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Polar vector to rectangular equation

  1. Oct 15, 2015 #1
    1. The problem statement, all variables and given/known data
    The problem I'm stuck on is to come up with an equation which will model the path of the projectile in an xy-plane.
    We are given:
    r(t)={ [(v0cos(θ))t], [h+(v0sin(θ))t-(½)gt2] }
    v0=100 ft/s
    θ=60°
    h=4 ft
    so basically,
    r(t)={[100cos(60)t], [4+100sin(60)t-16t2]}
    2. Relevant equations
    I know that:
    x=rcosθ
    y=rsinθ
    tanθ=y/x

    3. The attempt at a solution
    I tried to get the t out for both components.
    x=100cos(60)t
    x=50t
    t=x/50

    y=4+100sin(60)t-16t2
    y=4+86.6t-16t2
    I used the quadratic equation and ended up with
    t(0)=(-86.6±88.1)/8 =-21.83, 0.1875
    I thought that when I solved for the t's and set them equal to each other, I would get an equation in xy coordinates but I'm not sure where to go from here.

    I also tried to use x=rcosθ
    since x=50t,
    50t=rcos(60)
    r=100t
    x2+y2=r2
    so (50t)2+y2=(100t)2
    y2ends up equalling 7500t2
    and I'm back to being stuck.
     
  2. jcsd
  3. Oct 15, 2015 #2

    BvU

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    Hi ace, welcome to PF :smile: !
    Seems to me you mistake r(t) for a vector in polar coordinates ##\vec r(t) =(r(t),\theta(t))##.
    It is given to you already in cartesian (rectangular) coordinates: ##\vec r(t) = (x(t), y(t))##

    My guess is you now need to convert to y(x) to draw the trajectory (but I can't conclude that from the problem statement -- could also be the exercise wants ##(r(t),\theta(t)) ##

    [edit] when I read again, I think you are well underway writing ##x(t) = v_0\cos\theta t \Rightarrow t = x/(v_0\cos\theta)## and substitute that t in the expression for y.
     
  4. Oct 15, 2015 #3

    SteamKing

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    These equations in Section 2 are unnecessary.

    Why? Don't you think the path of a projectile depends on the time variable t?

    This is an example of what is known as a parametric equation, one where the values of x and y depend on, or are expressed, in terms of another variable called a parameter. In this case, the parameter is time, or t. Sometimes the parameter can be eliminated, but this is not always necessary nor desirable.

    You had your answer back at the beginning:

    r(t)={[100cos(60)t], [4+100sin(60)t-16t2]}

    All you needed to do was evaluate the terms of r(t) for which you were given values (like h and θ) and express r(t) in terms of t.

    x(t) = 50t
    y(t) = 4 + 86.6t-16t2

    If you want to plot r(t), all you need to do is substitute different values of the parameter t and you can determine x(t) and y(t) from those.
     
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