Could someone help me out please?

  • Thread starter honeydukes
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Hello!

I am doing a project with a few other people. It is sort of a mock trial. None of us are physics wiz's. We need to know the speed a body would impact the ground from a fall of 100 meters. We figured out that it would be 99 miles per hour/44 meters per second WITHOUT air resistance. But we need to know what it would be WITH air resistance. The person is about 5'10 and weighs about 150lbs/68kg. Your help would be much appreciated! Thanks!
 
honeydukes said:
Hello!

I am doing a project with a few other people. It is sort of a mock trial. None of us are physics wiz's. We need to know the speed a body would impact the ground from a fall of 100 meters. We figured out that it would be 99 miles per hour/44 meters per second WITHOUT air resistance. But we need to know what it would be WITH air resistance. The person is about 5'10 and weighs about 150lbs/68kg. Your help would be much appreciated! Thanks!
You'll need to add a drag term to your force balance equation. IIRC from my undergraduate mechanics, there can be a viscous drag, which is proportional to the velocity, or a quadratic drag, which is proportional to the square of the velocity. I think that above a few meters per second, the quadratic drag term is used in a typical approximate calculation. So, the drag force should be [tex] f = -cv^2 [/tex]. You can approximate the drag coefficient with something like [tex] c= \frac{1}{2} C_D S\rho [/tex]. [tex]C_D[/tex] is a dimensionless term related to the geometry of the falling object, S is the cross-sectional area of the falling obect and [tex]\rho[/tex] is the density of air.
 

Chi Meson

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As suggested by Geoff's post, the cross-sectional area is important here. This is the surface area that "meets" the oncoming wind and it will be very different for the position that the person takes. In the "pencil" position (straight up and down), a 68 kg person is not going to have experienced much significant air resistance (this is a 4.5 second fall, you know). My estimate has a maximum of 18 N and a minimum of 2 N of air resistance, so let's say 10 N of air resistance. This would make the acceleration atthe bottom about 9.6 m/s^2. The average acceleration would then be 9.7 m/s^2 (this is dirty estimation, please understand) so the maximum speed would be about 44.0 m/s instead of 44.3 m/s.

You'd really need to bring out calculus to do a better estimation if the person falls in a spread eagle position.
 

Chi Meson

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A quick estimate of a spread eagle fall still gets a maximum speed of no LESS than 40 m/s. This was found by overestimating just about every factor. I haven't worked in fluid dynamics for a while, so if anyone has a smaller maximum speed, I'd like to know what I missed.
 
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