(a) To find the speed of P when it is 10 m above the ground, use the equation of motion:

\(v^2 = u^2 + 2as\)

where \(u = 15 \text{ m s}^{-1}\), \(a = -g = -9.8 \text{ m s}^{-2}\), and \(s = 10 \text{ m}\).

Substitute the values:

\(v^2 = 15^2 - 2 \times 9.8 \times 10\)

\(v^2 = 225 - 196\)

\(v^2 = 29\)

\(v = \sqrt{29} \approx 5 \text{ m s}^{-1}\)

(b) For the collision, use the equations of motion for both particles:

For P: \(s_P = 15t - \frac{1}{2}gt^2\)

For Q: \(s_Q = 18 - \frac{1}{2}gt^2\)

Set \(s_P + s_Q = 18\) and solve for \(t\):

\(15t - \frac{1}{2}gt^2 + 18 - \frac{1}{2}gt^2 = 18\)

\(15t = gt^2\)

\(t = \frac{15}{g}\)

Substitute \(t\) back into \(s_P\):

\(s_P = 15 \times \frac{15}{g} - \frac{1}{2}g \left(\frac{15}{g}\right)^2\)

\(s_P = 10.8 \text{ m}\)