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anonymous
 5 years ago
an observer stands 200 meters from the launch site of a hotair balloon. balloon rises 4m/sec at a constant rate. how fast is the angle of elevation of the balloon increasing 30sec after the launch?
anonymous
 5 years ago
an observer stands 200 meters from the launch site of a hotair 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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anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0after 30sec, the balloon reaches a height of 120m. the angle is then: \[\alpha=\tan^{1} (120/200) = 31degrees\]

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0sorry 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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