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Projectile problem proof

  1. Feb 22, 2006 #1
    A projectile is fired up an incline (incline angle φ) with an initial speed vi at an angle θi with respect to the horizontal (θi > φ). (a.) Show that the projectile travels a distance d up the incline, where

    d = 2*vi^2*cosθi*sin(θi-φ) / g*cos(φ)^2

    (b.) For what value of θi is d a maximum, and what is the maximum value?


    I was given a hint to get d into the projectile equations by expressing x(final) and y(final) in terms of d and φ.

    So far I haven't been trying to use that hint, but I can't figure out how...

    Any help would be highly appreciated.
     
  2. jcsd
  3. Feb 22, 2006 #2
    Rotate the x-axis so that it's parallel with the incline
     
  4. Feb 22, 2006 #3

    nrqed

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    Write the usual equations for x and y ( [itex] x= v_i cos (\theta_i) t, y= v_i sin ( \theta_i) t - 1/2 g t^2 [/itex]).

    Now, simply use [itex] x= d cos (\phi) [/itex] and [itex] y = d sin ( \phi) [/itex]. (this is just saying that the coordinates of the landing point are up at a distance d along the inclined plane).

    Patrick
     
  5. Feb 22, 2006 #4
    That wouldn't mess up the acceleration on the y-axis? I was thinking that if I rotated my coordinate system the acceleration due to gravity would change direction in some way.
     
  6. Feb 22, 2006 #5

    nrqed

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    Yes, if you rotate the coordinate system, you have to watch for gravity. In the new coordinate system, you would have both an a_x and and an a_y due to gravity. It's not too hard, but it's probably easier to do it the way I describe in my post.

    Patrick
     
  7. Feb 22, 2006 #6
    Ok. Thanks, that has already been a big help.
     
  8. Feb 22, 2006 #7
    That's right. [tex]g_x=g\sin\theta[/tex] and [tex]g_y=g\cos\theta[/tex]
     
  9. Feb 22, 2006 #8

    nrqed

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    It depends on the side of the inclined plane. If the plane is at the right
    (as is usually used...I mean of the form "<"), then [itex] a_x= -g \cos (\phi) [/itex]. And it should be a [itex] \phi [/itex], to be aligned with the plane.

    Pat
     
  10. Feb 22, 2006 #9
    It depends on the direction of the coordinate you choose to be positive
     
  11. Feb 22, 2006 #10

    nrqed

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    LOL! You are right ;-)
     
  12. Sep 8, 2007 #11
    I've been working on this same problem for quite some time but I'm still stuck.

    Following Patrick's method, i used "the usual x and y equations" and plugged in x = d(cos(phi)) and y = d(sin(phi)) into them.

    I ended up with d(cos(phi)) = v_i(cos(θi))t
    and d(sin(phi)) = v_i(sin(θi)t - .5gt^2

    after this point, i'm not sure how to determine the time (t) that i should plug into the equation.

    also, once i find t, am i supposed to use the pythagorean theorem to find what point the projectile hits the incline?

    thanks in advance
     
    Last edited: Sep 8, 2007
  13. Sep 9, 2007 #12

    nrqed

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    Just isolate the time from the first equation and plug in the second equation. then isolate the distance d and you have your final answer: d expressed in terms of the angles phi and theta and the initial speed. To get the form quoted in the original post, you will ahve to play with trigh identities.

    Patrick
     
  14. Sep 9, 2007 #13

    nrqed

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    Just isolate the time from the first equation and plug in the second equation. then isolate the distance d and you have your final answer: d expressed in terms of the angles phi and theta and the initial speed. To get the form quoted in the original post, you will have to play with trigh identities. I haven`t double checked the equation quoted there but if there is no typo it should work

    Patrick
     
  15. Sep 9, 2007 #14
    if i isolate t from the first equation, i get t = vicos(θi)/dcos(phi)
    then i take this t and plug it into the 2nd equation,

    dsin(phi) = visin(θi)t - .5gt^2

    however, once i plug in my t value, i get a very complicated equation (especially because of the t^2) and i don't know how to isolate d from it. i'm not sure if i am quite following what you are saying.
     
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