How Do You Calculate Object Fall Times with Air and Water Resistance?

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Homework Help Overview

The discussion revolves around calculating the fall times of an object, specifically a bowling ball, under different conditions: in a vacuum, through air, and through water. The original poster seeks to understand how air and water resistance affect the time it takes for the object to fall a certain distance, particularly 1000 feet.

Discussion Character

  • Exploratory, Assumption checking, Conceptual clarification

Approaches and Questions Raised

  • The original poster attempts to apply a basic formula for free fall and questions how to adjust it for air and water resistance. Participants discuss the complexities of fluid resistance and the factors that influence it, such as object shape and velocity. Some participants suggest that the original poster's understanding of the problem may be overly simplistic.

Discussion Status

Participants are exploring various aspects of fluid dynamics and resistance. While some guidance has been offered regarding the complexities of air and water resistance, there is no consensus on specific calculations or estimates for the fall times through different mediums. The discussion remains open-ended with multiple interpretations being considered.

Contextual Notes

The original poster expresses a desire for a straightforward comparison of resistance between air and water, questioning whether water resistance could be quantified as significantly greater than air resistance. Participants note the challenges in making such comparisons due to the varying factors involved in fluid dynamics.

Sterigo
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Hey folks, new guy here!
I have a couple of, what I believe to be, pretty elementary physics questions. I hope you can help me adjust my formulas and help me figure this out!

I have this formula for calculating how long it takes an object to fall a certain distance:
Time(squared) = (2 x Distance) / Gravity

Assuming the formula is correct I can use it to calculate how long it would take a bowlingball to fall, say, 1000 feet, like so:

Time(squared) = 2x1000 / 32

Time to fall 1000 feet = 7,9 seconds

Am I doing it right so far?

Now, this fall would be in a vacuum with no air resistance at all.

How do I calculate the same fall including the air resistance? Is there a formula for this?

Also, let's imagine that the bowling ball would fall the same distance, but this time through water instead of air. How do I calculate how long the fall would take when you consider that water has greater resistance than air?

Is there a formula for this?

Any help would be greatly appreciated!

Sterigo
 
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air resistance is a complex matter because it depends on the shape of the object. water resistance is just as complex, and there would also be an upward buoyant force resisting gravitational force, so the acceleration of the object (if it moves to the bottom as well) would change.

elementary physics doesn't deal with this kind of stuff
 
Check out this Wikipedia page on air resistance. It isn't basic math though. It uses differential equations. I guess that's debatable whether that subject is "basic" or not. But anyway, here you go and good luck. :)

http://en.wikipedia.org/wiki/Air_resistance
 
Hmm, damn - I was hoping for a simple answer to my questions.
Thanks for the replies, though.

Can't one say anything in general about the resistance of air contra vacuum contra water. I am thinking of something like: "the resistance of water on a bowlingball is tenfold the resistance of air on a bowlingball".

I mean, if we have the same object in all examples, in this case a bowlingball, aren't the air resitance and the water resistance constant so that we can tell how much more resistance water would give?

What I am trying to find out is how much more resistance water would give compared to air. Is it tenfold? Hundredfold?

Sterigo
 
You could use the following two equations as a compare the motion of the bowling ball in fluids. You could basically plug the numbers in without the need for calculus.

Fluid friction in a liquid (viscous resistance) is given by;

F_{d} = -6\pi\mu av

Where a is the radius of the sphere and v is it's velocity, \mu is the viscosity of the liquid.

Air friction (at high velocities) can be found as follows;

F_{d} = -\frac{1}{2}\rho CAv^2

Where \rho is the air density, C is the drag co-efficent (which is 0.5 for a sphere), A is the cross-sectional area and v is the velocity.

~H
 
Hootenanny said:
You could use the following two equations as a compare the motion of the bowling ball in fluids. You could basically plug the numbers in without the need for calculus.

Fluid friction in a liquid (viscous resistance) is given by;

F_{d} = -6\pi\mu av

Where a is the radius of the sphere and v is it's velocity, \mu is the viscosity of the liquid.

Air friction (at high velocities) can be found as follows;

F_{d} = -\frac{1}{2}\rho CAv^2

Where \rho is the air density, C is the drag co-efficent (which is 0.5 for a sphere), A is the cross-sectional area and v is the velocity.

~H
Thank you very much for your answer. I forgot to explain that I am as far from a physicist as can be and that formulas like the one you gave here is way beyond me.

Let me put my question a little bit differently:

We have a bowling ball falling 1000 feet through a vacuum, through air and through water.

I imagine that the bowling ball would take at least 1 or 2 seconds longer to fall this distance through air than through a vacuum - would you agree?

