Two tall buildings are situated 200 m apart. With what speed must a ball be thrown horizontally from the window 540 m above the ground in one building, so that it will enter a window 50 m above the ground in the other ?

Answer

To solve the problem, we need to determine the speed at which a ball must be thrown horizontally from a height of 540 m in one building so that it enters a window 50 m above the ground in another building located 200 m away.

### Step-by-Step Solution:

1. **Identify the Heights:**
- The height from which the ball is thrown: \( h_1 = 540 \, \text{m} \)
- The height of the window it must enter: \( h_2 = 50 \, \text{m} \)
- The vertical distance the ball must fall:
\[ h = h_1 - h_2 = 540 \, \text{m} - 50 \, \text{m} = 490 \, \text{m} \]

2. **Calculate the Time of Flight:**
- The time \( t \) it takes for the ball to fall 490 m can be found using the equation of motion:
\[ h = \frac{1}{2} g t^2 \]
- Here, \( g \) (acceleration due to gravity) is approximately \( 9.8 \, \text{m/s}^2 \).
- Rearranging the equation gives:
\[ t^2 = \frac{2h}{g} = \frac{2 \times 490}{9.8} \]
- Calculating \( t^2 \):
\[ t^2 = \frac{980}{9.8} = 100 \quad \Rightarrow \quad t = \sqrt{100} = 10 \, \text{s} \]

3. **Determine the Horizontal Distance:**
- The horizontal distance \( d \) the ball must travel is given as \( 200 \, \text{m} \).
- The horizontal speed \( U \) can be found using the formula:
\[ d = U \cdot t \]
- Rearranging gives:
\[ U = \frac{d}{t} = \frac{200}{10} = 20 \, \text{m/s} \]

4. **Conclusion:**
- The speed at which the ball must be thrown horizontally is:
\[ U = 20 \, \text{m/s} \]

### Final Answer:
The ball must be thrown horizontally with a speed of **20 m/s**.