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Equations of Motion

  1. Sep 23, 2014 #1
    • Warning! Posting template must be used for homework questions.
    Hey guys, I'm wondering if I can get some help with a question in my homework. Here's the question:

    A workman on the scaffolding outside one of the physics classrooms drops a wrench. A pupil decides to time how long it takes for it to pass the classroom window. It was found that it took 0.6seconds to fall past the 2 metre tall window. Calculate the wrenches initial velocity as it appears at the top of the window.


    I am presuming that this requires an equation of motion, but I am really not sure on where to start/what equation to use.

    Any help would be greatly appreciated.
     
  2. jcsd
  3. Sep 23, 2014 #2

    CWatters

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    Try writing down

    a) The equations of motion you know.
    b) What you know from the problem using the same variable names. Do you know the acceleration? Initial velocity? Final Velocity? Time? Distance?

    Then have a go at choosing an appropriate equation. Post your choice and working and someone will let you know if that's right/wrong etc
     
  4. Sep 23, 2014 #3
    Okay, thanks for the reply! Here's what I've done:

    I rearranged s=ut+0.5at^2, so that I could find u. ( u= (s-0.5at^2)/t)

    So, when I substitute in the values, I get:
    u=(2-0.5x-9.8x0.6^2)/0.6, which gave me an answer of 6.27m/s.

    I'm not too confident on whether this is correct or not though.
     
  5. Sep 23, 2014 #4

    haruspex

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    Need to be careful with the signs. Which way are you measuring s and u? Is up positive or negative? What about g?
     
  6. Sep 23, 2014 #5
    For me, up is positive, and down is negative. Also, I wasn't aware that g was involved for that equation, aside from being used for the acceleration. Have I made a stupid mistake?
     
  7. Sep 23, 2014 #6

    haruspex

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    If up is positive, what signs should you have for the distance travelled and the initial velocity?
     
  8. Sep 23, 2014 #7
    They should be negative! Okay, I get it now! Thanks for the help guys!

    So, the answer will be -0.39?
     
  9. Sep 24, 2014 #8

    CWatters

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    That's what I get.

    Remember that an examiner may not define the same direction as +ve so always state that early in your working.
     
  10. Sep 24, 2014 #9

    Orodruin

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    One very important thing: -0.39 what? A velocity is a dimensionful quantity and if you do not specify the unit we do not a priori know if it is m/s, km/h, km/s, or parsec/Hubble time.

    I do not know about your teacher, but I typically make significant deductions of points if units are missing.
     
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