- #1
chengbin
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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.
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.