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Approximations would help too.

Thanks.

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- Thread starter bolini
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Approximations would help too.

Thanks.

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- #2

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bolini, you must show us some of your own work in order for us to help you. What are the relevant equations that determine the terminal velocity of an object? Tell us what you know.

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The equation for terminal velocity is the square root of 2mg/CpA. I used 9.8 m/s for gravity, .5 or the drag co on a ball and 1.29 kg/m3 for air density. So I come up with approx 33 m/s.

My question now is how do I come up with the drop height necessary to reach terminal velocity. Is there an acceleration formula that must be used?

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Not sure how to include resistance in the equation because it varies depending on the velocity which is changing until 33m/s is achieved.

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Not sure how to include resistance in the equation because it varies depending on the velocity which is changing until 33m/s is achieved.

Yeah, there are different models at different levels of complexity for air resistance. At the simple end is having it vary as the square of the velocity, and at the complex end (like what the battle tank computers use to compute trajectories) I'm sure there are many, many terms.

That's why I asked if you knew how they computed the simple terminal velocity formula. If you saw how they derived it, that would give you a clue as to which version of the air resistance force equation to use.

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I believe the first equation basically used air resistance as the square of velocity. I am curious now, even though we probably won't look for a building or bridge high enough, what the height actually would be. If you could help me incorporate air resistance into the kinematic equation I used, that would be helpful.

Thanks for your help up until this point. I forgot how interesting physics used to be for me.

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Actually, an object never reaches terminal velocity, it only approaches it. The solution to the time and distance it takes to 'approach' terminal velocity (say reach .99V_t) involves a non linear differential equation (mg - cpAv^2 = mdv/dt), the solution of which is beyond me at this point, but it involves a natural log term I think and probably 'e' raised to some power. But on a practical level, terminal velocity is ordinarily reached (approached) in less than 9 seconds (but much less for lighter objects like a feather or penny). So you are correct in that it would take a height much greater than 180 feet for the ball to attain a speed of 33m/s or so when you consider air resistance. I'm guessing that it might be around 500 feet or so.

I believe the first equation basically used air resistance as the square of velocity. I am curious now, even though we probably won't look for a building or bridge high enough, what the height actually would be. If you could help me incorporate air resistance into the kinematic equation I used, that would be helpful.

Thanks for your help up until this point. I forgot how interesting physics used to be for me.

I believe there is a site I found awhile back that calculates the speed of an object in air at various points in its fall, and gives the speed at impact based on the height it is dropped from (and the objects weight and projected area, etc.). I'll see if I can find it.

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this one is for rockets and only computes Vterm: http://exploration.grc.nasa.gov/education/rocket/termvr.html

I've thankfully only had one come in at terminal velocity, dug a hole and left a debris field of about 10 yeards radius. The circuit boards had all their components stripped off--not a pretty sight.

PS: the full eqn for terminal velocity=

sqrt(2mg/pACd)*{tanh(t)*sqrt[(g*p*Cd*a)/(2*m)]}

I've thankfully only had one come in at terminal velocity, dug a hole and left a debris field of about 10 yeards radius. The circuit boards had all their components stripped off--not a pretty sight.

PS: the full eqn for terminal velocity=

sqrt(2mg/pACd)*{tanh(t)*sqrt[(g*p*Cd*a)/(2*m)]}

Last edited:

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Cool! Very helpful, thanks for the link.

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