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)).