Two tall buildings are `30 m` apart. The speed with which a ball must be thrown horizontally from a window `150 m` above the ground in one building so that it enters a window `27.5 m` from the ground in the other building is.

Options

A. `2 m s^-1`
B. `6 m s^-1`
C. `4 m s^-1`
D. `8 m s^-1`

Explanation

To solve the problem, we need to determine the speed at which a ball must be thrown horizontally from a height of 150 m so that it enters a window located 27.5 m above the ground in a building 30 m away.

### Step-by-Step Solution:

1. **Determine the vertical distance the ball falls:**
- The ball is thrown from a height of 150 m and must enter a window at a height of 27.5 m.
- The vertical distance fallen (h) is:
\[ h = 150 \, \text{m} - 27.5 \, \text{m} = 122.5 \, \text{m} \]

2. **Calculate the time of flight (t):**
- We can use the second equation of motion for vertical 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 to solve for \( t \):
\[ t^2 = \frac{2h}{g} \implies t^2 = \frac{2 \times 122.5}{9.8} \]
- Calculating:
\[ t^2 = \frac{245}{9.8} \approx 25 \implies t \approx \sqrt{25} = 5 \, \text{s} \]

3. **Calculate the horizontal speed (v₀):**
- The horizontal distance (d) to be covered is 30 m.
- The relationship between distance, speed, and time is given by:
\[ d = v_0 \cdot t \]
- Rearranging to find \( v_0 \):
\[ v_0 = \frac{d}{t} = \frac{30 \, \text{m}}{5 \, \text{s}} = 6 \, \text{m/s} \]

4. **Conclusion:**
- The speed with which the ball must be thrown horizontally is \( 6 \, \text{m/s} \).

### Final Answer:
The speed with which the ball must be thrown horizontally is **6 m/s**.

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