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Answer Dispute! Help me out?

  1. Jan 6, 2005 #1
    Hey, my teach and I have a disagreemnt on the answer. Can you help us out?

    If I would to jump out of a stationary helicopter, various stages of fall are shown in positions a to f. Using Newton’s 2nd law. A= Fnet/m= W-R/m

    Find my acceleration at each position. My mass is 100 kg so my weight is a constant 1000N. Air resistance ® varies with speed and cross-sectional area as shown.
    (Sorry guys, but I can’t get the diagram up)

    Position a. R=0; W=1000N, my acceleration would be 20m/s squared, 10m/s squared.

    Position b. R= 400; W= 1000, my acceleration would be 12m/s squared, 6m/s squared.

    Position c. R= 1000N; W= 1000N, my acceleration would be 0m/s squared,0m/s squared.

    Position d. R= 1200N; W= 1000N, my acceleration would be –4m/s squared, -2m/s squared.

    Position e. R= 2000N; W=1000N, my acceleration would be – 20 m/s squared, –10m/s squared.

    Position f. R= 1000N; W= 1000N, my acceleration would be 0m/s squared, 0m/s squared.



    My answer is underline. My teacher’s is in bold. Am I right or wrong?
     
  2. jcsd
  3. Jan 6, 2005 #2
    I'd like to offer some help if I can, but I can't quite decipher your post. Could you:

    1. Clear up the meaning and/or form of your equations. I can see A = F/m, but what's the second equation? It's jammed up against the first and I see two equals signs. Are you asserting that A = W-R/m? If so what's R? I see a little "registered trademark" symbol in your post, but only once.

    2. I see no bold/underlined answers as your last sentence describes.

    3. Finally, I quote you here, "If I would to jump out of a stationary helicopter...find my acceleration at each position." I'm curious why your acceleration isn't consistenly negative, and for that matter almost constant in magnitude. Air resistance would change the net acceleration of gravity a bit, but not by that much until you reach really fast speeds...
     
  4. Jan 6, 2005 #3
    Sorry about the mix-up. Please view the attachment
     

    Attached Files:

  5. Jan 10, 2005 #4
    Hey, a little help here! Pretty Please

    Can anyone help me out with this?
     
  6. Jan 10, 2005 #5

    dextercioby

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    The only reasonable explanation i could come up with is that the teacher thinks you weigh 50Kg. :bugeye:

    Daniel.

    PS.Are u given the height from which the jump occurs??
     
  7. Jan 10, 2005 #6
    Stargate, are u falling vertically? Is your fall in a vertical motion? Or does it take a projectile motion?
     
  8. Jan 10, 2005 #7
    Or Dex, the teacher might think that Stargate meant to say 100 pounds and not 100 kg :)
     
  9. Jan 10, 2005 #8
    Stargate,

    Regardless of your weight, in postion a there is no air resistance (R=0) so your acceleration should be g or about 10m/s^2 as you say. But further down, your acceleration becomes negative, meaning your velocity is decreasing. I can't see why that would happen.
     
  10. Jan 10, 2005 #9

    HallsofIvy

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    I don't understand this:
    "Position a. R=0; W=1000N, my acceleration would be 20m/s squared, 10m/s squared."

    What are the two "accelerations"?
     
  11. Jan 10, 2005 #10

    dextercioby

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    [tex] 100 pounds\sim 45.3 Kg\neq 50Kg [/tex]

    Daniel.

    PS.I guess the teacher is stupid...It happens... :grumpy:
     
  12. Jan 11, 2005 #11
    Well, roughly 50kg :approve:
     
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