A ball is released from the top of a multistory tower. The ball taked ` 1 s` to fall pasta floor of the tower `8 m` height of a floor some distance from the top of thetower. Find the velocities of the ball at the top and at the bottom of the window.

Explanation

To solve the problem of finding the velocities of the ball at the top and bottom of the window, we can follow these steps:

### Step 1: Understand the problem
We know that a ball is released from the top of a tower and takes 1 second to pass through a window that is 8 meters high. We need to find the initial velocity (U) at the top of the window and the final velocity (V) at the bottom of the window.

### Step 2: Use the average velocity formula
The average velocity (V_avg) when the ball passes through the window can be expressed as:
\[ V_{avg} = \frac{U + V}{2} \]
Since the ball takes 1 second to pass through the window, we can write:
\[ V_{avg} = \frac{\text{Distance}}{\text{Time}} = \frac{8 \, \text{m}}{1 \, \text{s}} = 8 \, \text{m/s} \]
Thus, we have:
\[ \frac{U + V}{2} = 8 \]
Multiplying both sides by 2 gives:
\[ U + V = 16 \quad \text{(Equation 1)} \]

### Step 3: Use the equations of motion
We can use the second equation of motion to find the relationship between U, V, and time (t):
\[ V = U + g t \]
Here, \( g \) is the acceleration due to gravity, which we can take as \( 10 \, \text{m/s}^2 \) (downward), and \( t = 1 \, \text{s} \):
\[ V = U + 10 \cdot 1 \]
Thus, we have:
\[ V = U + 10 \quad \text{(Equation 2)} \]

### Step 4: Solve the equations simultaneously
Now we can solve Equations 1 and 2 simultaneously.

From Equation 2, we can express U in terms of V:
\[ U = V - 10 \]

Substituting this expression for U into Equation 1:
\[ (V - 10) + V = 16 \]
Combining like terms gives:
\[ 2V - 10 = 16 \]
Adding 10 to both sides:
\[ 2V = 26 \]
Dividing by 2:
\[ V = 13 \, \text{m/s} \]

Now substituting V back into Equation 2 to find U:
\[ U = 13 - 10 = 3 \, \text{m/s} \]

### Conclusion
The velocities of the ball are:
- At the top of the window (U): **3 m/s**
- At the bottom of the window (V): **13 m/s**