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Projectile Launched at an angle problem.

  1. Oct 10, 2010 #1
    1. The problem statement, all variables and given/known data
    A spring-loaded cannon aimed at 47 degrees above the horizontal is on the last car of a long train of flat cars. The train has an initial velocity of 54.3 km/h. At the moment the train begins to accelerate forward at 0.325 m/s2 , the cannon fires a projectile at 180 m/s. The cannon points in the direction that the train is moving.
    a.) What is the horizontal range observed by a person standing on the ground?
    b.) How far on the train from the cannon does the projectile land? Neglect air resistance


    2. Relevant equations
    x=Vi*t+ 1/2a(t)^2


    3. The attempt at a solution
    I know that x component is 180cos 47 and y component is 180sin 47. Not sure where to go from here
     
  2. jcsd
  3. Oct 10, 2010 #2

    MysticDude

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    Gold Member

    I'm also in this section of my Physics class but I can be of some help.

    I found this formula from my book: [tex]x - x_0 = v_0cos(\theta)t + \frac{1}{2}at^2[/tex]. So I THINK that for the first part, you have to somehow find the t and get the distance.

    EDIT: AS I LOOKED AROUND THE SECTION, THERE IS A SPECIAL FORMULA FOR THE HORIZONTAL RANGE

    The formula is: [tex]R = \frac{v_0^2}{g}sin(2\theta_0)[/tex]
    Plugging into the formula should give you the horizontal range.



    PS. What book are you using, if you are using a book at all?
     
    Last edited: Oct 10, 2010
  4. Oct 10, 2010 #3
    The book im using is holt physics
     
  5. Oct 10, 2010 #4

    MysticDude

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    Gold Member

    Well, you should be able to get the horizontal range, I'm not sure how to do the second part though, I mean I have an idea but I'm not sure if it is correct at all.
     
  6. Oct 11, 2010 #5
    you should first scetch yourself a quick visual of what the question is asking and then draw an x l y chart so that you can seperate your x and y variables. this makes it much easier to see what you have and what you need to find in terms of x and y
    also try using the equaton
    [tex]x - x_0 = v_0cos(\theta)t + \frac{1}{2}at^2[/tex]
     
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