If you're uncomfortable using the conclusion mrjeffy made, break the problem up into finer steps: 1) Going up, then 2) coming down.
Throughout each of these steps, acceleration is constant and remains a=g~=-9.8m/s^2
1a) Find the time (call it t_1) it takes for the rock to go as high as it will go:
When the rock starts it has v_i1=20m/s. When it stops at the top and turns to come back down, it has a v_f1=0m/s.
v_{f,1} = v_{i,1} + at_1
1b) using this t_1, v_i1, v_f1 and a: find the distance the rock traveled, call it d_1.
d_1 = v_{i,1} t_1 + \frac{1}{2}at_1^2
2a) Now the rock is still at the top of the path. We know it's d_1 m above the ground. It now has to fall d_2 = d_1 + 10 m. It starts out at v_i2 = 0m/s (suspended in the air momentarily, before it turns and heads down).
Use d_2, v_i2, and a to find t_2, the time it takes the rock to fall a distance of d_2 (from the top, down to the bottom of the hole).
d_2 = v_{i,2} t_2 + \frac{1}{2}at_2^2
2b) Now that we know the time it takes we can find v_f2, the velocity of the rock at the bottom of the hole.
v_{f,2} = v_{i,2} + at_2
