Determine the speed of the parachutist

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Discussion Overview

The discussion revolves around determining the speed of a parachutist after the parachute opens, focusing on the forces acting on the parachutist, specifically the weight and air resistance. The scope includes mathematical reasoning and the formulation of a differential equation to describe the motion.

Discussion Character

  • Mathematical reasoning, Technical explanation, Debate/contested

Main Points Raised

  • One participant proposes analyzing the net force on the parachutist, considering weight and drag, and formulates a differential equation based on Newton's second law.
  • Another participant questions the value of the constant \( k \) in the drag function, seeking clarification on its derivation.
  • There is a challenge regarding the formulation of the drag force, with participants suggesting different possible expressions for the drag, indicating ambiguity in the problem statement.

Areas of Agreement / Disagreement

Participants express disagreement regarding the correct formulation of the drag force and the value of the constant \( k \). The discussion remains unresolved with multiple competing views on these aspects.

Contextual Notes

There is ambiguity in the definition of the drag force, leading to uncertainty in the formulation of the differential equation. The assumptions regarding the constants and their derivations are not fully clarified.

kayella19
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A parachutist is falling with speed 176 ft/s when his parachute opens. If the air resistance is Wv2/225 lb where W is the total weight of the man and the parachute and v is in ft/s, find his speed as a function of time after the parachute opens.
 
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Re: Anyone can solve this?

I would begin by analyzing the net force on the parachutist as he falls after his parachute has opened. He has his weight $W$ pulling him down, and he has drag $D$ opposing his weight:

$$F_{\text{net}}=W-D$$

Using Newton's second law of motion, the given drag function, and the fact that acceleration is the time rate of change of velocity, this becomes:

$$m\d{v}{t}=W-kWv^2$$ where we are given $$k=\frac{1}{225}$$

$$m\d{v}{t}=mg-kmgv^2$$

$$\d{v}{t}=g\left(1-kv^2\right)$$

Can you identify the type of ODE we have?
 
But how how would K is equal to 1/225?
 
kayella19 said:
But how how would K is equal to 1/225?

There was some ambiguity in how the drag (air resistance) was given...is it:

$$D=\frac{1}{225}Wv^2$$

or:

$$D=\frac{2}{225}Wv$$

or something else?
 

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