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Help with Projectile motion- equation problem

  1. Feb 1, 2016 #1
    1. The problem statement, all variables and given/known data
    A projectile is launched horizontally with an initial velocity v0 from a height h. If it is assumed that there is no air resistance, which of the following expressions represents the vertical trajectory of the projectile? (A) h–gv0^2/2x^2 (B) h–gv0^2x^2 (C) h-gx^2/2v0^2 (D) h-gx^2/v0^2

    2. Relevant equations
    d=vi*t+1/2a*t^2

    3. The attempt at a solution
    t=d/v; a=g; d=vi*t+1/2g*(x/v)^2-->d=vi*t+(gx^2/2v^2)--->why is this subtracted from h? Also where does d=vi*t go?
     
  2. jcsd
  3. Feb 1, 2016 #2

    Student100

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    Your equation writing skills are a bit funky, so I might be off base here, for the one question "why is this subtracted from h" draw a picture using vanilla Cartesian coordinates. What direction is the acceleration pointing in? Explain what you're thinking/why you're doing what you're doing as well.
     
  4. Feb 2, 2016 #3

    CWatters

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    You are on the right lines but you should take more care writing equations. You appear to be using d for both the horizontal displacement (in the equation t=d/v) and the vertical displacement (eg in the equation d=vi*t+1/2g*(x/v)2)

    The object starts from the position (0, h). If you put x=0 into your equation does it give the answer y=h ?

    Vi is the initial vertical velocity. What is the initial vertical velocity if the object is "launched horizontally"?
     
  5. Feb 3, 2016 #4
    is x the horizontal range of your projectile?



    UchihaClan13
     
  6. Feb 3, 2016 #5

    haruspex

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    It would be the horizontal coordinate at an arbitrary point in the trajectory.
     
  7. Feb 3, 2016 #6
    Okay so all we need is an equation of trajectory for the projectile or rather the vertical motion of the projectile
    Right???



    UchihaClan13
     
  8. Feb 3, 2016 #7

    haruspex

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    The trajectory is what it asks for. I don't know why it specifies vertical.
    Vertical motion could just mean y as a function of t.
     
  9. Feb 3, 2016 #8
    Yes nor do i
    A projectile has a general trajectory which has both y and x coordinates in it,at a certain instant of time "t"

    UchihaClan13
     
  10. Feb 3, 2016 #9

    haruspex

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    The term trajectory refers to the shape of the path, without consideration of time.
     
  11. Feb 3, 2016 #10

    CWatters

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    Perhaps they mention vertical trajectory so you write an equation for y(x) rather x(y) ? Although it's a bit obvious they mean y(x) as they give the answer equation in the problem statement.
     
  12. Feb 4, 2016 #11
    i figured the whole thing out
    It's easy
     
  13. Feb 4, 2016 #12
    did you get an equation connecting y,x,g and v0?
     
  14. Feb 4, 2016 #13
    It's easy to see that if x=0 y=h(for the equation of trajectory)
    here I consider the point from which the particle is projected as my origin and work accordingly
    Now as there's a term for H-something in the options,one can clearly see that the height or the y-coordinate is assumed from the ground to the corresponding x-coordinate
    Now all you need is to subtract your equation of trajectory from the initial y-coordinate(which is?/) and voila
    You have your answer!
    :)




    UchihaClan13
     
  15. Feb 4, 2016 #14

    CWatters

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    Forum rules discourage just giving the answer. That's why we ask leading questions of the OP.
     
  16. Feb 4, 2016 #15
    I didn't give the OP the answer
    I just gave him a hint
     
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