From a point `A, 80 m` above the ground, a particle is projected vertically upwards with a velocity of `29 .4. m s^(-1)`, Five seconds later, another particle is dropped from a point `B, 34. 3 m` vertically below `A` Derermine when and where one voertakes the other. Take `g=9.8 m s^(-2)`.

Explanation

To solve the problem, we need to determine when and where one particle overtakes the other. We have two particles: one projected upwards from point A and another dropped from point B.

### Step-by-Step Solution:

1. **Identify the Initial Conditions**:
- Particle A is projected upwards from point A at a height of 80 m with an initial velocity \( u_A = 29.4 \, \text{m/s} \).
- Particle B is dropped from point B, which is 34.3 m below point A (thus at a height of \( 80 - 34.3 = 45.7 \, \text{m} \)).
- Particle B is dropped 5 seconds after Particle A is projected.
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