How long will a ball of paper take to fall from a height of 45 meters?

[Aureus]

Alright, so, first of all, I created this problem myself. I made this to study, so I considered it as my homework.
1. How long will a crumpled ball of paper that is 3 inches in diameter take to fall 45 meters IN FREEFALL, knowing that the sheet of paper used to create the ball was 210mm x 297mm and that its weight is 75g/m²?

OK, so, obviously, there MUST be some equation out there that would tell me this. But I still wanted to figure it out myself. So here's what I did:

1. Calculate the ball's terminal velocity with Vt = SQRT((2mg):(pACd)) , where m is the paper ball's mass; g is the gravitational force exerted upon it (in our case, 9.81m/s² or just 10m/s², to simplify); p is the density of the fluid the paper ball is falling through (air, with a density of 0.0013g/cm³); A is the surface area of the paper ball (we can figure that out with A = 4*π*r²(yes that's a pi symbol)); Cd is the drag coefficient (for a sphere, which our paper ball would approximately be, the drag coefficient is 0.47 (note that this does NOT have a unit)); and finally, Vt is the terminal velocity.

1.1 Figure out the mass of the paper ball:

0.297m * 0.210m = 0.06237m²
That's the supposed volume of a sheet of paper.
Now, with 75g/m² we can find out how much 0.06237m² weighs, with what we (from Brazil) call "The rule of three":

g_|_ m²
75| 1
x | 0.06237

Multiply 75 with 0.06237 and 1 with x. Yes, I did put parenthesis around the units so it wouldn't get confusing.
1(m²)x(g) = 75(g) * 0.06237(m²)
x(g) = 4.67775(g)
Thus, the mass of the sheet of paper (and supposedly also the paper ball) is 4.67775 grams.

1.2 p is already given (with research :P), Cd is already given and g is already given, so next we'll tackle A.
Formula for A:
A = 4*π*r²
where r is the radius of the sphere and A is the surface area. Easily solved like this:
A = 4*π*1.5²(in)
A = 12.566 * 2.25 (in²)
A = 28.274(in²)

1.3 So now we're ready for the big formula:

Vt = SQRT((2mg):(pACd))
Vt = SQRT((2*4.67775g*10m/s²):(0.0013g/cm³*28.274in²*0.47))
Vt = SQRT((93.555gm/s²):(0.017275414gin²/cm³)
Vt = SQRT(5415.4997) [NOTE: By here I lost myself with the units.]
Vt = 73.59

Thus, the terminal velocity for the paper ball would be 73.59m/s possibly (I'm guessing the units, sorry.)

2. OK, so, somehow, I got to 56.55m/s on the first try, on the second I got 73.59m/s (Once again, guessing the units.). With that, I calculated how long it would fall with common sense. Once it fell 1s, it would be with a speed of 10m/s, 2s it'd be with a speed of 20m/s, by then it would be 30m down; on the 3rd second, it'd be stopped halfway through (15m down) by the ground.
Thus, I had the theory: once I dropped the paper ball, 3½ seconds later it would touch the ground.

3. I tried it out. I dropped the ball and counted how long it took with the classical "One mississippi, two mississippi, etc.". Once I finished saying "Six mississippi" it touched the ground.

Now, please help me: Obviously it got slowed down because of air resistance, but what formula should I use for that?

Oh, and, P.S.: I'm 14, so please try to keep technical talk to a minimum (pretend you're writing on Simple English Wikipedia (just kidding)).

 

[LowlyPion]

Welcome to PF.

Your method is not going to be useful if you don't work in consistent units. So lose the inches. And keep everything in Standard Units - meters, kilograms, seconds ... because these are the units of gravity, etc, that you want to use.

Secondly with the terminal velocity your primary interest is in weight and the physical characteristics like cross section. Whether your wad of paper is crumpled to 5 cm or 2.5 cm it will weigh the same. But given the surface roughness of your wad of paper and the ease that it may deform in falling, I think determining the C of this sphere looks like it will need to be observationally determined in any event by judging the speed at different distances, or by measuring times to fall from different heights.

Here are some links that might be useful to you:
http://www.grc.nasa.gov/WWW/K-12/airplane/falling.html
http://www.grc.nasa.gov/WWW/K-12/airplane/termv.html

 

[Aureus]

Thanks :)

Alright, so I'll attempt again (losing the inches, as you said) and I'd try experimenting to see what the drag coefficient was, but I lost the paper when I dropped it.
So anyway, if I fix these two, should the whole equation work?
And, isn't there a simpler way to find the same result? I attempted Kinematic equations, but that also gave me 3 seconds (actually, 3.03). I'll keep searching for a formula that will work, I think there is one in my math book :)
Thanks again.

 

[LowlyPion]

Thanks :)

Alright, so I'll attempt again (losing the inches, as you said) and I'd try experimenting to see what the drag coefficient was, but I lost the paper when I dropped it.
So anyway, if I fix these two, should the whole equation work?
And, isn't there a simpler way to find the same result? I attempted Kinematic equations, but that also gave me 3 seconds (actually, 3.03). I'll keep searching for a formula that will work, I think there is one in my math book :)
Thanks again.

Unfortunately the only useful formula you have to work with, likely you are going to need to derive your drag coefficient from observation. Your problem is a bit of a chicken and egg kind of thing.

For instance you can observe what the terminal velocity becomes through careful testing and from that you can determine the coefficient that you want to use from the NASA site to then arrive again at a formula that satisfies your particular interest in a particular wad of paper. But that is hardly encouraging to the notion that you can figure how long it will take prior to dropping it for the first time.

 

[Aureus]

Well, alright then... Thanks for all your help, LowlyPion, but I guess I'm going to have to leave this one for later.
Thanks again :)

 

[negitron]

It's very easy to experimentally determine the terminal velocity: tie a thread (the thinner the better) to the ball of paper and attach the other end to the centerpoint of a protractor. Hang this out a car window on a windless day and have the driver accelerate until the thread hangs at a 45-degree angle. The speed of the car at this point equals the terminal velocity.

 

[KRA2008]

Negitron that is simplistically brilliant. I'm currently writing a computer program for throwing paper balls into trash cans and I need to know the drag coefficient of a paper ball (roughly). I'm going to go use your method.