# Forming differential equation with air resistance

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 $y'(t)$.

• You have 2 differential equations. The first applies in the time interval 0 to 5 and the second after t = 5. You have the initial height for the first DE. Question 1 gives you the initial height for the second DE. The resistance being proportional to $v^2$ is just a model of resistance - linear in $v$ is another one which may be more realistic at low velocity. – Paul Oct 29 '17 at 22:27
• What equation should I use to answer question #1? I'm not sure where to input 1000ft into the equation I have ($my''(t) = mg - ky'(t)$). Thank you for your help! – Henry Oct 30 '17 at 3:14
• it's your initial condition: $y(0) = 1000$ (or if you are going to call "down" the positive direction, as it seems you are inclined to do, then $-1000$ would be better) – Nick Pavlov Oct 30 '17 at 11:27
• You still seem to have the mistaken belief that the equation is linear. By the task, it is $my''=-mg+c(y')^2$. Notice that the gravitation force points down, toward zero height, and the air resistance works against the direction of the movement, properly it should be $-c|y'|\,y'$ which for negative $y'$ gives the plus sign. Solve via separation or as Bernoulli equation. – Lutz Lehmann Oct 30 '17 at 12:16

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).