Asked by Gladys Garcia
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?
Answers
Answered by
Steve
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,
2ā(100^2+160^2) ds/dt = 2(160)(2)
ds/dt = 16/ā89 m/s
s^2 = 100^2 + h^2
2s ds/dt = 2h dh/dt
we know dh/dt = 2, so when the balloon is 160m up,
2ā(100^2+160^2) ds/dt = 2(160)(2)
ds/dt = 16/ā89 m/s
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