A stone drops from the edge of the roof. It passes a window 2 m high in `0*1 s`. How far is the roof above the top of the window?

Explanation

To solve the problem of how far the roof is above the top of the window when a stone drops from the edge of the roof and passes a 2 m high window in 0.1 seconds, we can follow these steps:

### Step-by-Step Solution:

1. **Identify Given Data:**
- Height of the window (h_w) = 2 m
- Time taken to pass the window (t_w) = 0.1 s
- Initial velocity (u) = 0 (since the stone is dropped)
- Acceleration due to gravity (g) = 9.8 m/s² (we will consider it as positive in the downward direction)

2. **Calculate the Distance Covered by the Stone While Passing the Window:**
- The stone covers the height of the window in 0.1 seconds. We can use the second equation of motion:
\[ s = ut + \frac{1}{2} a t^2 \]
- For the window:
\[ s_w = u \cdot t_w + \frac{1}{2} g t_w^2 \]
- Since \( u = 0 \):
\[ s_w = \frac{1}{2} g t_w^2 = \frac{1}{2} \cdot 9.8 \cdot (0.1)^2 \]
- Calculate \( s_w \):
\[ s_w = \frac{1}{2} \cdot 9.8 \cdot 0.01 = 0.049 \text{ m} \]

3. **Calculate the Total Distance Fallen When the Stone Reaches the Bottom of the Window:**
- The total distance fallen when the stone reaches the bottom of the window (h + 2 m):
\[ s_{total} = \frac{1}{2} g (t + t_w)^2 \]
- Here, \( t \) is the time taken to fall from the roof to the top of the window, and \( t + t_w \) is the time taken to fall from the roof to the bottom of the window.

4. **Set Up the Equations:**
- Let \( h \) be the height from the roof to the top of the window. The distance fallen to the top of the window is:
\[ h = \frac{1}{2} g t^2 \]
- The distance fallen to the bottom of the window is:
\[ h + 2 = \frac{1}{2} g (t + 0.1)^2 \]

5. **Subtract the Two Equations:**
- Subtract the first equation from the second:
\[ (h + 2) - h = \frac{1}{2} g (t + 0.1)^2 - \frac{1}{2} g t^2 \]
- This simplifies to:
\[ 2 = \frac{1}{2} g [(t + 0.1)^2 - t^2] \]

6. **Use the Identity for Difference of Squares:**
- Using the identity \( a^2 - b^2 = (a + b)(a - b) \):
\[ (t + 0.1)^2 - t^2 = (t + 0.1 + t)(t + 0.1 - t) = (2t + 0.1)(0.1) \]
- Substitute back:
\[ 2 = \frac{1}{2} g (2t + 0.1)(0.1) \]

7. **Solve for \( t \):**
- Rearranging gives:
\[ 4 = g (2t + 0.1)(0.1) \]
- Substitute \( g = 9.8 \):
\[ 4 = 9.8 (2t + 0.1)(0.1) \]
- Solve for \( t \):
\[ 4 = 0.98(2t + 0.1) \]
\[ 2t + 0.1 = \frac{4}{0.98} \approx 4.08 \]
\[ 2t = 4.08 - 0.1 \approx 3.98 \]
\[ t \approx 1.99 \text{ seconds} \]

8. **Calculate the Height \( h \):**
- Substitute \( t \) back into the equation for \( h \):
\[ h = \frac{1}{2} g t^2 = \frac{1}{2} \cdot 9.8 \cdot (1.99)^2 \]
- Calculate \( h \):
\[ h \approx \frac{1}{2} \cdot 9.8 \cdot 3.9601 \approx 19.4 \text{ m} \]

### Final Answer:
The height of the roof above the top of the window is approximately **19.4 meters**.