MHB 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 from an initial speed of 176 ft/s. The net force acting on the parachutist is analyzed using Newton's second law, incorporating weight and drag, leading to a differential equation for velocity. There is some confusion regarding the correct formulation of the drag force, with participants questioning whether it should be proportional to \( Wv^2/225 \) or another form. The conversation highlights the importance of clarifying the drag equation to solve the problem accurately. Overall, the thread emphasizes the need for precise definitions in physics problems to derive the correct speed function over time.
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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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