Let's break down the problem and solve it step by step.

We are given:

1. A 3-meter ladder leaned against a wall, initially positioned with its foot 60 cm (0.6 meters) away from the base of the wall.
2. During the day, the foot of the ladder moved 80 cm (0.8 meters) further away from the base, meaning the total distance from the wall is now $60 \, \text{cm} + 80 \, \text{cm} = 140 \, \text{cm} = 1.4 \, \text{m}$.

We can model this situation as a right triangle, where:

• The ladder forms the hypotenuse (3 meters),
• The initial and final distances of the foot from the base of the wall are the horizontal legs, and
• The heights on the wall (before and after slipping) are the vertical legs.

Step 1: Calculate the initial height of the ladder on the wall

Using the Pythagorean theorem: $h_1^2 + 0.6^2 = 3^2$ $h_1^2 + 0.36 = 9$ $h_1^2 = 9 - 0.36 = 8.64$ $h_1 = \sqrt{8.64} \approx 2.94 \, \text{m}$

Step 2: Calculate the new height after the ladder slips

Now the foot of the ladder is 1.4 meters from the wall, so applying the Pythagorean theorem again: $h_2^2 + 1.4^2 = 3^2$ $h_2^2 + 1.96 = 9$ $h_2^2 = 9 - 1.96 = 7.04$ $h_2 = \sqrt{7.04} \approx 2.65 \, \text{m}$

Step 3: Calculate how far the ladder slipped down

The difference in height between the two positions is: $\Delta h = h_1 - h_2 = 2.94 - 2.65 \approx 0.29 \, \text{m} = 29 \, \text{cm}$

Thus, the ladder slipped down 29 cm.

Would you like further clarification on any part of the process?

Here are 5 related questions to expand your understanding:

1. How would the problem change if the ladder were longer or shorter?
2. What if the wall wasn't vertical—how would you adjust the calculations?
3. How could you use trigonometry instead of the Pythagorean theorem here?
4. What safety precautions might be involved in preventing such slips in real life?
5. What if the ladder slipped to a position where the angle to the wall was exactly 45 degrees?

Tip: Always keep track of units when solving word problems—especially when converting between centimeters and meters!