From a building two balls `A` and `B` are thrown such that `A` is thrown upwards and `B` downwards ( both vertically with the same speed ). If `v_(A) and v_(B)` are their respective velocities on reaching the ground , then

Options

A. `v_(B) gt v_(A)`
B. `v_(A) = v_(B)`
C. `v_(A) = v_(B)`
D. their velocities depend on their masses.

Explanation

To solve the problem, we need to analyze the motion of both balls A and B when they are thrown from the building.

### Step-by-Step Solution:

1. **Identify the Initial Conditions:**
- Let the initial speed of both balls (A and B) be \( u \).
- Ball A is thrown upwards, and Ball B is thrown downwards.

2. **Analyze Ball A:**
- When ball A is thrown upwards, it will first move up until it reaches its highest point and then fall back down to the ground.
- The height of the building is \( h \).
- At the highest point, the velocity of ball A becomes 0.
- The total distance it travels upwards is \( h + h' \) (where \( h' \) is the height it reaches above the building).
- Using the third equation of motion:
\[ v_A^2 = u^2 - 2g(h + h') \]
- When it falls back down to the ground, it will cover the height \( h + h' \) again.

3. **Calculate Final Velocity for Ball A:**
- The final velocity \( v_A \) when it reaches the ground can be expressed as:
\[ v_A^2 = u^2 + 2gh \]
- Therefore, we can write:
\[ v_A = \sqrt{u^2 + 2gh} \]

4. **Analyze Ball B:**
- Ball B is thrown directly downwards with the same initial speed \( u \).
- The distance it travels to reach the ground is \( h \).
- Using the third equation of motion:
\[ v_B^2 = u^2 + 2gh \]
- Therefore, the final velocity \( v_B \) when it reaches the ground can be expressed as:
\[ v_B = \sqrt{u^2 + 2gh} \]

5. **Compare the Velocities:**
- From the equations derived:
\[ v_A = \sqrt{u^2 + 2gh} \]
\[ v_B = \sqrt{u^2 + 2gh} \]
- This implies:
\[ v_A = v_B \]

### Conclusion:
The velocities of both balls A and B when they reach the ground are equal:
\[ v_A = v_B \]