Solution:

Concept: Motion under gravity and meeting condition.
Formula: \(h = ut + \frac{1}{2}gt^2\).
Solution: For ball 1 (dropped at \(t=0\)): \(h_1 = \frac{1}{2}g(18)^2 = 162g\). For ball 2 (thrown at \(t=6\text{ s}\), travels for \(12\text{ s}\)): \(h_2 = v(12) + \frac{1}{2}g(12)^2 = 12v + 72g\). When they meet, \(h_1 = h_2\): \(162g = 12v + 72g\). \(90g = 12v\). Using \(g=10\text{ m/s}^2\), \(900 = 12v\) => \(v = 75\text{ m/s}\).