Differential Equation Question

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Homework Help Overview

The problem involves a skydiver, Felix, who drops from a helicopter and experiences drag forces during his fall. The task is to determine the minimum altitude at which his parachute must open so that he slows to within 1% of a new terminal velocity by the time he reaches the ground. The context includes concepts from dynamics and differential equations related to motion under gravity and drag forces.

Discussion Character

  • Mixed

Approaches and Questions Raised

  • Participants discuss the application of Newton's second law and the forces acting on the skydiver, including gravity and drag. There are attempts to set up the equations of motion, but uncertainty remains about the correct formulation and the role of different drag coefficients before and after the parachute opens.

Discussion Status

Several participants are actively engaging with the problem, sharing their attempts to derive equations and clarify the forces involved. There is a mix of exploration of the equations and questioning of assumptions regarding the drag coefficients. Some guidance has been offered regarding the setup of the equations, but no consensus on the next steps has been reached.

Contextual Notes

Participants note the specific values provided for mass, gravity, and terminal velocities, which are essential for further calculations. There is an ongoing discussion about the implications of these values on the equations being formulated.

Ravi Ramdoolar
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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
 
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Ravi Ramdoolar said:

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.
 
Mark44 said:
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 anyone know how to work it out ?
 
Ravi Ramdoolar said:
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 anyone know how to work it out ?
This is what i did
 
Ravi Ramdoolar said:
kdx/dt=md2x/dt2

what about gravity? :redface:
 
tiny-tim said:
what about gravity? :redface:

i don't know i think the gravity will be the acceleration:confused:
 
gravity (times mass) is one of the forces on the skydiver

kv is the other force

now use Ftotal = ma :smile:​
 
tiny-tim said:
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:
 
Ravi Ramdoolar said:
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
haruspex said:
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
Ravi Ramdoolar said:
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
haruspex said:
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
Ravi Ramdoolar said:
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
Ravi Ramdoolar said:
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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