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Distance between 2 points on a projectile's path

  1. Aug 31, 2012 #1
    1. The problem statement, all variables and given/known data

    This doesn't seem like much more than a kinematics problem but it's for an upper division mechanics course so I chose to post it here. Feel free to move to intro if necessary.


    i6psz.jpg

    2. Relevant equations

    kinematics equations

    3. The attempt at a solution

    I focused only on the vertical component to find the speed at point 1:

    V2=Vy2 -2gh

    V=√(Vy2 -2gh)

    Now I found the time it takes to get from point 1 to the top of the path:

    0 = √(Vy2 -2gh) -gt

    t = √(Vy2 -2gh) /g

    To get from the top to point 2 would take the same amount of time so the total time to get from point 1 to 2 is 2*√(
    Vy2 -2gh) /g

    Since there's no acceleration in the x direction the distance between the two points is

    2Vx*√(Vy2 -2gh) /g

    I think this is the correct answer, however I can't get it to match the given solution. Furthermore even if this is correct it seems like it would take a lot of algebra to match the given solution, leading me to believe another approach was taken. Any help?
     
  2. jcsd
  3. Aug 31, 2012 #2
    There was another condition: the distance traveled must be maximized. That means a particular condition on Vx and Vy. Get it, then plug into your equation.
     
  4. Aug 31, 2012 #3
    As in theta should be 45 degrees?

    EDIT: This would mean the horizontal and vertical components are initially equal making the equations match. Thank you!
     
    Last edited: Aug 31, 2012
  5. Aug 31, 2012 #4
    Try looking at it this way:

    You're probably effectively solving the quadratic equation y=h. The distance between the two solutions is going to be twice the discriminant appearing in your quadratic formula.
     
  6. Aug 31, 2012 #5

    Doc Al

    User Avatar

    Staff: Mentor

    Express Vx & Vy in terms of V0. (Use the information in the problem statement.)

    Edit: voko beat me to it.
     
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