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Homework Help: Initial velocity

  1. May 17, 2010 #1
    What formula would I use to find the initial velocity?

    An object is thrown vertically into the air and reaches a height of 30.0 m. Neglecting air friction, what was the object’s initial velocity?

    I dont quite remember how to do this. Is it 30.0m / 9.81[m/s]^2?
     
  2. jcsd
  3. May 17, 2010 #2

    rock.freak667

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    There are mainly three equations of motion

    [tex]s= ut+\frac{1}{2}at^2[/tex]

    [tex]v=u+at[/tex]

    [tex]v^2=u^2+2as[/tex]

    u= initial velocity
    v= final velocity
    s= displacement.

    Which one has all of the information you need in it? (note: a is acceleration, which is just 'g' in your case)
     
  4. May 17, 2010 #3
    Hello :)
    I'm not given the time so I'm assuming its v^2=u^2+2as ?
     
  5. May 17, 2010 #4

    rock.freak667

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    Yes, also since the maximum height is given you have the final velocity.
     
  6. May 18, 2010 #5
    Thank you! :)
    one more question, how would I find the initial velocity? Would it be 0? or 9.81[m/s]^2? dont remember that part either :\
     
  7. May 18, 2010 #6

    rock.freak667

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    Remember m/s2 are the units for acceleration, so in your equation, your acceleration is -9.81.

    At the maximum height, the object stops rising. So the final velocity is?

    When you understand these two things, using the equation you told me it would be v2=u2+2as, you can can find 'u'.
     
  8. May 18, 2010 #7

    the final velocity would be (30.0m) (-9.81)?
     
  9. May 18, 2010 #8

    rock.freak667

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    no, remember the object's goes up, reaches the maximum height and then moves back down to Earth. So during this, the velocity goes from +ve to 'something' (at max) and then -ve as it moves back down to Earth. What is the 'something' ?
     
  10. May 18, 2010 #9
    displacement? :\
     
  11. May 18, 2010 #10

    rock.freak667

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    no no, I am trying to get you to understand what the final velocity is at the maximum height.

    Say you throw something up in the air. It does not continuously move upwards, eventually it fall back down, for it to change direction, what must it instantaneously do?


    For a similar analogy, say you are driving forwards, for you to move backwards, what should happen to your forward velocity before you start to move backwards?
     
  12. May 18, 2010 #11
    It would decrease velocity? =)
     
  13. May 18, 2010 #12

    rock.freak667

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    Right yes, as it moves up, it decrease velocity until it reaches what at the max height?

    (Because for it to change direction, it needs to stop moving up and then start moving down)
     
  14. May 18, 2010 #13
    final velocity?
     
  15. May 18, 2010 #14

    rock.freak667

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    If at the maximum height the object instantaneously stops, the final velocity is?
     
  16. May 18, 2010 #15
  17. May 18, 2010 #16

    rock.freak667

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    Right! So v=0

    v2=u2+2as

    v=0
    u2+2as = 0


    (remember 'a' is negative)
     
  18. May 18, 2010 #17
    Thanks! :)
    So..
    a=-9.81m/s
    s=30.0m
    v=0
    u2??

    How would I find the initial velocity now?
     
  19. May 18, 2010 #18
    30*-9.81?
     
  20. May 18, 2010 #19

    rock.freak667

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    u2-2as=0

    rearrange for u2 and then take the square root of both sides.
     
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