Let the upward direction be positive. For the first particle projected upwards:

Using the equation of motion: \(s_1 = ut - \frac{1}{2}gt^2\)

\(s_1 = 12.5t - \frac{1}{2}gt^2\)

For the second particle released from rest:

Using the equation of motion: \(s_2 = \frac{1}{2}gt^2\)

Since the total distance between the particles is 10 m, we have:

\(s_1 + s_2 = 10\)

Substituting the expressions for \(s_1\) and \(s_2\):

\(12.5t - \frac{1}{2}gt^2 + \frac{1}{2}gt^2 = 10\)

\(12.5t = 10\)

\(t = 0.8 \text{ s}\)

Substitute \(t = 0.8\) back into the equation for \(s_1\):

\(s_1 = 12.5 \times 0.8 - \frac{1}{2} \times 9.8 \times (0.8)^2\)

\(s_1 = 10 - 3.2\)

\(s_1 = 6.8 \text{ m}\)

Therefore, the height above O at which the particles meet is 6.8 m.