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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?

  • physics -

    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

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