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 analyze the motion of the ball falling from the building and apply the equations of motion.

### Step 1: Understand the Problem
The ball is dropped from the top of a building and falls through a window of length 3 meters in 0.5 seconds. We need to find the speeds at the top (Vt) and bottom (Vb) of the window.

### Step 2: Identify the Initial Velocity
Since the ball is dropped, its initial velocity (u) is 0 m/s.

### Step 3: Use the Equation of Motion to Find Vt
We can use the equation of motion:
\[ V_t^2 = u^2 + 2as \]
Where:
- \( V_t \) = speed at the top of the window
- \( u \) = initial velocity = 0 m/s
- \( a \) = acceleration due to gravity = 9.8 m/s²
- \( s \) = distance fallen = 3 m

Substituting the values:
\[ V_t^2 = 0 + 2 \cdot 9.8 \cdot 3 \]
\[ V_t^2 = 58.8 \]
\[ V_t = \sqrt{58.8} \]
\[ V_t \approx 7.67 \, \text{m/s} \]

### Step 4: Find Vb Using the Time of Fall
Now we will find the speed at the bottom of the window (Vb). We can use the equation:
\[ V_b = u + at \]
Where:
- \( t \) = time taken to fall through the window = 0.5 s

Substituting the values:
\[ V_b = 0 + 9.8 \cdot 0.5 \]
\[ V_b = 4.9 \, \text{m/s} \]

### Step 5: Calculate the Total Velocity at the Bottom of the Window
Now we can find the total velocity at the bottom of the window:
\[ V_t + V_b = 7.67 + 4.9 \]
\[ V_t + V_b \approx 12.57 \, \text{m/s} \]
This can be approximated to 12 m/s.

### Conclusion
The speeds at the top and bottom of the window are:
- \( V_t \approx 7.67 \, \text{m/s} \)
- \( V_b \approx 4.9 \, \text{m/s} \)
- The sum of the speeds \( V_t + V_b \approx 12 \, \text{m/s} \)