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Initial velocity in projectile motion

  1. Feb 9, 2005 #1
    I'm supposed to use g=9.81 m/sec and my measured values of s and h to find the initial velocity of a ball shot out of a gun i did in lab today. I was wondering what equation i use to find the initial velocity.
     
  2. jcsd
  3. Feb 9, 2005 #2

    dextercioby

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    I'm sure you meant [itex] g=9.8 ms^{-2} [/itex].
    This equation is the best:
    [tex] \vec{v}(t)=\vec{v}_{0}+\vec{g}t [/tex]

    Daniel.
     
  4. Feb 9, 2005 #3
    Could you explain what each of those letters represent?
     
  5. Feb 9, 2005 #4

    Integral

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    Perhaps one that relates the initial velocity to the the distance traveled and the time of flight? Seems like you need to examine the equations you have at your disposal. Maybe make and effort to look thorough your book and gather all of the motion equations together. Study the equations with emphasis on learning the meaning of each term.

    Here is the most basic equation, from which all can be derived.

    [tex] \ddot {x} = - g [/tex]

    Where x increases up.

    Did that help? Perhaps not, as you have not provided even so basic of information as to what level of course this is. Please provide more information. Could you show us the "candidate" equations? What are you given?
     
  6. Feb 9, 2005 #5
    Heres what the previous steps

    Since the initial velocity is only in the x-direction, v=vox, is vox =s/t and t can be determined from the y equation since the y displacemnt is equal to the height and is due solely to the acceleration of gravity: t=square root of (2h/g). Use g=9.81m/sec and your measured values of s and h to find the initial velocity of the ball.

    s= 284.5cm, h=76.5cm

    I'm also given h=1/2gt(squared)
     
  7. Feb 9, 2005 #6

    Integral

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    You are getting there. We are now even getting a glimmer of the experiment you did!

    Seems like you have the equation and the information requiered to compute the time.
    Or in our Latex capable fourm

    [tex] t = \sqrt { \frac {2h} g [/tex]

    (click on the equation to see what I typed to get that)

    Use that time to get your last result.

    Edit: Whoops I just recalled that you need [itex] V_0 [/itex]

    Surely you have some equations relating [itex] V_0 [/itex] to your other values? What is that equaition?
     
    Last edited: Feb 9, 2005
  8. Feb 9, 2005 #7

    I see 2 equations.

    Vox = s/t
    r = Vot + 1/2at(squared)
     
  9. Feb 9, 2005 #8
    Here's something else too

    The initial velocity components are:
    Vox = Vo(Cos angle) and Voy = Vo(Sin angle)
     
  10. Feb 9, 2005 #9

    dextercioby

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    Yeah,then we could say you got the picture right. :smile:

    Daniel.
     
  11. Feb 9, 2005 #10

    xanthym

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    However, you said before that the INITIAL velocity had no vertical component and was just horizontal. Thus, Voy=0. Try to focus on your original set of data and equations: (your words are quoted below)

    t = sqrt(2*h/g)
    Vox = s/t
    Do you see the technique??
     
  12. Feb 9, 2005 #11
    So is Vox = s/t the equation i need to find the initial velocity?
     
  13. Feb 9, 2005 #12

    xanthym

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    That equation requires values for "s" and "t".
    You measured "s".
    How would you determine "t" in that equation??
     
  14. Feb 9, 2005 #13

    use the equation t=2*h/g?
     
  15. Feb 9, 2005 #14

    xanthym

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    CORRECT!!
    Now substitute your measured values in the appropriate equations, and solve for Vox. Give it a try!!
    (Show all your work.)
    ~
     
    Last edited: Feb 9, 2005
  16. Feb 9, 2005 #15
    Vox = s/t
    284.5cm/3.95s
    i ended up getting 72, would the units be cm/sec?
     
    Last edited: Feb 9, 2005
  17. Feb 9, 2005 #16

    xanthym

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    Check your UNITS.
    (Hint #1: Are ALL your calculation units in cm?)
    (Hint #2: g)
     
  18. Feb 9, 2005 #17
    yah it's all in cm except for the gravity which is 9.81

    I'm not sure how to put the gravity into it...
     
  19. Feb 9, 2005 #18

    xanthym

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    The quickest method is to convert g to "cm/sec^2" from it current "m/sec^2":
    g = (9.8 m/sec^2) = (980 cm/sec^2)
    You need to repeat ALL calculations with everything the same EXCEPT now use g=(980) in the first equation. Please show all your work.
    ~
     
    Last edited: Feb 9, 2005
  20. Feb 10, 2005 #19
    t = sqrt(2*h/g)
    Vox = s/t

    so, h=1/2gtsqred, so h=1/2(981cm/s)(3.95s)squared which is 7653

    t=sqrt(2*7653/980cm/s) which is 3.95s
    Vox = s/t so, 284.5cm/3.95s which finally turns out to be 72
    is this right??
     
  21. Feb 10, 2005 #20

    xanthym

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    NO. It is not correct this time.
    You originally performed the calculations correctly except that "g" had the wrong units. Let's return to your original calculation technique:

    THE EQUATIONS:
    Equation #1 -----> t = sqrt(2*h/g)
    Equation #2 -----> Vox = s/t

    YOUR MEASURED VALUES:
    s= 284.5cm, h=76.5cm

    STEP #1:
    Calculate "t" from Eq #1 using your MEASURED value of h=(76.5cm) and the known constant g=(980 cm/sec^2)

    STEP #2:
    Calculate "Vox" from Eq #2 using your MEASURED value of s=(284.5cm) and the value of "t" calculated from Step #1.

    You are very close to the correct answer. Take your time and focus on the steps above.

    ~~
     
    Last edited: Feb 10, 2005
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