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 step by step, we will analyze the motion of the ball thrown horizontally from the window of one building to the window of another building.

### Step 1: Understand the problem
We have two buildings that are 40 meters apart. A ball is thrown horizontally from a height of 145 meters above the ground and needs to reach a window that is 22.5 meters above the ground in the other building.

### Step 2: Calculate the vertical distance
First, we need to determine the vertical distance the ball will fall. This is calculated as:
\[ \text{Vertical distance} = \text{Initial height} - \text{Final height} = 145 \, \text{m} - 22.5 \, \text{m} = 122.5 \, \text{m} \]

### Step 3: Use the equation of motion for vertical motion
Since the ball is thrown horizontally, its initial vertical velocity is 0. The only force acting on it in the vertical direction is gravity. We can use the second equation of motion:
\[ s_y = u_y t + \frac{1}{2} a_y t^2 \]
Where:
- \(s_y\) = vertical distance = 122.5 m
- \(u_y\) = initial vertical velocity = 0 m/s
- \(a_y\) = acceleration due to gravity = \(g = 9.8 \, \text{m/s}^2\)
- \(t\) = time of flight

Substituting the known values:
\[ 122.5 = 0 \cdot t + \frac{1}{2} \cdot 9.8 \cdot t^2 \]
This simplifies to:
\[ 122.5 = 4.9 t^2 \]

### Step 4: Solve for time \(t\)
Rearranging the equation gives:
\[ t^2 = \frac{122.5}{4.9} \]
Calculating this:
\[ t^2 \approx 24.98 \implies t \approx \sqrt{24.98} \approx 5 \, \text{s} \]

### Step 5: Use the horizontal motion equation
In the horizontal direction, the ball travels a distance of 40 meters. The horizontal motion can be described by:
\[ x = u_x t \]
Where:
- \(x\) = horizontal distance = 40 m
- \(u_x\) = horizontal velocity (the speed we need to find)
- \(t\) = time of flight (which we found to be approximately 5 s)

Substituting the known values:
\[ 40 = u_x \cdot 5 \]

### Step 6: Solve for horizontal velocity \(u_x\)
Rearranging gives:
\[ u_x = \frac{40}{5} = 8 \, \text{m/s} \]

### Conclusion
The speed at which the ball must be thrown horizontally is:
\[ \boxed{8 \, \text{m/s}} \]