A rock is tossed straight up with a velocity of 37.1 m/s. When it returns, it falls into a hole 18.8 m deep. What is the rock's velocity as it hits the bottom of the hole?
The gain in kinetic energy equals the loss of potential energy.
(1/2) M (V2^ - V1^2) = M g *(18.8)
The mass M cancels out and V1 = 37.1 m/s is the initial velcoity
Solve for V2, the final velocity
V2^2 = 2 g*18.8 + V1^2