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Transformations physics problem

  1. Nov 29, 2007 #1
    A boy stands at the peak of a hill which slopes downward uniformly at angle [tex] \phi [/tex]. At what angle [tex] \theta [/tex] from the horizontal should he throw a rock so that is has the greatest range.

    Ok, so this is a rotation of the normal [tex] x_{1} - x_{2} [/tex] plane right? So we can use the direction cosines [tex] \lambda_{ij} [/tex] to make this problem easier.

    So [tex] x'_{1} = x_{1} \cos \phi + x_{2} \cos \left(\frac{\pi}{2} + \phi \right) [/tex] and [tex] x'_{2} = \cos \theta + \cos \phi [/tex].

    Are these the right transformations? Is this the right way to set up the problem? Then just apply the equations of projectile motion? This problem seems pretty difficult if I didn't have these tools available. But basically I am using the following:

    [tex]A = \begin{bmatrix} \lambda_{11} & \lambda_{12} & \lambda_{13} \\ \lambda_{21} & \lambda_{22} & \lambda_{23} \\ \lambda_{31} & \lambda_{32} & \lambda_{33} \end{bmatrix} [/tex]
    Last edited: Nov 29, 2007
  2. jcsd
  3. Nov 29, 2007 #2
  4. Nov 29, 2007 #3

    D H

    Staff: Mentor

    You don't really need to do that. The primary reason for transforming frames is to make the problem easier to solve. This transformation makes the problem harder to solve, not easier. And why a three-dimensional transform? This is a 2D problem.
  5. Nov 29, 2007 #4
    At the risk of sounding like an uneducated fool:

    Isn't 45° always the most effective angle to use?
  6. Nov 29, 2007 #5

    D H

    Staff: Mentor

    No. At this point in time it would not be appropriate to derive the answer. Perhaps the original poster will. If he/she doesn't, I'll be glad to do so after a few days. For now, this is a homework thread.
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