Calculating Terminal Velocity: Physics Help for a Mock Trial Project

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SUMMARY

The discussion focuses on calculating the terminal velocity of a person falling from a height of 100 meters, specifically considering the effects of air resistance. The initial estimate without air resistance is approximately 100 miles per hour (44 meters per second). The formula provided for terminal velocity with air resistance is v(h) = (mg/k)√(1 - e^(-2kh/m)), where m represents mass, k is the air resistance proportionality constant, and h is the height fallen. The conversation emphasizes the significance of the drag coefficient, which varies based on the object's shape, in determining the impact of air resistance.

PREREQUISITES
  • Understanding of basic physics concepts, particularly forces and motion.
  • Familiarity with the concept of terminal velocity.
  • Knowledge of air resistance and drag coefficients.
  • Ability to manipulate and understand mathematical equations related to physics.
NEXT STEPS
  • Research the effects of drag coefficients on different shapes, such as spheres and arrows.
  • Learn how to derive terminal velocity equations in various scenarios.
  • Explore numerical methods for simulating falling objects with air resistance.
  • Investigate real-world applications of terminal velocity in sports and safety equipment design.
USEFUL FOR

This discussion is beneficial for physics students, educators, and anyone involved in projects requiring an understanding of motion dynamics, particularly in contexts like mock trials or safety assessments related to falling objects.

honeydukes
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Hi. I could use some help from someone who is pretty good at physics.

Me and a few others are doing a project (that has nothing to do with physics) and we need to know the velocity of a person at the time of impact who fell from 100 meters. We figured out that it would be about 100miles per hour/44meters per second without air resistance, but we need to know what it would be WITH air resistance. This person is about 5'10'' and weighs about 150lbs/68.04kg. It's for a sort of mock trial. If anyone could help, we'de really appreciate it!
 
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At that short of a distance, I doubt that air drag has much of any effect. Basically, air drag is proportional to the square of velocity. The proportionality constant (called drag coefficient) is dependent mainly on the shape of the object. Example - an arrow has minimal air drag compared to a sphere of the same volume.

Anyway, if you want to dig into it, Google searches with terms like 'terminal velocity', 'air drag', 'drag coefficient', will probably get you everything you wanted to know.
 
Assuming you jump off from rest, and that air resistance is modeled as proportional to the square of the velocity, it is relatively easy to derive the velocity dependence on h, the distance fallen from the jump off point:
[tex]v(h)=\frac{mg}{k}\sqrt{1-e^{-\frac{2kh}{m}}}[/tex]
where m is the object's mass, k the air resistance proportionality constant.
Note that in the limit [itex]\frac{kh}{m}\to{0}[/tex], we get [itex]v(h)=\sqrt{2gh}[/tex] as we of course should have.<br /> (This include the special cases of zero air resistance or infinite mass of the falling object)[/itex][/itex]
 
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