# Differential Equation Question

1. Jun 21, 2013

### Ravi Ramdoolar

1. The problem statement, all variables and given/known data

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?

2. Relevant equations

newtons second law F = ma

3. 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

2. Jun 21, 2013

### Staff: Mentor

Please show us what you did and we can go from there.

3. Jun 23, 2013

### Ravi Ramdoolar

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. Jun 23, 2013

### Ravi Ramdoolar

This is what i did

5. Jun 23, 2013

### tiny-tim

6. Jun 23, 2013

### Ravi Ramdoolar

i don't know i think the gravity will be the acceleration

7. Jun 23, 2013

### tiny-tim

gravity (times mass) is one of the forces on the skydiver

kv is the other force

now use Ftotal = ma

8. Jun 23, 2013

### Ravi Ramdoolar

ok will try something

9. Jun 23, 2013

### haruspex

Just checking: it's kv2 without parachute but Kv with? Not Kv2?

10. Jun 24, 2013

### Ravi Ramdoolar

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. Jun 24, 2013

### haruspex

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. Jun 26, 2013

### Ravi Ramdoolar

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. Jun 26, 2013

### Ravi Ramdoolar

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. Jun 26, 2013

### tiny-tim

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

what do you get?