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  • 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. amistre64
    • 5 years ago
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    Now heres a problem I can do :) think of it like a triangle problem: the base(x) is 200 m and the height(y) is increasing at a constant rate. And the angle(a) is changing at a rate of tan(a) The change in "y" with respect to time "t" is 4m per sec. So lets solve tan(a) = y/200 with repect to time and see what we get. tan(a) = y/200 (d/dt)(tan(a)) = (d/dt)(y/200) (da/dt) (sec^2(a)) = (dy/dt) (1/200) (da/dt) = (4)/(200 sec^2(a)) = 1/50sec^2(a) What does sec(a) equal at 30 sec? y=120m x=200 r=40sqrt(34) sec(a) = r/x = 40sqrt(35)/200 = sqrt(34)/5 (sqrt(34)/5)^2 = 34/25 (da/dt) = 1/(50(34/25)) (da/dt) = 1/(2(34)) (da/dt) = 1/68 radians per sec. But you might want to double check the work :)

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