Determine the speed of the parachutist

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The discussion focuses on determining the speed of a parachutist after the parachute opens, starting with an initial speed of 176 ft/s. The net force acting on the parachutist is defined by the equation \( F_{\text{net}} = W - D \), where \( D \) represents the drag force given by \( D = \frac{Wv^2}{225} \). Applying Newton's second law, the resulting ordinary differential equation (ODE) is \( \frac{dv}{dt} = g(1 - kv^2) \) with \( k = \frac{1}{225} \). The discussion also highlights ambiguity regarding the drag force formulation, questioning whether it should be \( D = \frac{1}{225}Wv^2 \) or \( D = \frac{2}{225}Wv \).

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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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