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The formula of 2D motion

  1. Feb 17, 2016 #1

    1. The problem statement, all variables and given/known data
    z0g83Q.jpg

    how to get
    R=Vi 2 sin2θ/g
    y=x*tanθ-[x2*g]/[2(Vi*cosθ)2]
    2. Relevant equations
    Vy=Vi*sinθ-gt
    y=Vi*sinθ*t-1/2gt^2
    Vx=Vi*cosθ
    x=Vi*cosθ*t
    3. The attempt at a solution
    I use this formula to find the expression for t
    Vy=Vi*sinθ-gt
    gt=Vi*sinθ
    t=Vi*sinθ/g

    I substitute it into xf=xi+Vi*t+1/2at^2 (a=0)
    xf=xi+Vi*cosθ*t
    I get
    R=Vi2*sinθcosθ/g
    But the correct one is R=Vi2*2sinθcosθ/g
    What's wrong with it?
    I also see this formula in my book
    y=xtanθ - [x2*g]/[2(Vi*cosθ)2]
    how to figure out it?
    thank you
     
    Last edited: Feb 17, 2016
  2. jcsd
  3. Feb 17, 2016 #2

    TSny

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    It would be a good idea when stating the problem to describe the situation you are dealing with. Apparently it is projectile motion and not just any 2D motion. You leave it to the reader to figure out whether the projectile returns to the same height it was fired, etc. You did not define the symbols in your equations.

    It would also be a good idea to include an x or y subscript on the velocities to indicate which component of velocity your are working with.

    Does the expression for t given above represent the time for the entire flight or for only part of the flight?
     
  4. Feb 18, 2016 #3

    TSny

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    Thanks for adding the figure. When you calculated the expression for t, what time during the flight does this represent?
     
  5. Feb 18, 2016 #4
    The time taken for the particle reach the highest point which is t
    And the time taken for R is 2t
    I forget it is 2t previously so I get some mistakes
    But I still don't know how to figured out
    y=xtanθ - [x2*g]/[2(Vi*cosθ)2]
    Thank you
     
  6. Feb 18, 2016 #5

    TSny

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    Work with the second and fourth equations.
     
  7. Feb 18, 2016 #6

    cnh1995

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    You have,
    and
    And there is no time variable in the required equation..
     
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