Skydiver's velocity with retarding forces

  • Thread starter rpcarroll
  • Start date
  • #1
7
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Homework Statement


An 80 kg skydiver jumps from an altitude of 1000m and opens his chute at 200m
(a) The total retarding force on the diver is a constant 50.0N without his chute open, it is constant 3600N with his chute open. What will be the skydiver's speed when he reaches the ground?


Homework Equations


(sum)F=ma
P.E.=mgy
K.E.=(1/2)mv^2
E(total)=P.E. + K.E.

The Attempt at a Solution


Well, frankly I've gone through several methods and it's all become a big mess. It seems like a straight forward problem but I think I've gotten tunnel vision from it and am missing something obvious. I have a hunch I should be using PE and KE rather than the following methods, but this was all I could come up with.

I got a velocity (at 200m)=122 m/s doing this:
(sum)F=ma
750N=80a
a=9.375 m/s/s
--> 200=1000+V(initial)t+.5at^2
=200=1000+.5*9.375*t^2
t=13s (appox.)
V(@200)=9.375*13=122 m/s


(sum)F=ma
-3500N=80a
a=-35 m/s/s

0=200+122t=.5*(-35)*t^2
t= 8.34 (or -1.3... obviously time can't be negative)
V(final)=V(@200)-35*8.34
V(final)=-170 m/s.... this does not seem right at all to me.

Any suggestions? I had tried using PE and KE to solve for it, but got lost along the way.
 

Answers and Replies

  • #2
Perillux
I don't think you need PE or KE. You know the force that is slowing him down and it is constant so just find the acceleration due to this force, and subtract it from the acceleration due to gravity.
This will be your overall acceleration now, and just use projectile motion equations. Then once he passes 200m you need to find the new overall acceleration because the force changes.

But keep in mind that when he opens his parachute he will have some initial velocity.
 
  • #3
7
0
Oh jeez, yeah I see that now, thank you.
 
  • #4
LowlyPion
Homework Helper
3,090
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Homework Statement


An 80 kg skydiver jumps from an altitude of 1000m and opens his chute at 200m
(a) The total retarding force on the diver is a constant 50.0N without his chute open, it is constant 3600N with his chute open. What will be the skydiver's speed when he reaches the ground?


Homework Equations


(sum)F=ma
P.E.=mgy
K.E.=(1/2)mv^2
E(total)=P.E. + K.E.

The Attempt at a Solution


Well, frankly I've gone through several methods and it's all become a big mess. It seems like a straight forward problem but I think I've gotten tunnel vision from it and am missing something obvious. I have a hunch I should be using PE and KE rather than the following methods, but this was all I could come up with.

I got a velocity (at 200m)=122 m/s doing this:
(sum)F=ma
750N=80a
a=9.375 m/s/s
--> 200=1000+V(initial)t+.5at^2
=200=1000+.5*9.375*t^2
t=13s (appox.)
V(@200)=9.375*13=122 m/s


(sum)F=ma
-3500N=80a
a=-35 m/s/s

0=200+122t=.5*(-35)*t^2
t= 8.34 (or -1.3... obviously time can't be negative)
V(final)=V(@200)-35*8.34
V(final)=-170 m/s.... this does not seem right at all to me.

Any suggestions? I had tried using PE and KE to solve for it, but got lost along the way.
When he jumps from the plane what is his PE?
How much work then goes into slowing him as he drops to 200m?
How much more work goes into slowing him to the ground?

Isn't the surplus what his KE must be?
 
  • #5
20
0
Here's how I would do it
from 1000m to 200m
upwards force=50N
downwards force=W=mg=784.8N
so net force is 734.8N downwards.
F=ma
solve to get a (from 1000m to 200m), which I found to be -9.185m/s^2
Vfy^2 - Viy^2 = 2a(yf-yi)
solve for Vfy, for which I got 121.2m/s (at 200m) However, if you are supposed totake into account approximate terminal velocity of a human, roughly 54m/s, you might simply need to replace 121.2m/s with that.
So now we can use the final velocity of that part as initial velocity in the next part
From 200m to 0m
upwards force = 3600N
downwards force=W=mg=784.8N
so net force = 2815.2N upwards
F=ma
solve for a (from 200m to 0m), which I found to be 35.19m/s^2 (upwards)
Vfy^2 - Viy^2 = 2a(yf-yi)
solve for Vfy, as we did previously for the first length, and there's your answer (~2.2m/s if you took into account my approximate value of human terminal velocity)
 
  • #6
LowlyPion
Homework Helper
3,090
4
Here's how I would do it
from 1000m to 200m
upwards force=50N
downwards force=W=mg=784.8N
so net force is 734.8N downwards.
F=ma
solve to get a (from 1000m to 200m), which I found to be -9.185m/s^2
Vfy^2 - Viy^2 = 2a(yf-yi)
solve for Vfy, for which I got 121.2m/s (at 200m) However, if you are supposed totake into account approximate terminal velocity of a human, roughly 54m/s, you might simply need to replace 121.2m/s with that.
So now we can use the final velocity of that part as initial velocity in the next part
From 200m to 0m
upwards force = 3600N
downwards force=W=mg=784.8N
so net force = 2815.2N upwards
F=ma
solve for a (from 200m to 0m), which I found to be 35.19m/s^2 (upwards)
Vfy^2 - Viy^2 = 2a(yf-yi)
solve for Vfy, as we did previously for the first length, and there's your answer (~2.2m/s if you took into account my approximate value of human terminal velocity)
Presuming Terminal veloicity is beyond the scope of the problem.

Besides you don't need any simplifying assumptions. If you bother to run the numbers you arrive at the same answer, that you can derive more simply.

PE = m*g*h = 80*9.8*1000 = 784,000 N*m
You have 50N acting over 800 m = 40,000 N*m
You then have 3600N acting over 200m = 720,000 N*m
Net PE into KE = 24,000 N*m = 1/2m*V2
V2 = 48,000/80 = 600
V = 24.5 m/s

Using the 121.16 Velocity at the opening of the chute and plugging in numbers yields:
V2 = (121.16)2 - 2*(35.2)(200) = 14,680 - 14,080 = 600
V = 24.5 m/s
 
  • #7
20
0
I don't know what you're saying; are you pointing out a way to solve it using PE? Is the way I used wrong?
 
  • #8
LowlyPion
Homework Helper
3,090
4
I don't know what you're saying; are you pointing out a way to solve it using PE? Is the way I used wrong?
I was pointing out that PE solves it much more directly with fewer steps, and is exactly equivalent to solving it with the various kinematic equations.

I was also pointing out that any concept of terminal velocity was outside the scope of the problem and that the OP should not be confused by considering any such effect.
 

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