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A hot-air balloon is rising vertically at a constant speed, an observer at a distant observes the elevation angle to be 30° at...Asked by helpppp
A hot-air balloon is rising vertically at a constant speed, an observer at a distant observes the elevation angle to be 30° at 10:00am, at 10:10am the elevation angle becomes 34°, then at 10:30am the elevation angle of the balloon should be closest to (using the table below) (A) 34° (B) 39° (C) 41° (D) 42° (E) 43°
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Answered by
Anonymous
change in angle / change in time = 4 deg/ 10 min
change in time = 10:30 - 10:10 = 20 min
so
change in angle if linear = (4 deg / 10 min) * 20 min = 8 degrees
34 + 8 = 42 degrees
change in time = 10:30 - 10:10 = 20 min
so
change in angle if linear = (4 deg / 10 min) * 20 min = 8 degrees
34 + 8 = 42 degrees
Answered by
oobleck
a constant vertical speed does not yield a constant change in θ.
It yields a constant change in tanθ
It yields a constant change in tanθ
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