Answer:
When the rock is dropped from a height of 9.0 m, its initial potential energy is given by
PE = mgh = (6.0 kg)(9.8 m/s^2)(9.0 m) = 534 J
We can solve for the height at which the kinetic energy of the rock is twice its potential energy by setting the kinetic energy equal to twice the potential energy and solving for h.
KE = 1/2 mv^2 = (1/2)(6.0 kg)(v^2)
2 * PE = 2 * (6.0 kg)(9.8 m/s^2)(h)
Substituting and solving for h yields
h = 15 m