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Homework Help: Maximum height and length on an incline.

  1. Feb 11, 2010 #1
    Hey I wondered what the best way to proceed about fiding the maximum lenght of a projectile when fired at an angle relative to an incline. Would a coordinate axis along the incline be the best choice? Then I could solve the equations for when y = 0, but it would also result in an acceleration in both dimentions.
     
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  3. Feb 11, 2010 #2

    tiny-tim

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    Welcome to PF!

    Hi center o bass! Welcome to PF! :smile:
    Nah … too complicated. :redface:

    The hillside is linear, but your acceleration equations are quadratic, so keep the acceleration equations as simple as possible. :wink:
     
  4. Feb 11, 2010 #3
    Thank you! :) I'm from Norway and I was looking for some good norwegian physics forums, but I couldnt find any. This forum however seems very good with lots of activity and lots of diffrent topics.

    Maximum height isn't really a problem when I think about it, but lenght is a bit tricky.

    So what if I assume that the projectile crosses the incline at the height y=h and i solve my second order equation for this time.. Then i subsitute this time into the component desciribing the x-position giving me x as a function of the angle? What do you think?
     
  5. Feb 11, 2010 #4

    tiny-tim

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    Yes, doing y first, finding t, then doing x (or the other way round), should be fine.

    What do you get? :smile:
     
  6. Feb 11, 2010 #5
    The y-component of the position is give by

    [tex]y(t) = v_0 \sin \theta t - \frac{1}{2} g t^2 = h[/tex]

    solving this for t i get

    [tex] t* = \frac{v_0 \sin \theta \pm \sqrt{v_0^2 \sin^2 \theta -2gh}}{g}[/tex]

    substituting this into [tex]x(t) = v_0 \cos \theta t[/tex] and assuming its the positive root that is valid i get

    [tex]x(t*) = \frac{1}{g} (v_0 \cos \theta \sin \theta + \cos \theta \sqrt{v_0^2 \sin^2 \theta - 2gh}) = \frac{1}{g} \left( \frac{1}{2} v_0 \sin 2\theta + \sqrt{\frac{1}{4} v_0^2 \sin^2 2\theta - 2gh}\right)[/tex]

    giving that the angle that maxemises the expression is [tex]\frac{\pi}{2}[/tex] which is reasonable and the same as for a projectile shot out from a straight line. Is this correct? :)
     
  7. Feb 11, 2010 #6

    tiny-tim

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    Sorry, I'm confused … where have you used y/x = slope of the hill? :confused:

    (and π/2 is vertical)
     
  8. Feb 11, 2010 #7
    Sorry. It is ofcourse [tex]\frac{\pi}{4}[/tex] that maxemises the expression. I have not used the slope I assumed that the angle was bigger than the angle of the hill so that the projectile would follow a path and hit the hill at a height h.

    In simplifying the expression I have used the trigonometric identity [tex]\sin \theta \cos \theta = \frac{1}{2}\sin 2\theta[/tex].

    Are you with me?
    How would one use the slope of the hill?
     
  9. Feb 12, 2010 #8
    I am unsure of this, so it would be great if anyone confirmed if this is true or not :)
     
  10. Feb 12, 2010 #9

    tiny-tim

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    If the slope of the hill is k, so that y = kx, then you need to get two equations, one for x and t, and one for y and t, and then eliminate t and put y = kx. :smile:
     
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