# Compute the velocity of a free-falling parachutist using Euler's method

## Homework Statement

For the second order drag model (Eq. 1.8), compute the velocity of a free-falling parachutist using Euler's method for the case where,

m = 80 kg
Cd = .25 kg/m

Perform the calculation from t = 0 to 20 with a step size of 1 s. Use an initial condition that the parachutist has an upward velocity of 20 m/s at t = 0. At t=10 s, assume that the chute is instantaneously deployed so that the drag coefficient jumps to 1.5 kg/m.

## Homework Equations

Eq. 1.8,

dv/dt = g-((Cd)/m)*v2

## The Attempt at a Solution

Used equation v(ti+1) = v(ti) + [g - (Cd/m)*v(ti)2](ti+1 - ti)v

(Used in example in book, unfortunately no example w/ an initial condition with an "upward velocity" though)

Plugged in the values to achieve,

t = 0........... V = 20 + [9.81 - (.25/80)(0)2] *1 = 29.81m/s
t= 1............V = 29.81 + [9.81 - (.25/80)2]*1 = 36.51m/s
t =2.............V=36.51 + [9.81-(.25/80)2]*1 = 42.15 m/s
...so on until t = 10 where Cd changes from .25 to 1.5

Am I doing this right? I don't know how the "upward velocity = 20" works into this. I assumed that it is the initial v(ti) as you can see from the first solution I have where t = 0, which may or may not be horribly wrong.

Thanks, much appreciated.

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D H
Staff Emeritus
Since you are representing an upward velocity as positive, you must have g as negative here. Try making g negative.

Making g negative makes sense to me but it begins to produce negative velocities. Also I made a mistake in the above post,

t = 0........... V = 20 + [9.81 - (.25/80)(0)2] *1 = 29.81m/s
t= 1............V = 29.81 + [9.81 - (.25/80)2]*1 = 36.51m/s
t =2.............V=36.51 + [9.81-(.25/80)2]*1 = 42.15 m/s
...so on until t = 10 where Cd changes from .25 to 1.5

t = 0........... V = 20 + [9.81 - (.25/80)(0)2] *1 = 29.81m/s
t= 1............V = 29.81 + [9.81 - (.25/80)(29.81)2]*1 = 36.51m/s
t =2.............V=36.51 + [9.81-(.25/80)(36.51)2]*1 = 42.15 m/s
...so on until t = 10 where Cd changes from .25 to 1.5

But in looking at that I realize perhaps that v(ti)2 should be 202 initially rather than 02.

Thoughts? Help?

Thanks

D H
Staff Emeritus
That is not enough to cover this scenario. The other problem is drag. Drag is always directed against the velocity vector. If your velocity is positive (regardless of sign convention), the drag acceleration must be negative. If your velocity is negative, the drag acceleration must be positive. You have drag acceleration as $- (c_d/m)v^2$, so it is always negative. This is incorrect for negative velocities. One way to correct this is to compute the drag acceleration as $- (c_d/m)(v^3/|v|)$.