(i) For particle P, using the equation of motion under gravity:

\(0 = V - gt\)

\(At t = 2 s, the velocity is zero, so:\)

\(0 = V - g \times 2\)

Solving for V gives:

\(V = 2g\)

\(Assuming g = 10 m s^-2, we find:\)

\(V = 20 \text{ m s}^{-1}\)

(ii) For particle Q, using the equation of motion:

\(v = u + gt\)

\(Since Q starts from rest, u = 0, and at t = 4 s:\)

\(v = 0 + 10 \times 4 = 40 \text{ m s}^{-1}\)

(iii) The height h from which Q was dropped can be found using:

\(h = \frac{1}{2}gt^2\)

\(Substituting t = 4 s:\)

\(h = \frac{1}{2} \times 10 \times 4^2 = 80 \text{ m}\)