substitute \( x = 24 \), \( y = 7 \), and \( \frac{dx}{dt} = 0.6 \) into the differentiated equation:
   \[
   24(0.6) + 7 \frac{dy}{dt} = 0
   \]
   \[
   14.4 + 7 \frac{dy}{dt} = 0
   \]
   Solving for \( \frac{dy}{dt} \):
   \[
   7 \frac{dy}{dt} = -14.4
   \]
   \[
   \frac{dy}{dt} = -\frac{14.4}{7}
   \]
   \[
   \frac{dy}{dt} = -2.057 \, \text{ft/s (approximately)}
   \]

### Conclusion
The top of the ladder is sliding down the wall at a rate of approximately **2.06 ft/s** when the foot of the ladder is 24 feet from the wall.

Would you like more details or have any questions?

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### Related Questions
1. What would happen to \( \frac{dy}{dt} \) if the ladder were longer?
2. How would the rate of change of \( y \) vary if \( \frac{dx}{dt} \) increased?
3. What if the ladder was not moving along a frictionless surface—how would this affect the calculations?
4. Can we find the angle between the ladder and the ground when \( x = 24 \) feet?
5. What is the significance of the negative sign in \( \frac{dy}{dt} \)?

### Tip
For related rates problems, always start by identifying the relationship (equation) between the variables, then differentiate with respect to time before plugging in known values.

A 25-foot ladder is leaning against a vertical wall when Jack begins pulling the foot of the ladder away from the wall at a rate of 0.6 ft/s. How fast is the top of the ladder sliding down the wall when the foot of the ladder is 24 ft from the wall?

Explore how to solve a related rates problem involving a 25-foot ladder leaning against a wall. Given that the ladder's foot is moving away from the wall at 0.6 ft/s, find the rate at which the top of the ladder slides down when the foot is 24 feet from the wall. This solution involves implicit differentiation and the Pythagorean theorem, making it suitable for high school students in Grades 11-12.

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