A steel ball is dropped from th roof of a building. `A` man standing in front of a `1-m` high window in the building notes tha the ball takes `0.1 s` to the fall from the top to the boottom of the window. The ball continues to fall and strikes the ground. On striking the ground, the ball gers rebounded with the same speed with which it hits the ground. If the ball reappears at the bottom of the window `2 s` after passing the bottom of the window on the way down, dind the height of the building.

Explanation

To solve the problem step by step, we will break down the information given and apply the appropriate kinematic equations.

### Step 1: Understanding the Problem
We have a steel ball dropped from the roof of a building. A man observes that the ball takes 0.1 seconds to fall from the top to the bottom of a 1-meter high window. After passing the bottom of the window, the ball continues to fall, strikes the ground, and rebounds with the same speed. It reappears at the bottom of the window 2 seconds after passing the bottom on the way down.

### Step 2: Calculate the Speed of the Ball at the Bottom of the Window
The time taken to fall through the window (1 meter) is 0.1 seconds. We can use the kinematic equation for uniformly accelerated motion:

\[ s = ut + \frac{1}{2} a t^2 \]

Here, \(s = 1 \, \text{m}\), \(u = 0\) (initial speed when dropped), \(a = g = 9.8 \, \text{m/s}^2\), and \(t = 0.1 \, \text{s}\).

Substituting the values, we get:

\[ 1 = 0 \cdot 0.1 + \frac{1}{2} \cdot 9.8 \cdot (0.1)^2 \]

Calculating this gives:

\[ 1 = \frac{1}{2} \cdot 9.8 \cdot 0.01 \]
\[ 1 = 0.049 \, \text{m} \]

This is incorrect; we need to find the final speed \(v\) at the bottom of the window. We can use the equation:

\[ v = u + at \]

Substituting \(u = 0\), \(a = g\), and \(t = 0.1\):

\[ v = 0 + 9.8 \cdot 0.1 = 0.98 \, \text{m/s} \]

### Step 3: Calculate the Time to Fall to the Ground
Let \(H\) be the height of the building. The ball falls from the top of the building to the ground. The time taken to fall from the top of the building to the bottom of the window (1 meter) is 0.1 seconds.

The time taken to fall from the top of the building to the ground can be calculated using:

\[ H = \frac{1}{2} g t^2 \]

Let \(t_1\) be the time taken to fall from the top of the building to the bottom of the window, and \(t_2\) be the time taken to fall from the bottom of the window to the ground. The total time to fall from the top to the ground is:

\[ t_{\text{total}} = t_1 + t_2 \]

### Step 4: Calculate the Time to Rebound
After hitting the ground, the ball rebounds with the same speed \(v\) it had just before hitting the ground. The time taken to rise back to the bottom of the window is the same as the time taken to fall from the bottom of the window to the ground.

Given that the ball takes 2 seconds to reappear at the bottom of the window after passing it, we have:

\[ t_2 + t_{\text{rebound}} = 2 \, \text{s} \]

### Step 5: Solve for Height of the Building
The time taken to fall from the bottom of the window to the ground is equal to the time taken to rise back to the bottom of the window. Therefore, we can express the total time as:

\[ t_{\text{total}} = 0.1 + t_2 + t_2 = 0.1 + 2 \]

This gives us:

\[ t_2 = 1.45 \, \text{s} \]

Now, substituting \(t_2\) back into the equation for height:

\[ H = \frac{1}{2} g (t_1 + t_2)^2 \]

### Final Calculation
Using \(g = 9.8 \, \text{m/s}^2\), \(t_1 = 0.1 \, \text{s}\), and \(t_2 = 1.45 \, \text{s}\):

\[ H = \frac{1}{2} \cdot 9.8 \cdot (0.1 + 1.45)^2 \]
\[ H = \frac{1}{2} \cdot 9.8 \cdot (1.55)^2 \]
\[ H = \frac{1}{2} \cdot 9.8 \cdot 2.4025 \]
\[ H \approx 11.8 \, \text{m} \]

### Conclusion
The height of the building is approximately **11.8 meters**.