(i) To find the time taken for particle P to return to the ground, we use the equation of motion:

\(s = ut + \frac{1}{2} a t^2\)

where \(s = 0\) (displacement is zero when it returns to the ground), \(u = 12\) m/s (initial velocity), and \(a = -g\) (acceleration due to gravity, \(g = 9.8\) m/s²).

Substituting the values, we get:

\(0 = 12t - \frac{1}{2} \times 9.8 \times t^2\)

Solving for \(t\), we find:

\(t = \frac{2 \times 12}{9.8} = 2.4 \text{ s}\)

(ii) For particle Q, it is projected at \(t = 1\) with speed 10 m/s from 5 m above the ground. We need to find when both particles are moving in the same direction.

Critical points occur when the velocity of each particle is zero. For P, this occurs at:

\(t = \frac{12}{9.8} = 1.2 \text{ s}\)

For Q, the critical point occurs at:

\(t = 1 + \frac{10}{9.8} = 2 \text{ s}\)

Thus, the particles move in the same direction in the intervals:

• \(1 < t < 1.2\)
• \(2 < t < 2.4\)