For the second particle, using the equation of motion:

\(s_2 = 20t - 0.5gt^2\)

For the first particle, which was projected 2 seconds earlier:

\(s_1 = 12(t + 2) - 0.5g(t + 2)^2\)

Setting \(s_1 = s_2\) to find when they collide:

\(12(t + 2) - 0.5g(t + 2)^2 = 20t - 0.5gt^2\)

Solving for \(t\):

\(t = \frac{1}{7} = 0.143 \text{ s}\)

For the height at which they collide:

Substitute \(t = \frac{1}{7}\) into the equation for \(s_2\):

\(s = 20 \times \frac{1}{7} - 5 \times \left(\frac{1}{7}\right)^2 \approx 2.755\)

Height is approximately 2.76 m.