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Projectile Question

  1. Sep 24, 2005 #1
    im completely puzzled, dont nkow what to do, so frustratiing :(

    heres the pic, ive done one like this beofre, but i had eth angle it was fired at, this time we dont

    [​IMG]
     
  2. jcsd
  3. Sep 24, 2005 #2
    Separate the velocity vector into its two components.

    [tex]{V_{x0}}={V_0}\cos{\theta}}[/tex]
    [tex]{V_{y0}}={V_0}\sin{\theta}}[/tex]

    You can then solve for the total time it will take to hit the ground by working with the y-component, and thus use the x-component to solve for the x displacement.
     
    Last edited: Sep 24, 2005
  4. Sep 24, 2005 #3
    lets say i want to to the v(x)
    id get
    v(x) = 5 cos ?
    i dont know the degree, usually i have 5 cos 60 degrees, then id have a V(x), this time id ont
     
  5. Sep 24, 2005 #4

    Päällikkö

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    As it's fired horizontally, what's the angle?
     
  6. Sep 24, 2005 #5
    Use one of your kinematic equations.... how about [tex]V_{yf}^2=V_{y0}^2+2gd[/tex]. All of those variables, and gravity, are given to you. What can you solve for now?
     
  7. Sep 24, 2005 #6
    180 degrees right, tried that and i got real close, just a 0.01 away, think i got it
    thx
     
  8. Sep 25, 2005 #7
    can i use 180 degrees?
     
  9. Sep 25, 2005 #8

    Päällikkö

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    I'd use 0 degrees. With 180 you have to hassle with directions (you get negative cosine).
     
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