easy physics question.

Gord.T

Gord.T

Still "Senior Member"
This just came up.

What is the minimal height that a person must jump from to reach terminal velocity at the point of impact on the ground.
Assume no air resistance, sea level I suppose.

And how long is the freefall?

I should know this, but not at this moment.

Assume a terminal velocity of 120 ft/sec (36.576m/s). That sound close ?

I'd like to see proof via mathematical calculation.



//Double check me on this...
So far I have time to impact = 3.73 seconds.
And a height of 68.17 meters. That seems pretty high. ???
 
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Assuming the drag from the air has a negligible effect on your acceleration, then your numbers are right. However, the drag will mean you are accelerating at less than 9.8 m/s/s (32 ft/s/s), so it will actually take a little longer and require a little more height.

How much more? I don't know and don't feel like dragging out my fluid dynamics textbook to figure it out. Call it 5 seconds and 120 meters, and you should be close.
 
Well, if there's no air resistance, there's no terminal velocity. You'll just keep accelerating and accelerating until impact.

Terminal velocity comes about when the air resistance causes enough drag that you can't accelerate any more.

On the Red Bull Stratos project, our jumper is going to exceed 700+ miles per hour, by jumping up near the stratosphere where there is practically no air resistance.

In practical terms when there is normal atmospheric pressure and "normal" air resistance, the terminal velocity is usually around 120mph, depending on how the jumper has oriented his/her body -- 120mph is for someone laying flat out; obviously if they were diving headfirst they'd encounter less resistance and would travel faster.
 
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It depends on the number of beers, minus the visits to the bathroom, times the amount of time to the next beer.


You guys think too hard.
 
OK, assuming terminal velocity is 120 ft/s,

I'm getting the time it would take to reach that velocity, with no other factors but gravity in mind, would be 3.726708 seconds, or ~3.73 like you have.

And the distance you would need to fall to reach that velocity would be 669.418609 feet.

I believe I did that right - that was just using integrals.

That's really fast and not very far to reach terminal because, like Barry said, you actually wouldn't ever hit terminal, since terminal is dependent on air resistance. The appropriate way to calculate terminal velocity for an object is through Force of air and Weight of object, etc. etc.
 
Well that was a refresher. My physics book doesn't go into friction of air, that I can find, it's mostly fluids.

I suppose I could approximate a value for friction assuming a given terminal velocity,
maybe, but I think I'll call that close enough for what I wanted to know.
 
Gord.T said:
Well that was a refresher. My physics book doesn't go into friction of air, that I can find, it's mostly fluids.
Click to expand...

For the purpose of the calculation you treat air as a low-density fluid. Its behavior is governed by the same set of dynamics.

If you were to ignore air resistance the equation would be a simple x(t) = 1/2 * g * t^2 + v0*t + x0, where g is the acceleration of gravity, (9.8 m/s^2), v0 is the initial velocity of the falling body, and x0 is the initial position of that body when you start the clock.

But of course you have to factor in air resistance (or, as observed, no terminal velocity...) so just treat this as friction. Kinetic frictional forces typically are proportional to velocity. At terminal velocity the 'downward' force of gravity (m*g, the mass of the falling body * 9.8 m/s^2) is exactly counteracted by the 'upward' force of air resistance (k*v, where v is the velocity and k is a number proportional to the coefficient of kinetic friction for the human body in flat or vertical fall). So:

m*g = k*VT,

where VT is the terminal velocity.

How do we figure out the distance traveled before reaching terminal velocity and the position as a function of time?

Now we have to deal with a non-homogeneous differential equation, starting with the classic F = ma. "Force" means the net force upon the falling body (the difference of the 'downward' force of gravity (m*g) and the 'upward' force of the friction (k*V). The acceleration, a(t), is the first derivative of the velocity with respect to time, dV(t)/dt. Therefore our equation F=ma becomes:

m*g - k*V(t) = m*dV(t)/dt

Or, substituting k=mg/VT and cleaning it up a bit:

-(g/VT)*V(t) + g = dV(t)/dt

The solution to this differential is:

V(t) = A*exp(-g*t/VT)+VT,

where A is an unknown constant, and exp(x) means e^x, where e is the natural number.

Knowing the initial condition that the velocity at t=0 is zero (assuming the falling body begins 'at rest'), we can solve V(0) = 0 to find that A=-VT.

Thus,

V(t) = VT*(1-exp(-g*t/VT))

or

V(t) = 2.4*(1-exp(-4.083*t)

To find position, x(t), we can simply integrate the velocity with respect to time:

x(t) = integral (V(t) dt)
x(t) = 2.4*integral(1-exp(-4.083*t) dt)
x(t) = 2.4*t+0.588*exp(-4.083*t) + x0,

Wherein x0 is the initial height. If you just want to know how far the object has fallen since it started, x0 = 0.

