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Let's break down the problem step by step to determine from which floor the bucket was dropped.

### Given Data:
- **Time to hit the ground after passing the 12th floor:** \( t = 1 \, \text{second} \)
- **Height of each floor:** \( 4.9 \, \text{meters} \)
- **Acceleration due to gravity:** \( g = 9.8 \, \text{m/s}^2 \)
- We neglect air resistance.

### Approach:

1. **Determine the distance fallen in 1 second after passing the 12th floor.**
   The distance fallen \( d \) under constant acceleration \( g \) for time \( t \) can be calculated using the equation of motion:
   
   \[
   d = \frac{1}{2} g t^2
   \]
   Plugging in the values:
   \[
   d = \frac{1}{2} \times 9.8 \times 1^2 = 4.9 \, \text{meters}
   \]
   
   Since the height of each floor is 4.9 meters, this means the bucket traveled **exactly one floor** in the final second before hitting the ground. Therefore, the bucket passed the 12th floor and hit the ground after traveling 1 more floor, meaning it hit the ground from the 11th floor.

2. **Find the initial height from where the bucket was dropped.**
   Let \( h \) be the distance the bucket fell before reaching the 12th floor. The total time for the fall before passing the 12th floor is \( T \), which we need to calculate.
   
   Using the equation of motion for the distance traveled before reaching the 12th floor:
   
   \[
   h = \frac{1}{2} g T^2
   \]
   The total height the bucket fell until it passed the 12th floor is 12 floors times 4.9 meters per floor:
   
   \[
   h = 12 \times 4.9 = 58.8 \, \text{meters}
   \]
   
   Substituting this into the equation:
   
   \[
   58.8 = \frac{1}{2} \times 9.8 \times T^2
   \]
   
   Simplifying:
   
   \[
   58.8 = 4.9 \times T^2
   \]
   
   \[
   T^2 = \frac{58.8}{4.9} = 12
   \]
   
   \[
   T = \sqrt{12} \approx 3.46 \, \text{seconds}
   \]
   
   So, the bucket was dropped approximately 3.46 seconds before passing the 12th floor.

3. **Determine the total distance fallen during the entire fall.**
   Now, we use the total time of the fall (which is \( T + 1 = 3.46 + 1 = 4.46 \, \text{seconds} \)) to calculate the total distance the bucket fell from the moment it was dropped until it hit the ground:
   
   \[
   d_{\text{total}} = \frac{1}{2} g \times (4.46)^2 = \frac{1}{2} \times 9.8 \times (19.89) \approx 97.96 \, \text{meters}
   \]
   
4. **Find the floor from which the bucket was dropped.**
   Since each floor is 4.9 meters high, we divide the total distance by the height of one floor:
   
   \[
   \frac{97.96}{4.9} \approx 20
   \]
   
   Therefore, the bucket was dropped from approximately the **20th floor**.

### Final Answer:
The bucket was dropped from the **20th floor**.

Let me know if you'd like further details, and here are a few follow-up questions to expand on this topic:

1. How would the result change if the height of each floor were different?
2. What if air resistance were considered in this calculation?
3. How would you calculate the impact velocity of the bucket just before it hit the ground?
4. Can you use a different equation of motion to solve this problem?
5. How does the assumption of constant acceleration influence the result?

**Tip:** When solving physics problems, always start by identifying the known quantities and relevant equations before diving into calculations.

A man on the 12th floor of a building sees a bucket (dropped by a window washer) passing his window and notes that it hits the ground 1 second later. Assuming a floor is 4.9 meters high (and neglecting air friction), from what floor was the bucket dropped? (Use the acceleration due to gravity g = 9.8 m/s².)

Learn how to solve a free fall problem involving the distance traveled by a bucket falling from a building. This example involves kinematic equations, including the formula for distance under constant acceleration. Suitable for students in Grades 10-12.

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