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Derivation of area of trajectory of projectile

  1. Jul 14, 2012 #1
    1. The problem statement, all variables and given/known data

    This is a problem about an aircraft flying above a cannon which is capable of firing in any direction. The plane's height, velocity and the speed of the projectile fired by the cannon are given (I am not trying to get help in solving the problem, so the numbers are not given). I am supposed to find the time duration when the plane is at risk of get hit. I understood the problem, but my query is about the derivaiton of the formula for getting the area of the projectile's trajectory

    2. Relevant equations

    [tex]y = x tan \Theta - gx^2 / 2 u^2 cos^2\Theta[/tex]

    3. The attempt at a solution

    I know that [tex]tan \Theta = y/x[/tex], but why are we reducing the second component from this area? Does this have something do with the shape of the path of the projectile being a parabola?
     
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  3. Jul 14, 2012 #2

    TSny

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    The equation is derived from y = voyt - (1/2)gt2.

    Since you are trying to get an equation relating y to x rather than y to t, you need to find an expression for t in terms of x and substitute it into this equation. So, think about how x and t are related in projectile motion.
     
    Last edited: Jul 14, 2012
  4. Jul 14, 2012 #3

    TSny

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    In the equation you want to derive, [itex]\Theta[/itex] is the initial angle of projection. It is not the angle such that [itex]tan \Theta = y/x[/itex] for a point on the trajectory.
     
  5. Jul 14, 2012 #4
    time of flight

    [tex]t = 2u sin\theta/g[/tex]

    x (horizontal range) = [tex]u^2 sin2\theta / g[/tex]

    How is this related to [tex] gx^2 / 2u^2cos^2\Theta ?[/tex]

    I used the equation [tex]S = ut - 1/2gt^2[/tex] and tried to substitute the value of 't', but I ended up with something else.
     
  6. Jul 14, 2012 #5
    OK, I got it.

    For a projectile motion, the distance travelled in x axis is

    [tex]x = ut + at^2/2[/tex]

    But since 'a' is zero along the x axis, [tex]x = ucos\theta.t[/tex]

    distance travelled in y axis is

    [tex]y = u sin\theta.t - 1/2 gt^2[/tex].

    Now I can substitute the value of [tex]t = x/u cos \theta[/tex] and I get the result. Thanks for the help.
     
  7. Jul 14, 2012 #6

    TSny

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    Nice work! :cool:
     
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