Determine the speed of the parachutist

[kayella19]

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.

 

[MarkFL]

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:

\(\displaystyle F_{\text{net}}=W-D\)

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:

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

\(\displaystyle m\d{v}{t}=mg-kmgv^2\)

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

Can you identify the type of ODE we have?

 

[kayella19]

But how how would K is equal to 1/225?

 

[MarkFL]

But how how would K is equal to 1/225?

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

\(\displaystyle D=\frac{1}{225}Wv^2\)

or:

\(\displaystyle D=\frac{2}{225}Wv\)

or something else?