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Skylark

weight / mass / freefall speed

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Ok, to settle a disagreement with someone I love (who just happens to be a school-teacher) can someone tell me why heavier jumpers fall faster than others, when, according to science books, a brick and a paper-clip should fall at the same speed and hit the ground at the same time?



"Into the dangerous world I leapt..." William Blake, Songs of Experience

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like someone else said, they would fall at the same rate in a vacuum where there are no variables. in the sky what makes the difference is the air. as our bodies fall through the air we create drag/wind resistance, and when we reach a certain point the amount of drag matches how fast we are falling and we have reached terminal velocity. when a heavier person is falling through the air it takes more drag to reach terminal velocity(ie the air has to pull harder to slow that person down). more drag is created by more speed. therefore fatasses fall faster!!! i know its probably not worded the best way possible but it should clear it up a little for you
History does not long entrust the care of freedom to the weak or the timid.
--Dwight D. Eisenhower

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LOL I just watched Myth Busters on Discovery. The guy did a tandem jump (200 km/hr?) then let go of a couple of coins, which went up. In a specially-build coin-windtunnel the coins would not go any faster than 100 km/hr.
No idea what myth they were proving/debunking though, didn't watch the rest.

ciel bleu,
Saskia

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Remind your teacher that to prove that 2 items of different weights fall at the same rate IN A VACUME one one of the moon missions they droped a feather and a hammer and they landed at the same time.
"We've been looking for the enemy for some time now. We've finally found him. We're surrounded. That simplifies things." CP

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In a specially-build coin-windtunnel the coins would not go any faster than 100 km/hr. No idea what myth they were proving/debunking though, didn't watch the rest.



That was the myth that a coin tossed off the top of the Empire State building would accelerate to such a high speed that it would go through someone's skull on the sidewalk down below.

Myth busted!

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On a side note, if you solve the equation of motion assuming a quadratic velocity dependence and no variation in atmospheric density (not true, I know), you end up with a velocity function that is:

sqrt(m*g/k)*tanh[sqrt(k*g/m)*t]

m: mass
g: acceleration of gravity
k: coefficent of friction
t: time

The hyperbolic tangent goes asymptotically to one, so your terminal velocity is sqrt(m*g/k). Note how an increase in mass leads to a higher terminal velocity.
HF #682, Team Dirty Sanchez #227
“I simply hate, detest, loathe, despise, and abhor redundancy.”
- Not quite Oscar Wilde...

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disagreement with someone I love (who just happens to be a school-teacher)



An example of a beginning physics question: "Assume the horse is in a vacuum and is a perfect sphere..."

"Real world" physics goes so far off the map for complexity that my brain starts to overheat just thinking about thinking about it.
--"When I die, may I be surrounded by scattered chrome and burning gasoline."

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Tell them to think about it this way: Back in the day, round parachutes were used. The round chutes did not create any 'lift' like a square does. But....deploying your round chute still slowed you down after it opened.

The mass of the system in this case doesn't change as a result of deploying the canopy, but the drag does. If what your friend is telling you is true then when you deploy the round chute, you will not slow down at all.....that would make for one hell of a hard laningB|.
Flying Hellfish #470

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On a side note, if you solve the equation of motion assuming a quadratic velocity dependence and no variation in atmospheric density (not true, I know), you end up with a velocity function that is:

sqrt(m*g/k)*tanh[sqrt(k*g/m)*t]

m: mass
g: acceleration of gravity
k: coefficent of friction
t: time

The hyperbolic tangent goes asymptotically to one, so your terminal velocity is sqrt(m*g/k). Note how an increase in mass leads to a higher terminal velocity.



PhD Student you may be, but this makes no sense what-so-ever. This is a simple mechanics problem if you ignore air density.

In the simple state of a body, which is stable and falling towards the earth, the force pulling the body down is just the force of gravity, so the force is mass times gravity.

This force of gravity must be counteracted by the drag force on the body, which goes as the square of the velocity times the surface area times the coefficient of drag (coefficient of friction is ignoring many many things which are extremely important to the problem) times some miscellaneous constants. If two bodies have the same type of jumpsuit (same fit, same material, etc) then they will practically have the same coefficitn of drag. If they end up having the same surface area, then they will have the same drag force. Now if one person weighs more, then they will have a larger force of gravity, so the only variable in the above expression for drag is the velocity, so velocity must increase to counteract the downward force. Hence the body must speed up.

There is absolutely no reason to use a hyperbolic fuction here, not to mention it is incorrect.



I got a strong urge to fly, but I got no where to fly to. -PF

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a brick and a paper-clip should fall at the same speed and hit the ground at the same time? <<<<

OK... now ask your girl to drop a brick and a feather. The feather will drift or float. Its the air resistance thingy.:)
Which weighs more a pound of water, or a pound of gas? neither they both weigh a pound.

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If you are going the trouble of bashing someone elses answer, then how about also giving the correct answer???

Vt = terminal velocity
g = acceleration of gravity
m = mass
Cd = coefficient of drag
P = air density (rho)
A = area

Vt = sqrt(2 * g * m / (Cd * P * A))
"There are only three things of value: younger women, faster airplanes, and bigger crocodiles" - Arthur Jones.

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Which weighs more a pound of water, or a pound of gas? neither they both weigh a pound.