How much longer would it at least take to fall through water? 5 seconds longer? 10 seconds longer? 20?

Can you possibly give me any indication as to the difference in time with regards to water and air?

I don't really need an exact number (although that would be fine). An estimate that people familiar with Physics would agree with would help a lot.
 
Could you perhaps tell me of the purpose of this thought experiment?

~H
 
Sterigo said:
Can you possibly give me any indication as to the difference in time with regards to water and air?

This is a very complicated business and many effects come in. The resistance a fluid offeres to a body moving through it is not always constant. There are things like turbulence at high speeds which complicate things even further. The resistance the fluid offers to the bowling ball depend on factors like the shape of the ball, the velocity of the ball, the type of liquid etc. If you just want the difference in times for a particular ball, it would be best to actually perform the experiment.
 
Hootenanny said:
Could you perhaps tell me of the purpose of this thought experiment?

~H

The question has to do with the collapse of the south tower of the WTCs on 911.

According to the official version, the 911 Commission Report, the building collapsed in 10 seconds. Utilizing the formula above we can determine that an object falling from the top of the tower to the ground would do so in a total of 9.2 seconds in a vacuum.

Some would argue that the actualy collapse should take longer than 10 seconds - because air resistance and the resistance provided by the floors below the impact point would slow the fall considerably, and certainly beyond 10 seconds.

Now, there are sources claiming that the actual collapse time of the south tower was longer than 10 seconds - some claim as long as 15 seconds.

What I am trying to figure out is whether or not the building could possibly have collapsed in 15 seconds when you take into account air resistance and the resistance of the supporting steel columns, the concrete and whatnot.

Obviously, air resistance provides some resistance, and, also obviously, steel and concrete provides much more resistance than air.

If we take a substance that has resistance greater than air, but less than concrete and steel we might be able to imagine this easier. This is why i am asking about water resistance - this is a substance with greater resistance than air and less resistance than concrete and steel.

Imagine putting your hand out the car window while the car is driving at 100 mph - you can literally feel significant air resistance. Now imagine that the air was water instead - the resistance would be much greater.

If the bowlingball would fall through water, say, 10 times slower than it would through air...might we not then conclude that the building would collapse at least 10 times as slow as if there was only air resistance - since concrete and steel would provide stronger resistance than water?

Is there sense to this reasoning, physically speaking, or is my thinking faulty?

P.s. Obvisouly the tower consisted not only of concrete and steel - there were lots of air in it. Still, I would think that all the concrete and steel would offer more resistance than if the tower was made of water...

Sterigo
 
  • #10
As, dx commented above this is an extreamly complex question, you cannot simply consider the tower as a falling object, because it is collapsing, it becomes even more complicated if we consider a liquid. It is also a complex question because the reisative force is not constant, it therefore becomes mathematically quite involved.

~H
 
  • #11
Hootenanny said:
As, dx commented above this is an extreamly complex question, you cannot simply consider the tower as a falling object, because it is collapsing, it becomes even more complicated if we consider a liquid. It is also a complex question because the reisative force is not constant, it therefore becomes mathematically quite involved.

~H

Ah, yes. Even with my severely limited knowledge of Physics I knew it would be very complex to calulate this in any precise way.
What I was hoping for was something like "Yes, resistance of water is much greater than air, and, although it is difficult to calculate precisely, it is obvious that a fall through water would take at least 5 times as long as through air".

I guess I won't find any definite answers turning to Physics.
Thanks for all your responses, though - you've been most helpful!

For anyone interested, this is the article that made me start thinking about this in the first place: http://www.911blimp.net/prf_FreeFallPhysics.shtml

Sterigo
 
  • #12
Someone tried to calulate the average speed of the fall of the building using 15 seconds as collapse time. This is the way he formulated this:

if we take a well-endowed number of 15 secs for the fall, that makes it:

400m (height) / 15 secs = 26.66 m/s overall speed

26.66 x 3600 / 1000 = ca. 96 km/h / 1.6 = 60 mph overall,

which is the final speed of a free falling fluffy cat for example after the cw (air resistance) is calculated in.

It can't be this simple, can it?

Sterigo
 
  • #13
Lets say you have some very gooey stuff in a box, and you overturn it on the table. It slowly looses its shape and 'falls' to the ground. The building case is similar to this. It doesn't fall in the usual sense, but it collapses. The last post of yours is definitely an oversimplification.
 
  • #14
dx said:
Lets say you have some very gooey stuff in a box, and you overturn it on the table. It slowly looses its shape and 'falls' to the ground. The building case is similar to this. It doesn't fall in the usual sense, but it collapses. The last post of yours is definitely an oversimplification.

Many thanks, that's what I suspected.

Sterigo
 

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