What is the effect of air resistance on a parachutist's acceleration?

AI Thread Summary
The discussion centers on calculating the acceleration of a parachutist with a mass of 60 kg experiencing an air resistance force of 1200 N. The initial calculations suggest a resultant force of 600 N, leading to an acceleration of 10 m/s². However, participants point out that with such a high air resistance, the parachutist would actually experience an upward acceleration of -10 m/s², indicating an error in the interpretation of the forces involved. The conversation also touches on the dynamics of parachuting, noting that parachutists can experience significant deceleration upon parachute deployment, often exceeding 1 g. Overall, the calculations and assumptions about air resistance and acceleration are critically examined, emphasizing the complexity of real-world physics in parachuting scenarios.
lila adel omar
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this is an exam question and i am not sure about my answer:

a man with a mass of 60 kg is a parachutist :

calculate the acceleration if the air resistance is 1200N

this is how i solved it:

we must take the wieght into consideration so his wieght is : w=mG G=approximatly 10ms^-2

60(10)=600Nso 1200-600= 600

600=60a

10=aplease tell me is this right?
 
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Um the question sounds kinda funny. You're solving it right (except that the resultant force will be 600-1200 since you took the value of g to be positive, so the answer should be -10) but with a 1200N air resistance force, the man would actually fly upwards at 10m/s^2! Are you sure you've copied the question correctly?
 
That was my thought also. The math is right, but I don't often see parachutes falling up!
 
DaleSpam said:
That was my thought also. The math is right, but I don't often see parachutes falling up!
yes i don't think so either
 
PWiz said:
Um the question sounds kinda funny. You're solving it right (except that the resultant force will be 600-1200 since you took the value of g to be positive, so the answer should be -10) but with a 1200N air resistance force, the man would actually fly upwards at 10m/s^2! Are you sure you've copied the question correctly?
yes i am 100% sure it is the question and i know it will be -10ms^-2 but i put it 10ms^-2 because it does sound wierd
 
DaleSpam said:
That was my thought also. The math is right, but I don't often see parachutes falling up!
It need not be falling up, merely accelerating up. Hint, how fast was the parachutist falling before releasing the chute?
 
haruspex said:
It need not be falling up, merely accelerating up. Hint, how fast was the parachutist falling before releasing the chute?
his acceleration is 10m/s^2 because of gravity
 
When you fall out of a plane, you accelerate -10 m/s2. After a while, you have picked up quite a bit of speed. Down (negative in physics parlance employing the usual orientation of + = up).

Parachute opens and decelerates you appreciably (this case +10 m/s2). Air resistance drops when speed drops and you hope to end up descending at a reasonable, as good as constant speed (i.e. air resistance 600 N for this bloke).
 
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haruspex said:
It need not be falling up, merely accelerating up. Hint, how fast was the parachutist falling before releasing the chute?
D'oh! Of course you are right.
 
  • #10
BvU said:
When you fall out of a plane, you accelerate -10 m/s2. After a while, you have picked up quite a bit of speed. Down (negative in physics parlance employing the usual orientation of + = up).

Parachute opens and decelerates you appreciably (this case +10 m/s2). Air resistance drops when speed drops and you hope to end up descending at a reasonable, as good as constant speed (i.e. air resistance 600 N for this bloke).
Imagining a person decelerate with a magnitude of made me wonder whether the question was correctly copied. I mean practically the decelerating effect is never close to g, since it takes much longer to reach terminal velocity than it takes for air resistance to become noticable. You rarely "slow" down as quickly as you "speed" up while falling. If however, 1200N is the maximum value air resistance reaches (instead of an average force as it appears to be in this case), then this logic is certainly plausible.
 
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  • #11
PWiz said:
You rarely "slow" down as quickly as you "speed" up while falling..

You certainly do when you hit the ground!
 
  • #12
"You rarely "slow" down as quickly as you "speed" up while falling.."
 
  • #13
PWiz said:
"You rarely "slow" down as quickly as you "speed" up while falling.."

I was going to say that I'd bet a skydiver does decelerate greater than 1g, and, for what it's worth, Wikipedia agrees:

"During a normal deployment, a skydiver will generally experience a few seconds of intense deceleration, in the realm of 3 to 4 g, while the parachute slows the descent from 190 km/h (120 mph) to approximately 28 km/h."

http://en.wikipedia.org/wiki/Parachuting
 
  • #14
Well I did say rarely. And I'd recommend that you try to avoid Wikipedia, anyone can edit the facts! (unless there's a quotation to a reputable site)
 
  • #15
I'd say that the number of times a parachutist experiences a deceleration in excess of 1 g from air resistance will tend to be least as high as his number of drops minus one.

With rare exceptions.
 
  • #16
How silly of me, I was constantly imagining the parachute to be closed!
 
  • #17
Pop the parachute and you're going to decellerate fast.

Fail to pop it and you are going to decellerate even faster. But that doesn't count as "air resistance".
 
  • #18
Yes you're right, the parachute definitely has to be open if the frictional force has a value causing such upward acceleration! If only they sold common sense boosters in the market...
 
  • #19
jbriggs444 said:
I'd say that the number of times a parachutist experiences a deceleration in excess of 1 g from air resistance will tend to be least as high as his number of drops minus one.

With rare exceptions.
That's a little unfair. PWiz was referring to a net deceleration greater than 1g. That would only be vital if free fall lasted longer than the time with parachute deployed.
That said, I have no trouble believing that the drag on the chute is more than double that on the person. If the chute is deployed when at near terminal velocity, the drag on the person is already mg, so the drag on the chute at that speed will exceed 2mg.
 

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