Solution:

First walnut (W1): Time to fall 5m from rest is \(t_1 = \sqrt{2s/g} = \sqrt{2 \cdot 5/10} = 1\) s. Total time for W1 to reach ground from 20m is \(t_{total} = \sqrt{2 \cdot 20/10} = 2\) s. Second walnut (W2) must fall 15m in \(t_{W2} = t_{total} - t_1 = 2-1 = 1\) s. Using \(s = v_0t + (1/2)gt^2\): \(15 = v_0(1) + (1/2)(10)(1)^2 \Rightarrow 15 = v_0 + 5 \Rightarrow v_0 = 10\text{ m/s}\).