Two tall buildings are 40 m apart. With what speed must a ball be thrown horizontally from a window 145 m above the ground in one building, so that it will enter a window 22.5 m above from the ground in the other?

Options

A. `5ms^(-1)`
B. `8ms^(-1)`
C. `10ms^(-1)`
D. `16ms^(-1)`

Answer

To solve the problem, we need to determine the speed at which a ball must be thrown horizontally from a height of 145 m so that it lands in a window 22.5 m above the ground in another building that is 40 m away.

### Step-by-Step Solution:

1. **Identify the vertical distance**:
The ball is thrown from a height of 145 m and needs to reach a height of 22.5 m. Therefore, the vertical distance it falls is:
\[ h = 145 \, \text{m} - 22.5 \, \text{m} = 122.5 \, \text{m} \]

2. **Calculate the time of flight**:
The time \( t \) it takes for the ball to fall 122.5 m can be calculated using the equation of motion for free fall:
\[ s = ut + \frac{1}{2} a t^2 \]
Here, \( s = 122.5 \, \text{m} \), \( u = 0 \, \text{m/s} \) (initial vertical speed), and \( a = g = 10 \, \text{m/s}^2 \) (acceleration due to gravity). The equation simplifies to:
\[ 122.5 = 0 + \frac{1}{2} \cdot 10 \cdot t^2 \]
\[ 122.5 = 5t^2 \]
\[ t^2 = \frac{122.5}{5} = 24.5 \]
\[ t = \sqrt{24.5} \approx 4.95 \, \text{s} \]

3. **Calculate the horizontal speed**:
The horizontal distance \( d \) the ball needs to cover is 40 m. The horizontal motion can be described by the equation:
\[ d = u \cdot t \]
Rearranging for \( u \):
\[ u = \frac{d}{t} = \frac{40}{t} \]
Substituting the value of \( t \):
\[ u = \frac{40}{4.95} \approx 8.08 \, \text{m/s} \]

4. **Final answer**:
The speed at which the ball must be thrown horizontally is approximately \( 8.08 \, \text{m/s} \).

### Summary:
The speed required to throw the ball horizontally from a height of 145 m to reach a window 22.5 m above the ground in the other building, which is 40 m away, is approximately \( 8.08 \, \text{m/s} \).