# Parachutist freefalling question

1. May 15, 2012

### sid9221

http://dl.dropbox.com/u/33103477/parachutist.png [Broken]

(This is from a mathematics not physics exam)

Well what I've done, is to compose the equation describing the velocity of the parachute once it has opened. I did this by setting up the differential equation as:

$$\frac{dv}{dt} = g-kv$$

I solved that to get:

$$v = \frac{g}{k} - \frac{Ce^{-kt}}{k}$$

Now the parachutist was travelling at v_0 when it deployed so I can say V(0)=v_0

hence solving I got:

$$v(t)= \frac{g}{k} - \frac{g-v_0k}{k}e^{-kt}$$

Now the point is If I know how far from the ground the parachute opened I can figure out the distance travelled since the parachute was opened ?

I'm confused so any direction on how I should get to x(t) from v(t) would be useful.

Last edited by a moderator: May 6, 2017
2. May 16, 2012

### ehild

The distance travelled is the integral of v(t). (V, the velocity, is the derivative of the displacement v=dx/dt. The parachutist travels vertically downward, so the displacement is the same as the distance travelled here.)

You made a small error in the first equation: dv/dt=-(k/m)v.

ehild

Last edited: May 16, 2012
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