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Finding when a projectile has a certain angle with the horizontal

  1. Aug 27, 2009 #1
    1. The problem statement, all variables and given/known data
    If a projectile is fired from the origin of the coordinate system with an initial velocity [tex] v_0 [/tex] and in a direction making an angle [tex] \alpha [/tex] with the horizontal, calculate the time required for the projectile to cross a line passing through the origin and making an angle [tex] \beta < \alpha [/tex] with the horizontal.


    2. Relevant equations
    [tex] \mathbf{F} = m \mathbf{\ddot{r}} [/tex]

    [tex] tan(\theta) = \frac{y}{x} [/tex]


    3. The attempt at a solution
    So, here I go. First, I treated the x-direction and y-direction separately.

    x-direction
    I know that there is no acceleration in the x-direction so,

    [tex] \ddot{x} = 0 [/tex]

    Assuming that [tex] x = y = 0 [/tex] at [tex] t = 0 [/tex] because the projectile is fired from the origin, we get (by integrating)
    [tex] \dot{x} = v_{0} cos(\alpha) [/tex]

    [tex] x = v_{0} t cos(\alpha) [/tex]

    y-direction
    We only have one acceleration in the y-direction (gravity).

    [tex] \ddot{y} = -g [/tex]

    By integrating , we get:

    [tex] \dot{y} = -gt + v_{0} sin(\alpha) [/tex]

    [tex] y = \frac{-gt^2}{2} + v_{0} t sin(\alpha) [/tex]



    Now is where my confusion starts. I suppose I need to solve something of the form [tex] arctan(\frac{y}{x}) < \alpha [/tex] for [tex] t [/tex].

    When substituting for [tex] y [/tex] and [tex] x [/tex] and doing some algebra, I get:

    [tex] arctan(\frac{2v_{0}sin(\alpha) - gt}{2v_{0}cos(\alpha)}) < \alpha [/tex]

    Is there any way to solve this for [tex] t [/tex]? Or, is there another way of working this problem that I am not seeing?

    It's been quite a long time since I've worked with mechanics, and this is my first upper-level mechanics course, so I don't really know what I'm dealing with here.

    Any help is appreciated. Thanks in advance.
     
  2. jcsd
  3. Aug 27, 2009 #2

    kuruman

    User Avatar
    Homework Helper
    Gold Member

    I have not checked your algebra, but if

    [tex]
    arctan(\frac{2v_{0}sin(\alpha) - gt}{2v_{0}cos(\alpha)}) < \alpha
    [/tex]

    is it not true that

    [tex]
    (\frac{2v_{0}sin(\alpha) - gt}{2v_{0}cos(\alpha)}) < tan\alpha ?
    [/tex]

    Can you now solve for the time?
     
  4. Aug 27, 2009 #3
    Wow, It's amazing how a senior math major can overlook simple things when he's afraid.

    So, that gives [tex] t > 0 [/tex]? I guess that makes sense actually. I kinda sketched the graphs of a bunch of projectile motions, and as long as they're concave down (which should always be the case assuming gravity) then the largest angle with respect to the horizontal is infact [tex] \alpha [/tex] at [tex] t = 0 [/tex], and any angle when [tex] t > 0 [/tex] is less than [tex] \alpha [/tex]. Does this sound like a good answer?

    "When [tex] t > 0 [/tex]."
     
  5. Aug 27, 2009 #4
    It's probably much easier if you take your [tex]+x[/tex] axis along the line through the origin inclined to the horizontal at the given angle [tex]\beta[/tex] and the [tex]+y[/tex] axis perpendicular to it, like in the figure.

    Then, we have the acceleration components:
    [tex]a_x=-g\sin\beta,\quad a_y=-g\cos\beta[/tex]
    and the initial velocity components:
    [tex]v_{0x}=v_0\cos(\alpha-\beta),\quad v_{0y}=v_0\sin(\alpha-\beta)[/tex]
    We require the time interval between the points where it crosses the x-axis. Obviously, it will be the same time during which the y coordinate returns to zero. Now the y coordinate of the particle
    [tex]y=v_{0y}t+\dfrac{1}{2}a_yt^2[/tex]
    So [tex]y=0[/tex] imply
    [tex]v_{0y}t+\dfrac{1}{2}a_yt^2=0[/tex]
    the roots of which are
    [tex]t_1=0,\quad t_2=-\dfrac{2v_{0y}}{a_y}=-\dfrac{2v_0\sin(\alpha-\beta)}{-g\cos\beta}=\dfrac{2v_0\sin(\alpha-\beta)}{g\cos\beta}[/tex]
    The required time is same as t2.
    Hence, the answer: [tex]\boxed{\dfrac{2v_0\sin(\alpha-\beta)}{g\cos\beta}}[/tex]
     

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  6. Aug 27, 2009 #5
    It's interesting how simple a problem becomes when you just change the coordinate system.

    Thanks a lot!
     
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