200 lb person is falling through the sky while attached to a parachute. The equa

Question

200 lb person is falling through the sky while attached to a parachute. The equation of the downward velocity as a function of time for this is:
Given: dv/dt=g-(k/m)(v(t))
G= Gravity = 32.2 ft/s^2
m= mass = 200lb = 6.22 slug
k= constant depending on the physical properties of the parachute = ?
Suppose the parachute is designed so that after several minutes in the air the person's velocity will be 10 ft/s.

I have a problem finding K due to me having problems with the integral constant.

So far I have this:

dv/dt=32.2-(k/6.22)v(t)
dv/dt+(k/6.22)v(t)=32.2
e^(kt/6.22)v(t)=e^(kt/6.22)*32.2
e^(kt/6.22)v(t)=(32.2)*(6.22)*e^(kt/6.22)/k+c
v(t)=(32.2)*(6.22)/k+c/e^(kt/6.22)
v(t)=(200.284)/k+c/e^(kt/6.22)
Now this is the part where I am having trouble with. I know that when I set the function as the limit of v(t) as t->inf I should get zero. I also tried v(o)=10 ft/s and solved for C with the k in there.
When I did that, I got V(t)= 200.284/k+(10-200.284/k)/e^(kt/6.22)
I set that function as t-> inf and get 0=10. Clarification would help a lot!!

Explanation / Answer

What I think is that the phrase "several minutes in the air itself symbolizes . Here there is nothing said whether the perso will reach the earth or not, so making this asssumption that t-> v=0 is supposedly wrong.

Instead the body reaches a terminal or constant velocity v=10ft/s

with zero acceleration at t->.

***HOPE IT HELPS***

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