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Final velocity with air resistance

  1. Jun 3, 2013 #1
    Ok so I am working on this problem and can only get through the first question, the second question keeps stumping me.

    I have uploaded the question along with my calculations for part a so you can see what i'm doing.


    -----
    A object of mass (m) is shot vertically up with V0 200 m/s.
    Air resistance R is proportional with the square root of the velocity, R= -mkv^2
    where k= 0.010m-1

    a) How high does the object reach.

    This one I have solved:

    h= (1/2k) * ln [(g+kv^2)/g]


    b)

    What is the speed the object has when it hits the ground?

    ------

    I just cant get the right answer to b, I keep getting the same velovity as V0 which cant be right.

    IF it is possible to hand write your work or write it mathematically I would really appreciate it.


    My work in attachment as both .pdf or .jpg, whatever you prefer.

    In advance, thank you very much!
     

    Attached Files:

  2. jcsd
  3. Jun 3, 2013 #2

    SteamKing

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    You looked like you understood what was going on in solving the first part. Why did you give up on the second part?

    Remember, the second part is slightly different from the first. From solving the first part, you know how far the object must fall, and you know its initial velocity. Can't you adjust your analysis from the first part to reflect this?

    And no, no one is going to write it out neatly for you; you must do the work yourself.
     
  4. Jun 3, 2013 #3
    Think a bit about the term -kmv2. It doesn't change sign when your object starts to descend.
     
  5. Jun 3, 2013 #4
    Do you mean the air resistance is proportional to the velocity squared. I think that is what you have written as an equation.
     
  6. Jun 3, 2013 #5
    Ok. I did not realize that no one will show me the solution here or I wouldn't have bothered posting. No point then is there.

    Thanks for the input everyone, I can't find a delete button maybe a mod can delete the thread for me?
     
  7. Jun 3, 2013 #6

    SteamKing

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    I'm sorry you feel that way. We can't help you if you are not willing to help yourself and put forth some effort.
     
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