Forming differential equation with air resistance

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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.

  1. At what height does the chute open?
  2. How long does it take to reach the ground?
  3. 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 $y'(t)$.

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The initial height does not appear in the differential equation itself. After you solve a differential equation, the solutions will contain one or more integration constants, which can then be determined by requiring that initial conditions be satisfied.

In other words, the differential equation has a whole family of solutions, and your initial conditions, $y(0) = 1000$ and $y'(0) = 0$, will select a specific one among them.

Watch out for the units in the problem. The use of lbs as units for both mass and force is, in my opinion, a rather blatant abuse of units. Your writing $m = 195/32$ does do the trick, I guess, but I would have said that what is implied in the statement of the question is that we should take $m = 195$ and $g = -32$ (the minus to show that acceleration is down - that would be consistent with $y$ measuring "height", i.e. positive is up; in other words, also watch out for how you set up your forces in the diff equation with respect to directions up/down), and write the equation like this $$ \frac{m}{|g|}y''(t) = -m + ky'(t) $$ which is the same as what you get by taking $m = 195/32$ (at least as far as unit scaling is concerned - yours has the wrong signs in some places).