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? (Round your answer to the nearest greater whole number. Use the acceleration due to gravity g 9.8 m/s².) number of floor: 

To solve this problem, we need to determine the floor from which the bucket was dropped. The key data points are:

1. The height per floor is 4.9 meters.
2. The acceleration due to gravity is \( g = 9.8 \, \text{m/s}^2 \).
3. The bucket hits the ground 1 second after passing the 12th floor window.

### Step 1: Find the distance the bucket falls after passing the 12th floor

Using the kinematic equation for distance under constant acceleration:
\[
d = \frac{1}{2} g t^2
\]
where:
- \( d \) is the distance fallen,
- \( g = 9.8 \, \text{m/s}^2 \),
- \( t = 1 \, \text{second} \).

Substitute the known values:
\[
d = \frac{1}{2} \times 9.8 \times 1^2 = 4.9 \, \text{meters}.
\]
So the bucket falls 4.9 meters after passing the 12th floor.

### Step 2: Determine the height of the 12th floor above the ground

Since each floor is 4.9 meters high, the height of the 12th floor above the ground is:
\[
h_{12} = 12 \times 4.9 = 58.8 \, \text{meters}.
\]

### Step 3: Determine the total distance the bucket fell

The total distance the bucket fell from the original point of release to the ground can be expressed as \( h \). So the height from the point of release to the ground is the sum of the height of the 12th floor and the additional 4.9 meters the bucket fell after passing the 12th floor:
\[
h = h_{12} + 4.9 = 58.8 + 4.9 = 63.7 \, \text{meters}.
\]

### Step 4: Find the total time of fall

We can now determine the total time, \( T \), it took for the bucket to fall from its original point to the ground using the equation:
\[
h = \frac{1}{2} g T^2.
\]
Substitute the known values:
\[
63.7 = \frac{1}{2} \times 9.8 \times T^2.
\]
Solve for \( T^2 \):
\[
T^2 = \frac{63.7 \times 2}{9.8} = \frac{127.4}{9.8} \approx 13.
\]
Thus:
\[
T = \sqrt{13} \approx 3.6 \, \text{seconds}.
\]

### Step 5: Calculate the height from which the bucket was dropped

The height from which the bucket was dropped is given by:
\[
h = \frac{1}{2} g T^2 = \frac{1}{2} \times 9.8 \times (3.6)^2 = 63.7 \, \text{meters}.
\]
Thus, the height is approximately 63.7 meters.

### Step 6: Determine the floor from which the bucket was dropped

Since each floor is 4.9 meters high, the number of floors corresponding to 63.7 meters is:
\[
\text{Number of floors} = \frac{63.7}{4.9} \approx 13.0.
\]
Rounding to the nearest greater whole number, the bucket was dropped from the **14th floor**.

---

Would you like more details on any part of this solution or have any questions?

Here are 5 related questions to explore:

1. What is the effect of air resistance on the time taken for an object to fall?
2. How would the result change if the bucket passed a lower floor, say the 10th floor?
3. How can you determine the velocity of the bucket when it passes the 12th floor?
4. How do you calculate the impact velocity when the bucket hits the ground?
5. What if the acceleration due to gravity were different on another planet?

**Tip:** When using kinematic equations, be mindful of the assumptions like neglecting air resistance, as it can significantly affect real-world outcomes.

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? (Round your answer to the nearest greater whole number. Use the acceleration due to gravity g 9.8 m/s².)

Learn how to determine the floor from which a bucket was dropped using kinematic equations. This problem involves calculating distances, time, and the effect of gravity. Ideal for students in Grades 9-12.

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