Time taken for a parachute to fall

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    Fall Parachute Time
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Discussion Overview

The discussion revolves around an experiment investigating whether the area of a parachute affects the rate of fall, with a focus on deriving a formula to calculate the time taken for a parachute to reach terminal velocity and subsequently fall to the ground. The scope includes theoretical considerations of drag and air resistance, as well as practical experimentation with parachute design.

Discussion Character

  • Exploratory
  • Technical explanation
  • Homework-related
  • Mathematical reasoning

Main Points Raised

  • One participant proposes a formula to calculate the total time taken for a parachute to reach terminal velocity and fall to the ground, considering various parameters such as mass, gravitational force, drag coefficient, air density, cross-sectional area, and distance.
  • Another participant questions the meaning of 's' in the context of the formula and emphasizes the importance of using consistent units, suggesting MKS (meters, kilograms, seconds) as a standard.
  • A different participant clarifies that 's' refers to the distance traveled by the parachute while reaching terminal velocity.
  • Concerns are raised about the mathematical validity of the proposed formula, particularly regarding the subtraction of time from distance, indicating a potential misunderstanding in the formulation.
  • There is a request for further explanation on how the participant derived their equations and conclusions.

Areas of Agreement / Disagreement

Participants express differing views on the correctness of the formula and the interpretation of variables. There is no consensus on the validity of the proposed approach or the units used, and the discussion remains unresolved regarding the mathematical formulation.

Contextual Notes

Limitations include potential misunderstandings in the formulation of the equations, unclear definitions of variables, and the need for consistent unit usage throughout the calculations.

Andrew Howe
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Hi.

I'm doing a KS3 coursework experiment whereby I need to investigate the question, 'Does the area of a parachute affect the rate of fall?' Obviously the answer is yes, however the theoretical physicist within me wanted to take this much further, so I attempted to pull together my basic understanding of drag, air resistance etc. in order to come up with a formula. I am using square parachute material made out of standard polyphene shopping-bags attatched to a spherical 1g weight of plasticine via. cotton. I am not taking into account the drag of the weight or cotton or the weight of the parachute material or the cotton, each of which I believed were neglieable. I came up with the formula seen in 'parachute2.JPG'. What I was trying to find, as explained better in 'Basicfor.JPG', was the length of time it takes for the parachute to reach its terminal velocity plus the time it takes for the parachute to travel the rest of the distance to the floor.

In the formula:
b = total time taken (seconds?)
m = mass of weight attactched to parachute (in kg?)
g = gravitational force (9.81?)
C = numerical drag coefficient (1.28 for a flat plane?)
p = air density (0.0022 approx.)
A = crossectional area (cm sqaured?)
'9.81' = acceleration under the affect of gravity (metres/second squared?)
d = total distance from where parachute was dropped from to floor (cm?)
u = initial velocity of parachute (zero m/s)
t = time taken
a = acceleration (9.81 m/s squared?)


Firstly, is this formula correct? And secondly, are the units correct? I would appreciate any help in this matter.
 

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Please identify what 's' means.
Terminal velocity is when the drag force exactly equals the gravitational force.
Its best to work in MKS, meters kilograms seconds.
Whatever units you decide, stick with them throughout.

Also tell us how you got those equations.
 
'S' (as in m/s) means 'seconds', as in m/s means metres per second.

The equation basically works out the time taken for the parachute to reach its terminal velocity added to the time taken for the paracute to traverse the remaining distance to the floor.
 
Your work makes no sense.

In picture "BasicFor" you are subtracting 's' with units of time from 'd' with units of distance.

I asked you to explain how you got to your conclusions but I still don't see any explaining.
 
Sorry. I appear to have misunderstood your previous enquiry.

'S' represents the distance traveled by the parachute whilst it reached mits terminal velocity.
 

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