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anonymous

  • 5 years ago

an observer stands 200 meters from the launch site of a hot-air balloon. balloon rises 4m/sec at a constant rate. how fast is the angle of elevation of the balloon increasing 30sec after the launch?

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  1. anonymous
    • 5 years ago
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    after 30sec, the balloon reaches a height of 120m. the angle is then: \[\alpha=\tan^{-1} (120/200) = 31degrees\]

  2. anonymous
    • 5 years ago
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    1.36

  3. anonymous
    • 5 years ago
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    sorry about the 1.36- I had a bad connection and couldnt finish. \[d \theta/dt=4m/(200\sec ^{2}\theta)\] and \[\sec ^{2}\theta=1.36\]

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