Differential Equation Question

  • #1

Homework Statement



Felix, a Red bull skydiver, (mass m) who drops out of a hovering helicopter, falls long enough without a parachute(so the drag force has strenght kv2 ) to reach his first terminal velocity ( denoted v1). When his parachute opens, the air resistance force has a strenght Kv.

At what minimum altitude must his parachute open so that he slows to within 1% of his new
(much lower) terminal velocity (v2) by the time he hits the ground?

Homework Equations



newtons second law F = ma

The Attempt at a Solution


I used newton second law of motion and think i have to use the time t = 0 and not sure what to do again could someone help me
 

Answers and Replies

  • #2
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Homework Statement



Felix, a Red bull skydiver, (mass m) who drops out of a hovering helicopter, falls long enough without a parachute(so the drag force has strenght kv2 ) to reach his first terminal velocity ( denoted v1). When his parachute opens, the air resistance force has a strenght Kv.

At what minimum altitude must his parachute open so that he slows to within 1% of his new
(much lower) terminal velocity (v2) by the time he hits the ground?

Homework Equations



newtons second law F = ma

The Attempt at a Solution


I used newton second law of motion and think i have to use the time t = 0 and not sure what to do again could someone help me
Please show us what you did and we can go from there.
 
  • #3
Please show us what you did and we can go from there.

F=ma
kv=md2x/dt2

kdx/dt=md2x/dt2

from here i don't know where to go i don't even know if this is correct does any one know how to work it out ???
 
  • #4
F=ma
kv=md2x/dt2

kdx/dt=md2x/dt2

from here i don't know where to go i don't even know if this is correct does any one know how to work it out ???
This is what i did
 
  • #7
tiny-tim
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gravity (times mass) is one of the forces on the skydiver

kv is the other force

now use Ftotal = ma :smile:
 
  • #8
gravity (times mass) is one of the forces on the skydiver

kv is the other force

now use Ftotal = ma :smile:
ok will try something:approve:
 
  • #9
haruspex
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without a parachute(so the drag force has strenght kv2 ) .... When his parachute opens, the air resistance force has a strenght Kv.
Just checking: it's kv2 without parachute but Kv with? Not Kv2?
 
  • #10
Just checking: it's kv2 without parachute but Kv with? Not Kv2?

yes when the parachute opens it is Kv and before it is Kv2 i am mostly sure we have o use the Kv since the question asked when the parachute was open
 
  • #11
haruspex
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yes when the parachute opens it is Kv and before it is Kv2
That's not quite what it says in the OP. There you wrote kv2, not Kv2. I can cope with the change from v2 to v as the parachute opens, but it seems very unlikely the coefficient would be the same. For a start, its dimensions would be different.
However, since you're given the first terminal velocity as v1, it's irrelevant what the drag was before the parachute opened. So ignore my questions and concentrate on getting the equation when drag = Kv.
 
  • #12
That's not quite what it says in the OP. There you wrote kv2, not Kv2. I can cope with the change from v2 to v as the parachute opens, but it seems very unlikely the coefficient would be the same. For a start, its dimensions would be different.
However, since you're given the first terminal velocity as v1, it's irrelevant what the drag was before the parachute opened. So ignore my questions and concentrate on getting the equation when drag = Kv.

this is what i did so far

ma = mg - kv (since kv is the opposite direction as the gravitational force)

md2x/dt2 = mg - k dx/dt
 
  • #13
this is what i did so far

ma = mg - kv (since kv is the opposite direction as the gravitational force)

md2x/dt2 = mg - k dx/dt

this was given

Mass of Felix = 70 kg
Gravity = 9.8ms^-2
Air resistance proportional constant K = 110 kgs^-1
terminal velocity without parachute, v1 = 52ms^-1
Terminal velocity with parachute v2 = 6.2ms^-1

what do i do from here ???????????
 
  • #14
tiny-tim
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ma = mg - kv (since kv is the opposite direction as the gravitational force)

ok, now rewrite that as mdv/dt = mg - kv, and integrate …

what do you get? :smile:
 

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