A rock is tossed straight up with a velocity of 32.1 m/s. When it returns, it falls into a hole 20.1 m deep. What is the rock's velocity as it hits the bottom of the hole?
Physics - drwls, Monday, October 4, 2010 at 12:46pm
The rock gains kinetic energy equal to the potential energy lost falling down the hole.
(M/2)[(Vfinal)^2 - 32.1)^2] = M*g*(depth)
Solve for Vfinal
The answer would be the same whether the rock was thrown up or down, initially.