Why justify an illegitimate attack with a legitimate response?

Once again, acceleration will go down with respect to time.

Nevermind, we are essentially saying the same thing. The object's acceleration will go up, which is what you are talking about, and the delta acceleration will go down which is what I was talking about. I still don't agree with your numbers though.

Quote from: cbarnett97 on August 21, 2008, 05:28:59 AMQuote from: TheEngineer on August 20, 2008, 12:29:05 AMRelative to the Earth? When your upwards acceleration equals the acceleration of the Earth.But, I've said this already.Made this using LaTex and I forgot to resize the page so sorryYour 'FE' equation is wrong. Since you let us all have a good laugh, I'll go ahead and post the derivation for you.The drag force on a body in a fluid is given by F = .5*C_{d}*v^{2}*A*rho (1)where C_{d} is the drag coefficient, v is the velocity of the object, A is the cross sectional area of the object and rho is the density of the fluid.We know from Newton's Second Law of Motion thatF = m*a_{o} (2)where F is force, m is the mass of the object and a_{o} is the acceleration of the object. Rearranging (2) gives usa_{o} = F/m (3)Combining (3) and (1) gives a_{o} = C_{d}*v^{2}*A*rho/(2*m) (4)Now, an object that has reached terminal velocity, v_{t}, has no relative acceleration to the Earth. Applying this to the FE, that means that the object must have an upwards acceleration equal to that of the Earth's. It follows then, thata_{e} - a_{o} = 0 (5)where a_{e} is the acceleration of the Earth and a_{o} is the acceleration of the object. Rearranging (5) givesa_{e} = a_{o} (6)which, by examination, is correct for all objects whose height is not changing relative to the Earth (aircraft for example). Combining (6) and (4) leaves us with the equationa_{e} = C_{d}*v_{t}^{2}*A*rho/(2*m) (7)Performing simple algebra on (7) yields the equationv_{t} = sqrt(2*a_{e}*m/(Cd*A*rho) (8)which can easily been seen to be the exact same equation as the one for the RE, with the exception that a_{e} refers to the acceleration of the Earth and a in the RE equation refers to the acceleration due to gravity.Using your numbers, we see that the terminal velocity on the FE is 22.1426 m/s.p.s. Your RE equation is wrong.

Quote from: TheEngineer on August 20, 2008, 12:29:05 AMRelative to the Earth? When your upwards acceleration equals the acceleration of the Earth.But, I've said this already.Made this using LaTex and I forgot to resize the page so sorry

Relative to the Earth? When your upwards acceleration equals the acceleration of the Earth.But, I've said this already.

I got 0 m/s^{2} relative acceleration at t=2.26

Quote from: TheEngineer on August 21, 2008, 09:53:44 AMQuote from: cbarnett97 on August 21, 2008, 05:28:59 AMQuote from: TheEngineer on August 20, 2008, 12:29:05 AMRelative to the Earth? When your upwards acceleration equals the acceleration of the Earth.But, I've said this already.Made this using LaTex and I forgot to resize the page so sorryYour 'FE' equation is wrong. Since you let us all have a good laugh, I'll go ahead and post the derivation for you.The drag force on a body in a fluid is given by F = .5*C_{d}*v^{2}*A*rho (1)where C_{d} is the drag coefficient, v is the velocity of the object, A is the cross sectional area of the object and rho is the density of the fluid.We know from Newton's Second Law of Motion thatF = m*a_{o} (2)where F is force, m is the mass of the object and a_{o} is the acceleration of the object. Rearranging (2) gives usa_{o} = F/m (3)Combining (3) and (1) gives a_{o} = C_{d}*v^{2}*A*rho/(2*m) (4)Now, an object that has reached terminal velocity, v_{t}, has no relative acceleration to the Earth. Applying this to the FE, that means that the object must have an upwards acceleration equal to that of the Earth's. It follows then, thata_{e} - a_{o} = 0 (5)where a_{e} is the acceleration of the Earth and a_{o} is the acceleration of the object. Rearranging (5) givesa_{e} = a_{o} (6)which, by examination, is correct for all objects whose height is not changing relative to the Earth (aircraft for example). Combining (6) and (4) leaves us with the equationa_{e} = C_{d}*v_{t}^{2}*A*rho/(2*m) (7)Performing simple algebra on (7) yields the equationv_{t} = sqrt(2*a_{e}*m/(Cd*A*rho) (8)which can easily been seen to be the exact same equation as the one for the RE, with the exception that a_{e} refers to the acceleration of the Earth and a in the RE equation refers to the acceleration due to gravity.Using your numbers, we see that the terminal velocity on the FE is 22.1426 m/s.p.s. Your RE equation is wrong.

You seem to have forgotten to tell us all the part where my derivation is wrong. How can you be confident in your numbers when I have shown them to be wrong already? Besides, your RE equation is wrong.

A_{net}=A_{E}-A_{D}=9.8-.02(2.26*9.^{2}=00.02 is all the stated variables other than v simplified.

Stupid smiley faces, that is supposed to be 9.8 )

That isn't a constant, it is v=at at t=2.26.

Our existentialist, relativist, nihilist, determinist, fascist, eugenicist moderator hath returned.

objectively good

It seems that cbarnett suffers from sokarulitis as well; everyone else is wrong, and he is right.

Where did you educate the biology, in toulet?

Quote from: Robbyj on August 23, 2008, 07:33:09 PMThat isn't a constant, it is v=at at t=2.26.but you cannot just the RE answer and show that it is the same in FE you need to show why it is the same

I've noticed that. I wonder if he lives in the same little world as sokarul or if they each have their own little worlds to live in.

Quote from: cbarnett97 on August 23, 2008, 07:35:49 PMQuote from: Robbyj on August 23, 2008, 07:33:09 PMThat isn't a constant, it is v=at at t=2.26.but you cannot just the RE answer and show that it is the same in FE you need to show why it is the sameV=at regardless of FE or RE.A_{net}=A_{Earth}-A_{Object}What, in your opinion am I missing?

Quote from: Robbyj on August 23, 2008, 07:44:07 PMQuote from: cbarnett97 on August 23, 2008, 07:35:49 PMQuote from: Robbyj on August 23, 2008, 07:33:09 PMThat isn't a constant, it is v=at at t=2.26.but you cannot just the RE answer and show that it is the same in FE you need to show why it is the sameV=at regardless of FE or RE.A_{net}=A_{Earth}-A_{Object}What, in your opinion am I missing?You are messing up with regards to the acceleration of the object, which is of course my whole point

Quote from: cbarnett97 on August 23, 2008, 07:46:36 PMQuote from: Robbyj on August 23, 2008, 07:44:07 PMQuote from: cbarnett97 on August 23, 2008, 07:35:49 PMQuote from: Robbyj on August 23, 2008, 07:33:09 PMThat isn't a constant, it is v=at at t=2.26.but you cannot just the RE answer and show that it is the same in FE you need to show why it is the sameV=at regardless of FE or RE.A_{net}=A_{Earth}-A_{Object}What, in your opinion am I missing?You are messing up with regards to the acceleration of the object, which is of course my whole pointBut we are using the same equation you used in your model.

reverse drag

I am assuming you are going to tell me...