Two tall buildings are 100 m apart with what speed must a ball be thrown horizontally
from the window 500 m above the ground in one building. So that it will enter a window 100 m
above the ground in the other ?

Options

A. `11 m//s`
B. `11.2 m//s`
C. `34 m//s`
D. `15 m//s`

Explanation

To solve the problem of determining the speed at which a ball must be thrown horizontally from a height of 500 m to enter a window 100 m above the ground in another building 100 m away, we can follow these steps:

### Step 1: Understand the Problem
We have two buildings 100 m apart. The ball is thrown horizontally from a height of 500 m and needs to reach a window that is 100 m above the ground in the other building. Therefore, the vertical distance the ball must fall is \(500 m - 100 m = 400 m\).

### Step 2: Calculate the Time of Flight
The time \(t\) it takes for the ball to fall 400 m can be calculated using the equation of motion under gravity:
\[ y = \frac{1}{2} g t^2 \]
where \(y\) is the vertical distance (400 m), and \(g\) is the acceleration due to gravity (approximately \(10 m/s^2\)).

Rearranging the equation gives:
\[ t^2 = \frac{2y}{g} = \frac{2 \times 400}{10} = 80 \]
\[ t = \sqrt{80} = 4\sqrt{5} \text{ seconds} \]

### Step 3: Calculate the Required Horizontal Speed
The horizontal distance \(x\) the ball needs to cover is 100 m. The horizontal speed \(u\) can be calculated using the formula:
\[ u = \frac{x}{t} \]
Substituting the values we have:
\[ u = \frac{100}{4\sqrt{5}} = \frac{100}{4 \times 2.236} \approx \frac{100}{8.944} \approx 11.18 \text{ m/s} \]

### Step 4: Final Answer
Thus, the speed at which the ball must be thrown horizontally is approximately:
\[ u \approx 11.2 \text{ m/s} \]

### Summary of Steps
1. Determine the vertical distance the ball must fall (400 m).
2. Use the equation of motion to calculate the time of flight (approximately \(4\sqrt{5}\) seconds).
3. Calculate the required horizontal speed using the distance and time (approximately \(11.2 m/s\)).