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` Determine when and where one overtakes the other. Take `g=9.8 m s^(-2)`.

Explanation

To solve the problem step by step, we will analyze the motion of both particles and determine when and where one overtakes the other.

### Step 1: Understand the motion of the first particle
The first particle is projected vertically upwards from point A, which is 80 m above the ground, with an initial velocity \( u = 29.4 \, \text{m/s} \). The acceleration due to gravity \( g = 9.8 \, \text{m/s}^2 \) acts downwards.

The position of the first particle at time \( t \) can be given by the equation of motion:
\[
s_1 = u t - \frac{1}{2} g t^2
...