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Projectile Motion - Elevated Target

  1. Oct 8, 2005 #1
    I was told by a professor that there was no analytic solution (it's a numerical methods course) when there is a elevated target. However, I can't find anything wrong with the analytic solution I came up with. Would anyone mind confirming that my formula (attached gif) for the launch angle when the initial velocity, range, target altitude are known is correct?


    In addition, I understand that there are two angles for hitting the same target. One is the theta in the attached formula, the other is: PI/2 - THETA. I just don't know how to show this mathmatically. Any pointers? Is there something I am missing with the arcsin of a double angle?

    Thanks in advance,
    Matt
     

    Attached Files:

  2. jcsd
  3. Oct 8, 2005 #2

    hotvette

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    This is true only if [itex]x < x_{max}[/itex]. Using two equations of motion for a projectile [itex]y(\theta,t)[/itex] and [itex]x(\theta,t)[/itex]. You you can combine them to eliminate t, resulting in [itex]y(\theta,x)[/itex]. Solving this for [itex]\theta[/itex] will result in 2 values.

    Actually, I'd be willing to bet that if [itex]x = x_{max}[/itex] the solution is two identical roots, [itex]\frac{\pi}{4}[/itex].

    Re the first item, I didn't check your math, but there is no reason an analytical solution isn't possible for an elevated target. In fact, my Intro Dynamics book has a problem solution for kicking a football for a field goal, the goal post cross bar being an elevated target.
     
    Last edited: Oct 8, 2005
  4. Oct 8, 2005 #3

    I believe this is what I did in the attachment. Is it not? Yet, I only got one value... Not sure what I am missing.

    Sorry it isn't in TEX. I did it in OpenOffice.org which is almost not completely TEX-like.
     
  5. Oct 8, 2005 #4

    Physics Monkey

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    I have to agree with hotvette, it's certainly possible find an analytic solution. This is basic intro physics parabolic motion. Perhaps the professor was including air resistance? I don't know what he/she could have meant otherwise.
     
    Last edited: Oct 8, 2005
  6. Oct 8, 2005 #5

    hotvette

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    Re two angles, I thought you were asking in general as opposed to an elevated target. The approach I mentioned works for a non-elevated target, but I'm not sure about elevated target. Hmmm, I need to think and reason a little.
     
  7. Oct 8, 2005 #6

    hotvette

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    A couple of things seem strange in what you did. In (3), you eliminate [itex]\theta[/itex] from the equation only to use it later in (4) to solve for [itex]\theta[/itex]. It may be valid, I've just never seen a derivation like this before, especially having it in terms of [itex]v_{yf}[/itex]. Secondly, in (3), you choose to use only the 2nd solution.

    Think of this. Take (4) and solve for t. Substitute this result into (1). Thus, you'll end up with a quatratic equation in theta. If you know x & y, there should be two solutions for theta. The only catch is the x,y position of the target must lie inside the trajectory at [itex]\theta = \frac{\pi}{4}[/itex]. Otherwise, there is only 1 solution or no solution.

    I may see the numerical analysis part. If you do as I suggest, you end up with an equation with terms like [itex](cos\theta)^2[/itex] and [itex]tan\theta[/itex]. This may not be solveable analytically for aribtrary x & y.

    Another thing I noticed. Your solution has a [itex]v_f[/itex] term in it. In reality, [itex]v_f[/itex] is a function of [itex]\theta[/itex]. Think about it.
     
    Last edited: Oct 8, 2005
  8. Oct 10, 2005 #7

    hotvette

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    I'm afraid some or most of what I said before is bunk. I've been playing with this for a while. Re 2 angles resulting in the same ending location (at least for non-elevated), the answer is quite simple, actually. Since we're dealing with periodic functions (sin, tan), multiple angles produce the same value of trig function. In the case of the simple trajectory, [itex]\theta_1 + \theta_2 = \frac{\pi}{2}[/itex]. If [itex]\theta = \frac{\pi}{4}[/itex], then the 2 solutions are identical because [itex]\frac{\pi}{4} + \frac{\pi}{4} = \frac{\pi}{2}[/itex].

    If you do what I said before, that is, combine [itex]y(\theta,t)[/itex] and [itex]x(\theta,t)[/itex] into [itex]y(\theta,x)[/itex], set y = 0 (completed trajectory for non elevated target), do some manipulation and using the trig identity [itex]sin(2\theta) = 2sin\theta cos\theta[/itex] you get the followng:

    [tex]sin(2\theta) = \frac{gx}{v_0^2}[/tex]

    As long as the right hand side is a valid value for [itex]sin(2\theta)[/itex], there are 2 values of [itex]\theta[/itex] that satisfy the equation. The same would be true for an elevated target, but the math is much more complex (I didn't go through it all), and I don't believe there is a closed form solution, thus the need for numerical analysis.
     
    Last edited: Oct 10, 2005
  9. Oct 10, 2005 #8

    Indeed it is. I probably should have not skipped so many steps in the orignal post... [itex]v_f = \sqrt{v_0^2 - 2gy_f}[/itex]

    Thanks for the replies. Still processing your later posts...

    EDIT... What I said above didn't make much since... there was no theta... so you have something here... looking into it. Thanks!
     
    Last edited: Oct 10, 2005
  10. Oct 10, 2005 #9

    hotvette

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    OK, finally got it. It was right in front of my eyes all along and I didn't see it. If you combine the equations for x(t) and y(t) into one equation to eliminate t, you get something that looks like:

    [tex]y = \frac{a}{cos^2 \theta} \ + \ b \ tan \theta \ \Longleftrightarrow \ y = a \ sec^2 \theta \ + \ b \ tan \theta[/tex]

    Using the trig identity [itex]sec^2 = tan^2 + 1[/itex] you end up with an equation in [itex]tan \theta[/itex] and [itex]tan^2 \theta[/itex], which is a quadratic equation in [itex]tan \theta[/itex] that can be readily solved. Attached thumbnail illustrates the solution for specific values of [itex]x, y, v_0[/itex] that I chose.

    The challenge question is to find the equation of the curve that bounds the valid values of x,y.

    EDIT: I now follow what you did, and why you picked the second t. Everything you did was valid except getting from (4) to (5). First, there is a math mistake, plus you solve for [itex]\theta[/itex] in terms of something that is a function of [itex]\theta[/itex]. If you go from (3) to (4) w/o the last simplifying step in (3), you'll get an equation in [itex]sin \theta[/itex], [itex]cos \theta[/itex], and [itex]\sqrt{asin^2 \theta \ - \ b}[/itex]. Solving this for [itex]\theta[/itex] (looks complicated) should yield the same answer as the approach I took. All you did was solve for t first, then substitute, whereas I substituted to eliminate t, then solve for [itex]\theta[/itex]. Both should yield the same result.
     

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    Last edited: Oct 11, 2005
  11. Oct 13, 2005 #10
    Thanks for all your help. Definately helped me sort through the solution.

    I ended up implementing the numerical solution (just using the brute force method through all values of theta basically root finding). But it's nice to know that I wasn't completely off base with my assertion that there was an analytical solution.
     
  12. Oct 13, 2005 #11

    hotvette

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    It was fun. Thanks for the challenge!:smile:
     
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