The equation for the velocity is a decaying exponential - the falling body would actually have to fall an infinite distance before reaching 'true' terminal velocity. *But* you can numerically calculate how long it would take to reach, say, 99.9% of that terminal velocity. Plugging that time into the equation for x(t) would then yield the distance necessary to reach that speed.

Good luck. For my part I'd rather go skydiving with a stopwatch.
 
Nice notes.

Just to be sure I'm clear, the line...

x(t) = 2.4*t+0.588*exp(-4.083*t) + x0,

is a typo, right? Shouldn't it read

x(t) = 2.4*t+0.588*1-exp(-4.083*t) + x0


//Cool. I'll put those in my notes and take a run through it. Good stuff.
 
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Pookie said:
It depends on the number of beers, minus the visits to the bathroom, times the amount of time to the next beer.


You guys think too hard.
Click to expand...

Yeah, boy.......

I'd help out, but I lost the ability to do math at a Pink Floyd concert back in the day!!!
 
I got out of mechanical engineering school in 1978. I've forgotten so much, and couldn't solve a differential equation now to save my PE license.

But my read on this is that you're assuming that you can model drag as friction. I don't believe that is valid. It looks to me as if you are assuming that your friction will be linear, IOW a linear function with velocity. Unfortunately atmospheric drag is a function of the cube of velocity. That's why gas mileage in cars decreases so rapidly after 55mph.

Atmospheric drag is also very dependent on the density of the atmosphere, which changes non-linearly with altitude because air is a compressible medium.

All of this just reinforces what my favorite fluids professor taught me. When it's a fluids problem, the best way to get accurate results is to perform an experiment. So I agree with your assessment -- skydiving with stopwatch and some way to measure velocity is perhaps not such a bad idea. But you'll reach terminal velocity fairly quickly which won't give you much time to make notes or plot curves. :D
 
Yeah, I think Bruce is right; benefit of a formal education. And I guess that you would have to take air temperature into account as well for really accurate modeling.

I'm almost jealous. I never had a favorite fluids professor. Although I do have a number of favorite fluids.
 
Gord.T said:
Well that was a refresher. My physics book doesn't go into friction of air, that I can find, it's mostly fluids.
Click to expand...

Air is a fluid. Fluids include gasses and liquids. Gasses completely fill their containers while liquids form a free surface if their volumes are less than their containers' volume. Gasses are typically easy to compress, while liquids are typically difficult to compress (liquids are often treated as incompressible). But both gasses and liquids have the common characteristics of fluids.
 
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That's interesting; I hadn't realized that liquids (water, anyway) were merely treated as uncompressible - I had thought that they actually are uncompressible.
 
Bruce Watson said:
All of this just reinforces what my favorite fluids professor taught me. When it's a fluids problem, the best way to get accurate results is to perform an experiment. So I agree with your assessment -- skydiving with stopwatch and some way to measure velocity is perhaps not such a bad idea. But you'll reach terminal velocity fairly quickly which won't give you much time to make notes or plot curves. :D
Click to expand...

Yeah, and screaming like a little girl and soiling my self wouldn't be conducive to experimental observation either!!! :thumbsup:
 
What? Skydiving is fun!

Although admittedly I wouldn't care to try and take measurements + make notes on the way down.
 
Sad Max said:
Yeah, I think Bruce is right; benefit of a formal education. And I guess that you would have to take air temperature into account as well for really accurate modeling.
Click to expand...

Yep. I shoulda caught that -- I said I was rusty. Changing temperature will effect athmosperic pressure which in turn effects density. I wonder what else we didn't nail down? With fluids, it's always something. Which is why meteorologists can't easily predict the weather -- the atmosphere changes too quickly and unpredictably, and the changes ripple all around in 3D. One of the reasons for the rise of Chaos Theory IIRC.

Sad Max said:
I'm almost jealous. I never had a favorite fluids professor. Although I do have a number of favorite fluids.
Click to expand...

He was a good guy. Actually interested in teaching the students. Pretty rare actually.

It's perhaps better to have favorite fluids. Certainly more useful in the video biz. But watch out for those volatile organic compounds. I hear they can make you dizzy.
 
Sad Max said:
That's interesting; I hadn't realized that liquids (water, anyway) were merely treated as uncompressible - I had thought that they actually are uncompressible.
Click to expand...

Water is weird. Don't go there unless you want a headache.

In the greater scheme of things, water is fairly incompressible. Not as good as some of the hydraulic fluids used in things like elevators and construction equipment however.
 
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