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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 - ,

    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.

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