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How can I find Horizontal and Vertical velocity, given distance and time?

  1. Oct 3, 2012 #1
    1. The problem statement, all variables and given/known data
    I need to find the horizontal vectors and vertical vectors at launch, maximum height and at final catch of a projectile. I know that the distance between point A and point B is 5 and the total time is 1.


    2. Relevant equations
    How can I also find the maximum height and the acceleration?


    3. The attempt at a solution
    Vx (would this be horizontal vector?) = d/t = 5 m/1 s = 5 m/s
    Vy (would this be vertical vector?) = Vinitial + aΔt = (and i got stuck here because i'm not sure what v initial is equal to.)
     
  2. jcsd
  3. Oct 3, 2012 #2

    tiny-tim

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    welcome to pf!

    hi kayte! welcome to pf! :smile:

    (try using the X2 button just above the Reply box :wink:)
    yup! :smile:
    you don't need vinitial, that's what you're trying to find!! :wink:

    you have a and t and (y - yinitial) = 0, and you want vinitial

    so choose one of the standard constant acceleration equations that uses a t y and v …

    show us what you get :smile:
     
  4. Oct 4, 2012 #3
    okay... so i chose to use the equation Δy=vinitialΔt+(1/2)a(Δt)2

    knowing the given:
    Δt=1 s
    a=0 m/s2 (because it's constant)

    i substitute:
    Δy=vinitialΔt+(1/2)a(Δt)2
    ∆y=vinitial(1)+(1/2)(0)(1)2
    ∆y=vinitial+0
    ∆y=vinitial

    and i'm stuck here... :frown: would i need to know max height of the path to find ∆y?
     
  5. Oct 4, 2012 #4

    tiny-tim

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    hi kayte! :smile:

    (just got up :zzz:)

    you're confused between v constant and a constant …

    a is -g :wink:

    (and if A and B are at the same height, then t = 1 isn't at y = maximum, is it?)
     
  6. Oct 4, 2012 #5
    oh O:

    so...
    Δt=.5 s (because the max height is in the middle)
    a=-9.81 m/s2

    Δy=vinitialΔt+(1/2)a(Δt)2
    ∆y=vinitial(.5)+(1/2)(-9.81)(.5)2
    ∆y=(.5)vinitial+1.22625
     
  7. Oct 5, 2012 #6

    tiny-tim

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    hmm … 2 variables, only 1 equation :frown:

    try using ∆t = 1, ∆y = 0 o:)
     
  8. Oct 5, 2012 #7
    alrighty, so...

    Δy=vinitialΔt+(1/2)a(Δt)2
    0=vinitial(1)+(1/2)(-9.81)(1)2
    0=vinitial-4.905
    4.905=vinitial

    this will be the vectors at catch point because time is 1 and height is 0?
     
  9. Oct 5, 2012 #8

    tiny-tim

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    looks ok :smile:

    now you can find ∆y for t = 1/2 ! :wink:
    i don't understand :confused:
     
  10. Oct 5, 2012 #9
    ohh, like why did you tell me to use y=0? is that like the start/end of the path of the projectile?
     
  11. Oct 5, 2012 #10

    tiny-tim

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    because you know both y and t at that point …

    you didn't know both y and t at the top point :wink:
     
  12. Oct 5, 2012 #11
    hmm... oh! i see

    haha :wink:

    one more question. what does the phrase "acceleration at end" mean and how can i find it?
     
  13. Oct 5, 2012 #12

    tiny-tim

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    where does it come from? :confused:

    (it's not in the question, and every projectile has the same acceleration anyway)
     
  14. Oct 5, 2012 #13
    it's a sub question dealing with trends as distance increases
     
  15. Oct 5, 2012 #14

    tiny-tim

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    are we still talking about projectiles?

    can you give a quote? :confused:
     
  16. Oct 5, 2012 #15
    yes we are

    "As distance increases, what trend(s) do you see in the values for 'acceleration at end'?"
     
  17. Oct 5, 2012 #16

    tiny-tim

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    the acceleration is constant :confused:
     
  18. Oct 7, 2012 #17
    alrighty. thank you so much for your help! :biggrin:
     
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