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Projectile motion: Bi level equation

  1. Sep 19, 2009 #1
    This isn't really a homework question, but rather general interest.
    I cam across a curious equation the other day,[tex]tan\alpha =\frac{a+b}{a}[/tex]

    Where alpha is the launch angle of the projectile b is the height of the platform its being shot onto and a is the distance from the launch to the platform.
    For example if you where playing golf and trying to hit the ball over a 1m high fence which was 5m away b=1 and a=5.

    I was wondering if anyone knew the proof to this equation, i cannot find it anywhere.
    It is not exact but it is a very very close approximation.
    Thanks everyone.
     
    Last edited: Sep 19, 2009
  2. jcsd
  3. Sep 19, 2009 #2

    Doc Al

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    Staff: Mentor

    A close approximation to what?

    I've never seen this and it doesn't seem to make much sense. What is it's purpose? What if a = b? It gives a zero launch angle.

    Where did you see it? Do you have a reference?
     
  4. Sep 19, 2009 #3
    For example if you were playing baseball, and needed to hit it 21.260 Meters. The launch height is 1.3 meters.
    using the formula;
    b=-1.3 a=21.260
    [tex]\alpha=tan^-1 (\frac{21.260-1.3}{21.260}[/tex]

    [tex]\alpha=43.194 (5sf)[/tex]

    This is a very close approximation for the launch angle, which is 43 degrees
    A colleague showed me this curious equation, she also had no idea why it worked.

    Lets try again; You're launching at 43 degrees and needed a range of 21.260, whats required initial height?
    [tex]b=(a)tan\alpha -a[/tex]
    [tex]b=(21.260)tan(43) -21.260[/tex]
    [tex]b=-1.435[/tex] (4sf)
    Hence height required is 1.435m which is close to the 1.3 proven by traditional means.

    If a=b then there is an obvious limitation.
     
  5. Sep 19, 2009 #4

    Doc Al

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    How did you calculate the launch angle of 43 degrees?
     
  6. Sep 19, 2009 #5
    I assumed a launch speed of 14ms-1 and used equations of motion.
    I shall write it up for you in latex tomorrow morning (10 hours from now, I'm Australian.)
     
  7. Sep 19, 2009 #6

    Doc Al

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    No need. As long as you realize that the launch angle depends on the speed of the projectile.
     
  8. Sep 19, 2009 #7
    Thats why it is so strange that the equation works!!!!
     
  9. Sep 19, 2009 #8

    Doc Al

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    Staff: Mentor

    Well, change your assumed speed and see what happens.
     
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