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Tough projectile question

  1. Dec 16, 2009 #1
    1. The problem statement, all variables and given/known data
    A gun is located at the origin. The gun is aimed 36.87 degrees above the horizontal. A vertical wall is located at x = 18m. The flat and level floor is at y=0m. Ignore air friction.

    a) What is the minimum initial speed necessary for the bullet to reach the wall without hitting the floor?

    b) Suppose the initial speed is 15 m/s. Does the bullet first hit the floor or the Wall? Where does it hit? (I can do this part, just thought I'd throw it in for some reason.)

    c) What initial speed is required in order to hit the wall 3.00m above the floor?


    2. Relevant equations
    d=vt
    Vf^2 = Vo^2 +ad
    d = Vot + 1/2at^2


    3. The attempt at a solution

    a)

    I'm really having troubles with this.

    I don't know the initial or final velocity or time. I thought that since that since the bullet will hit the wall at the same height that it was fired from, then half way along (x=9m) the bullet would peak, so I could use that point knowing the Final velocity in the y would be 0... then I could solve for the initial velocity in the Y, then use trig to find the initial velocity.

    Vfy^2 = Voy^2 + 2ad


    then: d = v*(2)t

    but unfortunately I don't know what the displacement (in the y) would be at this point.

    I'm pretty sure I need to find the initial velocity in the Y, and then time before I can do anything else, but I'm totally stuck.

    I might be on the complete wrong track

    b)

    I can do this, it's for fun if you want.. hah yeah right?

    c)

    this is basically the same as a) i think...

    Please help, my final is tomorrow and this is the only type of projection question I can't figure out.
     
  2. jcsd
  3. Dec 16, 2009 #2
    a)

    A way to solve any projectile question is to first list theses 6 equations:

    [tex] a_{x} = 0 \ \ \ \ \ \ \ \ \ \ \ \ \ \\ \ \ \ \ \ \ a_{y} = -g [/tex]
    [tex]v_{xf} = v_{i}cos \theta \ \ \ \ \ \ \ \ \ \ v_{yf} = v_{i}sin \theta - g*t [/tex]
    [tex]d_{x} = v_{i}cos \theta * t \ \ \ \ \ \ \ H = v_{i}sin \theta* t - 1/2*g*t^2 [/tex]

    In this case you only need the last two equations. (Most often).

    So you have 2 equations with 2 unknowns( t and vi). Then you can solve for.
     
  4. Dec 16, 2009 #3
    okay, thanks for your help. that doesn't really get me much farther though, i'm pretty sure those are a few of the equations i had but with the trig worked into them, can you show me the next step? like I said, I don't know vi or t, as far as i can tell...
     
  5. Dec 16, 2009 #4
    You know the horizontal distance travelled, so find an equation involving this distance. It will be in terms of t and V_0. So now you can get an expression for the time at which the particle hits the wall in terms of V_0. Now, what can you say about the vertical distance travelled when it hits the wall, specifically the case of the least velocity needed?
     
  6. Dec 16, 2009 #5
    d = Vo(cosTheta)*t
    Vo(cosTheta) = d/t

    I don't know the vertical distance, the vertical displacement is 0.
     
  7. Dec 16, 2009 #6
    Exactly, so now you have enough figured out to solve it, combine d_x with d_y.
     
  8. Dec 16, 2009 #7
  9. Dec 16, 2009 #8
    nevermind! i got it. wooooo. thanks for your help!
     
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