Parachutes and Differential Equations

[Ithryndil]

Homework Statement

A skydiver jumps out of an airplane...

Height = 1200m
Mass = 75kg
Acceleration = 9.80 m/s/s
Force of air resistance is proportional to the velocity w/ k1 = 14 kg/s without chute and with k2 = 160 kg/s with chute
The chute is deployed instantaneously.

The skydiver can land safely if the impact velocity is below 5.2m/s. When is the last possible moment the skydiver can pull the ripcord and land safely?

Homework Equations

m (dv/dt) = -mg - rv
v(t) = Ce^(/rt/m) - mg/r
x(t) = (-mC/r)e^(-rt/m) - mgt/r + A

In the above C and A are constants of integration.

The Attempt at a Solution

Well I have gotten the following set of equations:
Without Chute
v(t) = 52.5e^(-14t/75)-52.5
x(t) = -281.25e^(-14t/75) -52.5t + 1481.25

With Chute
v(t) = (147/32)e^(-32t/15) -147/32
x(t) = -2.1533e^(-2.133t) - 4.594t + 1202.15

Both sets of equations make sense and follow what I would intuitively think would happen. I am having a problem with figuring out how to connect the two in a manner that allows me to figure out the last possible moment. I am not sure where to go beyond that point. I thought that if I plugged in that 5.2 for the terminal velocity I could figure something out, but I can't figure out how to get a t value from that. Thanks for your help.

 

[gabbagabbahey]

m (dv/dt) = -mg - rv

This doesn't look right: you are essentially saying that the force of gravity points in the same direction as the resistive force.

I would adopt the sign convention that the speed is positive when it is downwards, so that the force due to gravity is positive and the resistive force is negative.

\implies m\frac{dv}{dt}=+mg-rv

You will need to correct this sign error.

Well I have gotten the following set of equations:
Without Chute
v(t) = 52.5e^(-14t/75)-52.5
x(t) = -281.25e^(-14t/75) -52.5t + 1481.25

After you correct the negative sign you will get slightly different constants

With Chute
v(t) = (147/32)e^(-32t/15) -147/32
x(t) = -2.1533e^(-2.133t) - 4.594t + 1202.15

You seem to have applied the same initial conditions as the non-chute case; but(!) the chute isn't used until the rip cord is pulled. Instead, call that time t_0 your initial conditions should be that v_{\text{withchute}}(t_0)=v_{\text{withoutchute}}(t_0) and x_{\text{withchute}}(t_0)=x_{\text{withoutchute}}(t_0) since when the rip cord is pulled, the initial height and velocity are those obtained by already falling for a time t_0 without the chute.

 

[Ithryndil]

Ok...what is in my book is this:

F = -mg + R(v) where R(v) is the is the resistance force and equals -r(v)v if it depends only on the v. Therefore, I suppose in the above equations I am using a negative value for v, which would then make

m (dv/dt) = -mg - rv
become
m (dv/dt) = -mg - r(-v)

I understand you wanting me to pick positive to be down, but I have already picked it as negative, and have some work riding on that.

 

[gabbagabbahey]

Ok...what is in my book is this:

F = -mg + R(v) where R(v) is the is the resistance force and equals -r(v)v if it depends only on the v. Therefore, I suppose in the above equations I am using a negative value for v, which would then make

m (dv/dt) = -mg - rv
become
m (dv/dt) = -mg - r(-v)

I understand you wanting me to pick positive to be down, but I have already picked it as negative, and have some work riding on that.

Okay, that works fine. But my comment on the initial conditions for your 'with chute' solution still stand.

 

[Ithryndil]

Ack, I left out an important part. The equation for the "with" chute is assume the chute is deployed immediately. I should have mentioned in the first post those equations are to set bounds to the problem, that way I know if I get an answer outside those bounds, well then, I am wrong. Maybe that is my problem. However, that being said, so far my answers are within the time interval which I calculated to be:

Between 28.19s and 261.69s.

 

[gabbagabbahey]

Okay, but the chute isn't deployed immediately; so you need to adjust your solution to reflect that.

You will end up with solutions for v(t) and x(t) that are functions of both t and t_0 (the time that the chute is deployed). you can then set x(t)=0 to find the time 't' (your solution will be in terms of t_0) at which the skydiver hits the ground. Then substitute this result into v(t) and set it equal to -5.2m/s and solve for t_0.

 

[Ithryndil]

Ok, I will work on it an get back to you.

 

[HallsofIvy]

This doesn't look right: you are essentially saying that the force of gravity points in the same direction as the resistive force.

No, it doesn't. It says that the force of gravity point opposite to the positive axis and that the resistive force points in the direction opposite v. If v is itself downward (negative) then -rv is upward.

 

[Ithryndil]

I don't know if I am going to be able to finish this. I would like to, but it will be after the time it is due...probably tomorrow or Thursday. There's something I am not getting, but I have my final for this class today and need to study for that. This is worth less than the final and I have answered 5 out of the 7 parts. Thanks for your help so far, and if I can't get to it before the final (because I will be studying for the final) I will aim to work on it after my finals. Can anyone recommend a good differential equations book to me? I don't like the book I am using nor did my teacher do a good job of teaching Laplace Transforms (he rushed through 7 sections in about two weeks, 4 class periods).