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Dynamics skydiver +drag force

  1. Dec 13, 2008 #1
    1. The problem statement, all variables and given/known data
    A sky diver and his parachute weigh 890N. He is falling vertically at 30.5m/s when his parachute opens. With the parachute open, the magnitude of the drag force (in Newton) is 0.5v2.
    What is the magnitude of the sky diver's acceleration at the moment the parachute opens?

    2. Relevant equations

    3. The attempt at a solution
    So I figured the drag force is acting up and =0.5v2, weight acts down and = 890N, the sky diver accelerates downwards at (W/g)a
    So I got to an equation:

    I just tried plugging in the velocity at which he's falling (because at the instant the parachute opens, the velocity doesn't change yet) and got a=4.89. But the answer is supposed to be 24g. I don't really get that, at all. Help?
  2. jcsd
  3. Dec 13, 2008 #2
    First of all 24g sounds way off. The drag force will not be greater than that of gravity, and the acceleration due to gravity will be 1g. Therefore, your answer should be >0g, and <1g. 4.89 Sounds entirely reasonable. Below is how I would do it. A little different with the signs but will still yield the same answer.

    Begin with a free body diagram W-.5v^2=M*a

    so something like [890-.5(30.5^2)]/(890/9.8)=a


    slightly different, however you reversed some negative signs I would recommend drawing a diagram and seeing what works best.
  4. Dec 13, 2008 #3
    So you just picked the negative direction as positive, I guess. Also, I get 4.68, it's just that calculation I must have entered something wrong. Glad to hear it's not me being an idiot, but maybe the book being wrong. Thanks!
  5. Dec 13, 2008 #4
    (24/9.81) * 2 = 4.89
  6. Dec 13, 2008 #5
    Is that random number play or is there logic behind it you would like to explain?
  7. Dec 13, 2008 #6


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    Something doesn't make sense in this problem. The terminal velocity of a parachutist is something like 8m/s or so. If the skydiver is initially travelling 30.5m/s as he opens his chute, that's much greater than terminal velocity, such that the the initial net force must be up, and the acceleration initially must be directed up, at a rather large rate, thus slowing down his initial speed. Check the problem again for the drag equation: did you type it incorrectly?
  8. Dec 13, 2008 #7
    No see the net force will NEVER be up in this situation. Unless there is an external force such as a gust of wind or a jet pack. We only have in this model 2 forces, force due to gravity, and air resistance. This resistance is dependent on the downward falling motion. Given what we are given I feel we can not make a truly concise answer. So here are some additional equations to think about.


    In which C is the drag coefficent, p is density of air, v is velocity and A is the area of the cross section. (For the record let me clarify this can be a vector as all forces can)


    In our case we again do not know area or C but, 2mg we know to be 2*890. p somebody surely has memorized, so really we have no basis for this question. In cases of really small C and A we have very large V and very small F. Your example of what the terminal velocity is for parachutists although apt in some cases is not a constant. Therefore these numbers may or may not be completely correct. I am however very interested to see the final result.
  9. Dec 13, 2008 #8


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    The net force initially MUST be up if the diver is falling initially at a speed greater than the terminal speed of the parachute/diver system. If he jumped from rest with the chute open, then, yes, the net force must always be down. Such is not the case here. The diver, travelling at 30m/s initially, opens his chute, and he feels a sudden jolt from the upward acceleartion. He is still travelling down, but the acceleration, initially, is up. The problem statement has some errors in it, or it was copied down incorrectly. I did some quick math, and using the drag force equation, its more like F_drag = 15v^2, not .5v^2.
  10. Dec 13, 2008 #9
    Jay, that's exactly how it appears in the problem, actually. I was able to rearrange the equation in a different way:

    And from that, a would equal 24 g if v=210.95m/s, or W was smaller. There is a similar example in the book which yields 24g, but in that case units were american (v=100ft/s, W=200lb) which are basically the same numbers if you convert to metric. So I'm thinking maybe it's the drag force formula? I guess the book might not have been entirely converted to metric, or something. I don't know. Thanks for all the help!
  11. Dec 13, 2008 #10


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    Yes, it could be that the problem messed up the units. Using F_drag = 0.5v^2, we know that at terminal velocity, the drag force equals the weight. Setting 890 = 0.5(v_term)^2, that yields a terminal velocity of about 42m/s, which, in USA units, is about 100mph! You might as well just jump without the chute, the terminal velocity of a person jumping without a chute is about 120mph or so, depending on his position as he falls. That's one heck of a lousy chute; either way, its lights out when you hit the ground (unless you're lucky enough to fall in a field of marshmallows :wink:).
  12. Dec 14, 2008 #11


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    Checking the numbers again, you are exactly correct, there was a textbook error in the conversion: they forgot to adjust the drag formula accordingly. Using the units of v_o = 100ft/s, and W=200 lbs, and F_drag = .5v^2, then F_drag = 5000 lbs, F_net = 5000 - 200 = 4800 lbs up, and the instantaneous accelertaion is F_net/m = 4800(g)/200 = 24g's up. That's a big acceleration, but it doesn't last long; the speed rapidly reduces and the drag force reduces accordingly, such that in just a few seconds, terminal velocity is reached ('a' goes down to 0), v_term = 20ft/s, or about 15mph, equivalent to free fall jumping from a 7 foot high set of stairs.
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