A balloon is moving vertically upward with constant acceleration (g/2) in upward direction Particle 'A' was dropped from the balloon and 2 sec later another particle 'B' was dropped from the same balloon. Assume that motion of the balloon remains unaffected. Find the separation distance between 'A' and 'B', 6 sec after dropping the particle 'B'. None of the particles reaches the ground during the time interval under consideration (g=10 m/`sec^2` )

Explanation

To solve the problem, we will analyze the motion of both particles A and B dropped from the balloon, which is moving upward with a constant acceleration of \( \frac{g}{2} \) (where \( g = 10 \, \text{m/s}^2 \)).

### Step-by-Step Solution:

1. **Determine the Balloon's Motion**:
- The balloon is moving upward with an acceleration of \( \frac{g}{2} = \frac{10}{2} = 5 \, \text{m/s}^2 \).
- The initial velocity of the balloon is \( u \) (which we will assume is unknown for now).

...