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a balloon is rising vertically at the rate of 2 m/s. an observer is standing on the ground 100m away from a point directly below the balloon. at what rate is the distance between the observer and the balloon changing when the balloon is 160m high?

  • Math -

    the distance s of the balloon at height h is given by

    s^2 = 100^2 + h^2

    2s ds/dt = 2h dh/dt
    we know dh/dt = 2, so when the balloon is 160m up,

    2√(100^2+160^2) ds/dt = 2(160)(2)
    ds/dt = 16/√89 m/s

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