Ahhhh... but which weighs more, a pound of gold or a pound of feathers? To make it interesting... also say which weighs more; an ounce of gold or an ounce of feathers? :S

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>Which weighs more a pound of water, or a pound of gas?
>neither they both weigh a pound.

A trick question! An amount of gas and an amount of water that weighs the same on the moon will not weigh the same on earth. In other words, although both have equal mass, other effects (bouyancy, in this case) will cause the gas to weigh less when measured in an atmosphere. (This is easy to prove; see how much ten grams of helium in a balloon weighs on a local scale.)

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Which weighs more a pound of water, or a pound of gas? neither they both weigh a pound.



Ahhhh... but which weighs more, a pound of gold or a pound of feathers? To make it interesting... also say which weighs more; an ounce of gold or an ounce of feathers? :S



Avoirdupois? Troy?
If you can't fix it with a hammer, the problem's electrical.

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On a side note, if you solve the equation of motion assuming a quadratic velocity dependence and no variation in atmospheric density (not true, I know), you end up with a velocity function that is:

sqrt(m*g/k)*tanh[sqrt(k*g/m)*t]

m: mass
g: acceleration of gravity
k: coefficent of friction
t: time

The hyperbolic tangent goes asymptotically to one, so your terminal velocity is sqrt(m*g/k). Note how an increase in mass leads to a higher terminal velocity.



PhD Student you may be, but this makes no sense what-so-ever. This is a simple mechanics problem if you ignore air density.

In the simple state of a body, which is stable and falling towards the earth, the force pulling the body down is just the force of gravity, so the force is mass times gravity.

This force of gravity must be counteracted by the drag force on the body, which goes as the square of the velocity times the surface area times the coefficient of drag (coefficient of friction is ignoring many many things which are extremely important to the problem) times some miscellaneous constants. If two bodies have the same type of jumpsuit (same fit, same material, etc) then they will practically have the same coefficitn of drag. If they end up having the same surface area, then they will have the same drag force. Now if one person weighs more, then they will have a larger force of gravity, so the only variable in the above expression for drag is the velocity, so velocity must increase to counteract the downward force. Hence the body must speed up.

There is absolutely no reason to use a hyperbolic fuction here, not to mention it is incorrect.



The velocity vs. time function for an object freefalling at high Reynolds No. under gravity under conditions of constant air density IS the hyperbolic tangent.

dv/dt = A-B*v^2 where A and B are constants.

separate the variables and integrate --> v = C*tanh(...)

Jeez!! Even my cat can do that!
If you can't fix it with a hammer, the problem's electrical.

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The velocity vs. time function for an object freefalling at high Reynolds No. under gravity under conditions of constant air density IS the hyperbolic tangent.


I see that I should have paid more attention to this thread, but that it has been looked well after by others :)
@larsrulz: I was making an observation. If you think it irrelevant, feel free to not comment.

@ryoder: Where did you find that factor of 2? My k is by your definition equal to Cd*P*A

@yall: Should we stop sounding like the geeks everybody by now know that we are? :P
HF #682, Team Dirty Sanchez #227
“I simply hate, detest, loathe, despise, and abhor redundancy.”
- Not quite Oscar Wilde...

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Rather than go through all the discussion of the most-misquoted physics law of all time. I do it this way.

Fallrate is affected by weight and air-friction.

In a vacuum with no air friction, a person with a closed parachute and a person with an open parachute fall at the same rate. People can get their brains around that example easily.

Obviously, that is different in reality.

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@ryoder: Where did you find that factor of 2? My k is by your definition



It's been some years since I worked it all out. IIRC, it appeared due to taking the derivative of an equation that contained "V^2". I have a lot of pages in a notebook I would need to wade through to figure it all out again.
"There are only three things of value: younger women, faster airplanes, and bigger crocodiles" - Arthur Jones.

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You were right ryoder, stand your ground!!!

In the simplest form,

Drag = 1/2 * ro * V^2 * S * CD

Where ro is air density, V is relative velocity, S is wetted area, and CD is coefficient of drag of the body of interest. Calculating velocity necessary for drag to equal weight (m * g) gives the exact equation which Ryoder gave. The 1/2 itself does not appear from an equation anywhere, it was originally included because aerodynamic testing determined a 2-D flat plate to have a Cd value of 2, so 1/2 was included to scale this down to unity.

The equation itself is not derived from anywhere, but instead uses simple fluid mechanics and Buckingham's Pi Theorem to determine

F ~ ro^a * V^b * L^c, which a=1, b=2, and c=2.

The fact that a 1/2 is included matters not because the coefficient is completely arbitrary, at least for the most part, particularly our case.

The equation you give for dv/dt~V^2 is how a bodies velocity changes under friction; not how a body reaches equilibrium under friction, not to mention the many components of drag you fail to include such as form drag, roughness, interference, 3-D effects, etc. Your equation is part of it, but it is far from the whole story in this situation. And it ends of complicating the problem far more than necessary for the incorrect answer.

And anyone who is actually still reading this thread must also be a geek or be so completely bored that they subject themselves to it anyway! :ph34r:



I got a strong urge to fly, but I got no where to fly to. -PF

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Wait a sec....what do you mean by "big"? I know of some tall/skiny people and they fall slower than short/skiny people. isn't the tall/skinny personal "big"ger than the short/skinny person? :ph34r::P



I got a strong urge to fly, but I got no where to fly to. -PF

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