Calculus

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

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  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

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  2. Thank You!!

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