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Motion in a Plane

  1. Dec 26, 2006 #1
    1. The problem statement, all variables and given/known data

    Hi, I have some answers to the following questions but I would just like to know if Im understanding the concepts correctly.

    A firefighter climbs up a 100 m ladder leaning against a verticle wall. The ladder makes an angle of 25 degrees with the wal. The firefighter reaches th eroof in 15 seconds.
    (a) what is the height of the wall?
    (b) how far is the base of the ladder from the wall?
    (c) what is the firefighters average verticle velocity?

    2. Relevant equations
    Pythagorean Theorum - (a^2) + (b^2) = c^2
    velocity (v) = distance/time


    3. The attempt at a solution

    (a) COS = adjacent/hypotenuse
    Cos25 degrees = X/10 m
    0.91 = x/10m
    X= 9.06 m

    (b) a^2 +b^2 = c^2
    a^2 + (9.06)^2 = 10^2
    a^2 +82.08 = 100
    a^2 = 17.92
    a= 4.2 m

    (c) given distance up ladder = 10 m
    time = 15.0 seconds
    Required: velocity
    Analysis: velocity = distance/time =10m/15s = 0.666 m/s

    what is the difference between average verticle velocity and resultant velocity???
    THanks for the help
    Pharm 89
    1. The problem statement, all variables and given/known data



    2. Relevant equations



    3. The attempt at a solution
     
  2. jcsd
  3. Dec 26, 2006 #2

    cristo

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    Since you've used 10m in your calculations, I presume this is a typo.

    Method correct.. I dont have a calculator to check your maths though. Also, this is correct, if we make the asumption that the ladder reaches the very top of the building, which we must make here.

    Again, correct method

    You have calculated the resultant velocity. The vertical velocity will be v=(vertical distance)/(time)
     
  4. Dec 26, 2006 #3
    Thanks for your help. So therefore for part c they are asking for the average verticle velocty...would that be the same answer though??
    verticle distance = 10 m/15 seconds = 0.666m/s
     
  5. Dec 26, 2006 #4

    cristo

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    No, the vertical distance is what you calculated in (a) i.e. the height of the building. Draw a diagram, you'll see that the length of the ladder, 10m, is not a vertical distance!
     
  6. Dec 26, 2006 #5
    Yes, that makes sense...verticle distance = 9.06 m/15 seconds = 0. 6 m/s
    Thanks for the assistance.
    Pharm 89
     
  7. Dec 26, 2006 #6
    10m? Wasn't it supposed to be 100m?

    I took

    [tex]

    sin(25)*100m=40m

    [/tex]

    [tex]

    cos(25)*100m=90m

    [/tex]

    Instead of doing it the hard way without going through Pythogarous and keeping the idea of significant digits in my head.

    And no tough work involved here...

    [tex]

    40m/15s = 3 m/s

    [/tex]

    Got 3 m/s, which isn't the same as you'lls (The Texan accent triumphs again)

    But pharm89, where did the 10m instead of the 100m come from?

    Please be careful in calculations.
     
    Last edited: Dec 27, 2006
  8. Dec 27, 2006 #7

    Hootenanny

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    :bugeye: A 100m ladder :bugeye:
     
  9. Dec 27, 2006 #8

    radou

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    There is no reason to round off 100*sin(25) = 42.26 to 40.
     
  10. Dec 27, 2006 #9

    cristo

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    I presumed (and was apparently right, as the poster didn't correct me) that there was a typo in the original post, and it should in fact be 10m. And, as Hootenanny points out, 100m is a little long for a ladder!
     
  11. Dec 27, 2006 #10

    DaveC426913

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    At first I wondered why he shrunk the ladder from 100m to 10m in his calcs.

    But 10m has to be the correct size. Can you imagine climbing 100m in 15s?
     
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