Question

A hot air balloon is attached to a spool of rope that is 125 ft away from the balloon when it's on the ground. The hot air balloon rises straight up in such a way that the length of rope increases at a rate of 15 ft/sec. How fast is the hot air balloon rising 20 seconds after it lifts off?

Answer 1

s= 125 ft, s' = 15 ft/sec, t = 20 sec
v' ?

Answer 2

What formula do I use?

Answer 3

I don't get it.

Answer 4

S is distance, v is velocity, t is time!
V = ?

Answer 5

Oh,v= d/t sorry

Answer 6

Oh okay

Answer 7

v'=(t-s)/t^2

Answer 8

Yeah, [g(x)*f'(x)-f(x)*g'(x)] / g(x)^2 Right?

Answer 9

Where's your f'(x) and g'(x) in: v'=(t-s)/t^2?

Answer 10

Oh, sorry. [t(s')-s(t')]/t^2

Answer 11

Yeah, but then I don't have the value for t'

Answer 12

Oh, I get it. Thanks. :)

Answer 13

So the answer would be v'=0.4375ft^3/sec?

Answer 14

v'=0.4375ft/sec

Answer 15

Okay, I see now. Thank you.