The question I'm having trouble with is:
A paratrooper steps out of an airplane at a height of 1000 ft and after 5 seconds opens her parachute. Her weight (with equipment) is 195 lbs. Let $y(t)$ denote her height above the ground after $t$ seconds. Assume the force due to air resistance is $0.005y'(t)^2 $ lbs in free fall and $0.6y'(t)^2$ lbs with the chute open.
- At what height does the chute open?
- How long does it take to reach the ground?
- At what velocity does she hit the ground?
So far I know that I need to use the equation $my''(t)=mg-ky'(t)$
I have $m=195/32$ and $ky'(t)$ a piecewise function depending on t (greater than or less than 5 sec). I'm wondering if I'm missing something because I don't have the initial height anywhere in my equation, and am confused on how to answer the first question without the initial height. I was also hoping someone could help me understand why the resistance in the prompt has $y'(t)^2$ and not $y'(t)$.