A ball is dropped from the top of a building. The ball takes `0.5s` to fall the `3m` length of a window some distance from the to of the building. If the speed of the ball at the top and at the bottom of the window are `v_(T)` and `v_(T)` respectively, then `(g=9.8m//s^(2))`

Options

A. `v_(T)+v_(B)=12ms^(-1)`
B. `v_(T)-v_(B)=4.9ms^(-1)`
C. `v_(B)+v(T)=1ms^(-1)`
D. `(v_(B))/(v_(T))=2`

Answer

To solve the problem step by step, we will use the equations of motion and the given information.

### Step 1: Identify the known values
- Distance of the window (S) = 3 m
- Time taken to fall past the window (T) = 0.5 s
- Acceleration due to gravity (g) = 9.8 m/s²

### Step 2: Calculate the average velocity during the fall
The average velocity (V_avg) when the ball falls past the window can be calculated using the formula:
\[ V_{avg} = \frac{S}{T} \]
Substituting the known values:
\[ V_{avg} = \frac{3 \, \text{m}}{0.5 \, \text{s}} = 6 \, \text{m/s} \]

### Step 3: Relate average velocity to the velocities at the top and bottom of the window
The average velocity can also be expressed in terms of the velocities at the top (V_T) and bottom (V_B) of the window:
\[ V_{avg} = \frac{V_T + V_B}{2} \]
Setting this equal to the average velocity we calculated:
\[ 6 = \frac{V_T + V_B}{2} \]
Multiplying both sides by 2:
\[ V_T + V_B = 12 \, \text{m/s} \] (Equation 1)

### Step 4: Use the equations of motion to find the relationship between V_T and V_B
Using the second equation of motion, we can relate the final velocity (V_B), initial velocity (V_T), acceleration (g), and distance (S):
\[ V_B^2 = V_T^2 + 2gS \]
Substituting the known values:
\[ V_B^2 = V_T^2 + 2 \times 9.8 \times 3 \]
\[ V_B^2 = V_T^2 + 58.8 \] (Equation 2)

### Step 5: Express V_B in terms of V_T
From the first equation (Equation 1), we can express V_B in terms of V_T:
\[ V_B = 12 - V_T \]
Now substitute this expression into Equation 2:
\[ (12 - V_T)^2 = V_T^2 + 58.8 \]
Expanding the left side:
\[ 144 - 24V_T + V_T^2 = V_T^2 + 58.8 \]
Subtracting \( V_T^2 \) from both sides:
\[ 144 - 24V_T = 58.8 \]
Rearranging gives:
\[ 24V_T = 144 - 58.8 \]
\[ 24V_T = 85.2 \]
Dividing both sides by 24:
\[ V_T = \frac{85.2}{24} \approx 3.55 \, \text{m/s} \]

### Step 6: Calculate V_B using V_T
Now substitute V_T back into Equation 1 to find V_B:
\[ V_B = 12 - V_T \]
\[ V_B = 12 - 3.55 \approx 8.45 \, \text{m/s} \]

### Conclusion
Thus, the speeds of the ball at the top and bottom of the window are:
- \( V_T \approx 3.55 \, \text{m/s} \)
- \( V_B \approx 8.45 \, \text{m/s} \)