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Terminal velocity help

  1. Mar 3, 2009 #1
    terminal velocity help!!

    1. The problem statement, all variables and given/known data
    ok here is the problem. i found parts (a) and (b), but dont know part c.
    You drop a single coffee filter of mass 1.8 grams from a very tall building, and it takes 45 seconds to reach the ground. In a small fraction of that time the coffee filter reached terminal speed.

    (a) What was the upward force of the air resistance while the coffee filter was falling at terminal speed?
    Fair = N

    (b) Next you drop a stack of 5 of these coffee filters. What was the upward force of the air resistance while this stack of coffee filter was falling at terminal speed?
    Fair = N

    (c) Again assuming again that the stack reaches terminal speed very quickly, about how long will the stack of coffee filters take to hit the ground? (Hint: Consider the relation between speed and the force of air resistance.)
    Fall time is approximately



    2. Relevant equations
    a is .01764 N
    b is .0882

    3. The attempt at a solution
    not really sure what to do. at first i thought oh 5 times bigger means 5 times faster
    WRONG lol
    any help would be great
    thanks
     
  2. jcsd
  3. Mar 3, 2009 #2
    Re: terminal velocity help!!

    (Hint: Consider the relation between speed and the force of air resistance.)

    force of air resistance is proportional to velocity^2
     
  4. Mar 3, 2009 #3
    Re: terminal velocity help!!

    i know the faster an object moves the more air particles it hits with means more air resistance force. but dont know what to do with this force.
    would force = kenetic?
    if so i could solve for velocity but i dont know height of the building so dont know how that will help.
     
  5. Mar 3, 2009 #4
    Re: terminal velocity help!!

    force of air resistance is proportional to velocity^2

    And you know the force increased by 5 times so the velocity would need to increase by...
     
  6. Mar 3, 2009 #5
    Re: terminal velocity help!!

    but how is velocity going to give me drop time (in seconds) if i dont know the height of the building. (velocity is in m/s and i dont know (m) of building)
     
  7. Mar 3, 2009 #6
    Re: terminal velocity help!!

    But the height of the building does not change. Velocity = distance / time
    Since the distance does not change the only thing that can alter velocity is time.
     
  8. Mar 3, 2009 #7
    Re: terminal velocity help!!

    ok i still got the wrong answer. can you show me what to do?
     
  9. Mar 3, 2009 #8
    Re: terminal velocity help!!

    Fair is proportional to v2
    v is proportional to 1/t

    Fair changes by a factor of 5.
    So velocity will change by a factor of [tex]\sqrt{5}[/tex]
    And time will change by a factor of 1/[tex]\sqrt{5}[/tex]

    So your answer should be t = 45 x 1/[tex]\sqrt{5}[/tex]
     
  10. Mar 4, 2009 #9
    Re: terminal velocity help!!

    thanks i had that my only problem was sig figs.
    got to hate them lol
    problem has 2 sig. figs. but the answer wanted
    4 dont know why ???
     
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