a rock is dropped from a height of 20m above the ground when it hits the ground it leaves a hole 10cm deep. With what speed does the rock hit the ground, what is the magnitude of the rock's deceleration after it hits the ground

Answer 1

a. V^2 = Vo^2 + 2g*h. Vo = 0, g = 9.8 m/s^2, h = 20 m, V = ?.

b. V^2 = Vo^2 + 2a^d = 0. V = 0, Vo = V calculated in part a, d = 0.10 m, a = ?.