The way you worded the question,
"How quickly is it moving away from the observer at that moment" , implies to find the rate at which the distance between the balloon and the observer changes.
So we could just use Pythagoras
after t sec, height of balloon is 8t ft
so if d is the distance between them, (the hypotenuse)
d^2 = 150^2 + (8t)^2
2d dd/dt = 2(8t)(8) = 128t
dd/dt = 64/d
when 8t = 250
t = 250/8 = 125/4
d^2 = 22500 + 62500
d = √8500 = 10√85
dd/dt = 64/(10√85) = appr .694.. ft/sec
An observer 150 feet from point C, where a weather balloon has been released. The balloon rises vertically at a rate of 8ft/sec/. At the very instant the balloon is 250 above ground how far is it from the observer? How Quickly is it moving away from the observer at that moment?
All answers I see for problems like this include sin tan cos but we did not learn this in class so I do not know how to solve this problem.
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