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a bungee cord jumper jumps off the new river bridge. the bridge is 876 feet high. How fast is the jumper falling when he reaches 200 feet above the ground?

  • physics -

    V^2 = Vo^2 + 2g*d
    V^2 = 0 + 64*(876-200)
    V^2 = 43,264
    V = 208 Ft/s

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