We can solve this problem using the equations of motion under constant acceleration. Given that the acceleration due to gravity $g = 9.8 \, \text{m/s}^2$ and each floor is 4.9 meters high, we will use the following equation of motion to find the initial height from which the bucket was dropped:

$d = v_0 t + \frac{1}{2} g t^2$

Where:

• $d$ is the distance fallen,
• $v_0 = 0 \, \text{m/s}$ is the initial velocity (since the bucket is dropped, not thrown),
• $t = 1 \, \text{second}$ is the time it takes to hit the ground,
• $g = 9.8 \, \text{m/s}^2$ is the acceleration due to gravity.

Substituting the values into the equation:

$d = 0 \cdot 1 + \frac{1}{2} \cdot 9.8 \cdot (1)^2 = 4.9 \, \text{meters}$

This means the bucket falls a total distance of 4.9 meters in the final second.

Now, since the man is on the 6th floor, which is 6 floors * 4.9 meters/floor = 29.4 meters from the ground, we add this to the distance the bucket fell. So the total distance from where the bucket was dropped to the ground is:

$d_{\text{total}} = 29.4 + 4.9 = 34.3 \, \text{meters}$

To find the floor from which the bucket was dropped, we divide this total distance by the height of each floor:

$\text{Floor number} = \frac{34.3}{4.9} \approx 7$

Since we're asked to round up to the nearest greater whole number, the bucket was dropped from the 8th floor.

Would you like any more details or have any questions?

Here are some related questions you may find interesting:

1. How would air resistance affect the fall time of the bucket?
2. What if the window washer was moving upwards when they dropped the bucket?
3. How would the problem change if the gravity was different on another planet?
4. Can you calculate the time it takes for the bucket to fall from the 8th floor?
5. What is the potential energy of the bucket just before it was dropped?

Tip: Always pay attention to units when solving physics problems. Converting heights, times, and velocities into consistent units helps prevent errors.