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Say I drop an object from a height z0. How long does it take the object to reach the ground? There are two frameworks to do this problem in, a simple framework and a more complex framework. I have answers for both, but I'm much more tentative about my answer for the complex framework.

Simply and ideally, we can assume the only force acting on the object is the gravitational field strength. Then the object's z position is given by

z(t) = -0.5gt^2 + V0t + z0

The initial velocity in the z-direction is 0, so

z(t) = -0.5gt^2 + z0

We want the time it takes for the object to fall from (x,y,z0) to (x,y,0). So we solve:

0 = -0.5gt^2 + z0

z0 = 0.5gt^2

2(z0) = gt^2

2(z0)/g = t^2

[2(z0)/g]^(1/2) = t

So, if I drop an object from rest at 6750 meters up and ignore everything but gravity, it would take about 37.12 seconds to reach the ground. (Am I right about this? Did I mess something up?)

Now, the more complex framework involves gravity and air resistance. I am unsure about my calculations here, because I had to derive everything myself and so things could be very wrong (I am basically feeling around in total darkness here, hoping to get something right). Anyway, the drag force acting on an object is given by Fd = 0.5CpAv^2 (I did not derive this, obviously!) where C is the drag coefficient, p is the desnity of the fluid (for air, p = 1.2 kg / m^3), A is the object's cross-sectional area, and v is the velocity at which the object is falling.

If I drop the object vertically from rest, the object's terminal speed (when it's no longer accelerating) would be when the force due to gravity is equal to the force due to air resistance (Am I right about this? Am I using the right terms?). The (magnitude of the?) force due to gravity is given by mg (Am I right about this!?). So

mg = 0.5CpA(vT)^2

2mg = CpA(vT)^2

2mg / (CpA) = (vT)^2

This would be the terminal velocity for the object (vT represents terminal velocity - sorry for the messy symbology!) Now if I play around with Newton's second law...

Fnet = ma = -mg + 0.5CpAv^2

That is, the net force on the object is the drag pushing it up and gravity pulling it down, if we restrict movement to only the z axis. To get the terminal velocity in there, I can rewrite this equation as:

ma = -mg + mg[CpA / (2mg)]v^2

ma = -mg + mg[1 / (vT)^2]v^2

ma = mg[-1 + (v^2 / (vT)^2)]

Now, to make the function nature more explicit...

a(t) = g[(v(t)^2 / (vT)^2) - 1]

This is the acceleration in the z axis at some time t. g and vT are constants, a(t) and v(t) are functions of time. I won't write down all the calculus after this part (I don't know how to show it on these forums), but basically

dv/dt = g[(v^2 / (vT)^2) - 1]

dv/[(v^2 / (vT)^2) - 1] = g dt

dv/[1 - (v^2 / (vT)^2)] = -g dt

Integrating both sides (the left from v(0) to v(t) and the right from 0 to t), I got

v(t) = (vT) tanh [(-gt)/(vT)]

Only one more integration to get to the position function! Integrating both sides with respect to t, this time indefinitely...

s(t) = -[(vT)^2 / g]ln [cosh[(-gt)/(vT)]] + C

At t = 0 in my model, the object should be at position (x,y,z0). So s(0) = z0, and

z0 = -[(vT)^2 / g]ln [cosh[0]] + C

z0 = 0 + C

C = z0

So the z-position function is

s(t) = -[(vT)^2 / g]ln [cosh[(-gt)/(vT)]] + z0

Now what makes me nervous is the negative sign in the cosh argument... Because eventually, this would lead to a negative number of seconds, which is nonsense. So I'm sure I messed up somewhere, either in my calculus or in my understanding of physics. That said, setting s(t) = 0 and solving for t:

[(vT)^2 / g]ln [cosh[(-gt)/(vT)]] = z0

g(z0) / (vT)^2 = ln [cosh[(-gt)/(vT)]]

e^[g(z0) / (vT)^2] = cosh[(-gt)/(vT)]

And so on (it's getting too messy to continue!). Anyway, let's return to that object I dropped from 6750 meters up. Let's just say it has a terminal velocity vT = 60 m / s (or we could say what its shape, mass and cross-sectional area is, and find vT that way). Using the stuff I derived, with gravity and air resistance acting on it, it looks like it would take -116.74 seconds to reach the ground from 6750 meters up. Now obviously, as I said, this is wrong (a negative second?), but assuming the problem is just in my calculus and a sign got mixed up in there, does the figure of ~ 117 seconds sound right? This is about pi times the original estimate that didn't factor in air resistance (I do not know why it is about pi times the original estimate, this frightened me deeply). Is that a bit too big? Or can air resistance have that significant an effect?

Sorry for posting so much disorganized junk - maybe in a few months my physics will be a bit more beautiful.

Thanks for any and all help!