The balloon leaves the ground 80 ft. from an observer and rises vertically upward at 5 ft/s.

Question

(a) How fast is the balloon receding from the observer after 12 seconds?
(b) find the corresponding acceleration?

How do I sketch it?
and what equation should I use?
Please show the solution/s

Thank You~!

Answer 1

draw a right-angled triangle (side view)
mark the base as 80, and the height of the balloon as h

a) you want to know how fast the balloon is receding, so we are talking about how fast the hypotenuse is changing.
let the hypo be x

x^2 = 80^2 + h^2
2x dx/dt = 0 + 2h dh/dt, but we know dh/dt = 5

when t = 12, h = 12(5) or 60 ft
x^2 = 80^2 + 60^2
x = 100

so in 2x dx/dt = 2h dh/dt
2(100) dx/dt = 2(60)(5)
dx/dt = 3

a) 3 ft/s

Answer 2

Thank You!!