What Is the Optimal Angle for Maximum Range in Projectile Motion on a Slope?

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Homework Help Overview

The discussion revolves around determining the optimal angle for throwing a rock from the peak of a uniformly sloped hill to achieve maximum range in projectile motion. The problem involves understanding the effects of the slope angle on the projectile's trajectory.

Discussion Character

  • Exploratory, Assumption checking, Conceptual clarification

Approaches and Questions Raised

  • The original poster attempts to apply transformations using direction cosines to set up the problem, questioning if their approach is correct. Some participants suggest that the transformations may complicate the problem rather than simplify it. Others raise the common assumption that a 45° angle is optimal for maximum range, prompting further discussion on its validity in this specific context.

Discussion Status

The conversation is ongoing, with participants exploring different interpretations of the problem setup and questioning the appropriateness of certain mathematical transformations. There is no explicit consensus on the best approach yet, and the original poster's method is still under scrutiny.

Contextual Notes

Participants note that this is a homework thread, indicating a focus on learning rather than providing direct solutions. There is an emphasis on not deriving answers prematurely, which may influence the direction of the discussion.

tronter
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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]
 
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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.
 
At the risk of sounding like an uneducated fool:

Isn't 45° always the most effective angle to use?
 
sephirothrr said:
Isn't 45° always the most effective angle to use?

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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