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Average vertical velocity

  1. Oct 30, 2007 #1
    Thanks in advance for reading this thread. Here goes the question;

    Q: A firefighter climbs up a 10.0m ladder leaning against a vertical wall.
    The ladder makes an angle of 25.0 degrees with the wall. The firefighter reaches
    the roof in 15.0s

    a) What is the height of the wall? 9.10m
    b) How far is the base of the ladder from the wall? 4.23m
    c) What is the firefighter's average vertical velocity?

    I have completed a) & b) as those were straight-forward questions by using Trig.
    However, I got stuck on c). I understand that the average vertical velocity is
    [tex]\Delta[/tex]d/t, t would be 15.0s but I'm not sure what the value of the d would be.

    Could anybody please direct me or give me some advice?

    Thank you
  2. jcsd
  3. Oct 30, 2007 #2


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    Staff Emeritus
    Science Advisor
    Gold Member

    Thats a good question. I think you're probably confused as to whether the distance is the length of the ladder or the height of the wall. Personally I'd say height of the wall but this is a subjective issue of semantics. If you have time I'd clarify it with whoever set you the question.
  4. Oct 30, 2007 #3
    If it says average vertical velocity then surely it has to be the height of the wall.
  5. Oct 30, 2007 #4
    Thanks guys, I really appreciate it.
    This is a course that I'm taking from adult learning centre.
    It's an independent learning so I have to get this done on my own :(

    but after getting help from you guys, it really helps me to re-direct from wrong to right.

    ANYWAYS!!! after squeezing out my brain I just put a conclusion.
    Because its asking for "vertical" meaning the "wall" in this case,
    I've had the height of the wall as D and divide it by 15 seconds to get the answer.
    I think that makes sense

    Many thanks guys :)
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