Modeling an equation for parachute landing

[chengbin]

I'm trying to prove why it is impossible for James Bond to open the parachute 2 seconds before impact and still land relatively unharmed. My argument is that there is no way a parachute can slow Bond from 190km/h to about 40-50km/h in 2 seconds.

I tried searching the web for information on the amount of upward force a parachute brings, but no luck.

The web gives me equations on descent velocity and stuff, but that calculates the terminal velocity when you have a parachute, so that's not useful.

The closest I've found is stoke's drag, where if I solve this conditional differential equation for v of the equation mdv/dt = mg - Bv, and when t = 0, v = 50m/s (terminal velocity of a human), but I can't find a B value for a parachute.

Can anyone fill in some missing pieces or suggest a better way to prove this? I'm in grade 11 physics, so my physics knowledge is very limited. I can learn new physics concept if they're not too complicated, so please dumb it down a little bit in your explanations. Thank you.

 

[hotvette]

Several of these links should help

http://www.google.com/search?hl=en&...i=g1&aql=&oq=drag+coefficient+for+a+parachute

 

[chengbin]

So for stokes drag calculation, would the drag coefficient be the "b" variable?

 

[Dr Lots-o'watts]

You know the terminal velocity for a human.

I was told landing with a parachute was like jumping of a chair. With the approximate height of a chair you can calculate Landing speed, which would be close to terminal velocity with a parachute.

I know for a fact it takes 3 to 5 seconds for a regular parachute to open, because when i I made my first jump, they made me count to 5 before looking up to check if it opened ok. So J.B.'s is surely better.

You should be able to calculate anything you need with this data.

 

[chengbin]

You know the terminal velocity for a human.

I was told landing with a parachute was like jumping of a chair. With the approximate height of a chair you can calculate Landing speed, which would be close to terminal velocity with a parachute.

I know for a fact it takes 3 to 5 seconds for a regular parachute to open, because when i I made my first jump, they made me count to 5 before looking up to check if it opened ok. So J.B.'s is surely better.

You should be able to calculate anything you need with this data.

Could you please explain how?

Drag is not taught in grade 11 physics, so I'm not too familiar with its calculation.

I want to calculate if someone is falling at 50m/s, and he releases his parachute at t = 0, what is his velocity after 2 seconds.

I'm guessing the drag coefficient has to be involved somehow. Right now I'm confused between the drag coefficient, Cd, and the b value from stoke's drag (see http://en.wikipedia.org/wiki/Drag_(physics)). Are they the same? Since you said I have all the information to calculate, how do I do that?

 

[Ranger Mike]

all the way - Dr Lots-o'watts..airborne..
we were jumping with static line and it took that long for the chute to deploy from the Dbag and break the static line tie at the bridal loop at the apex of the chute. no way can any chute be deployed in two seconds...

 

[chengbin]

The chute was open, and they were airborne with the chute open for 2 seconds before impact.

I hope someone can answer my question of stoke's drag and if the variable b is the same as drag coefficient. If not, what is b, and how can I find a value of b for a parachute?