Rifleman's Rule: Best angle for a projectile up a hill [ ]

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SUMMARY

The discussion focuses on optimizing the angle of projectile launch (alpha) to maximize the range (L) when fired up a hill with slope (beta). The equation governing this relationship is L=(2Vo^2)/g*(cos(a)/cos(b))*(sin(a)-cos(a)tan(b)). The solution involves taking the derivative of L with respect to alpha and applying trigonometric identities to simplify the equation. The user also raises a question about the potential errors in elevation when shooting on steep slopes.

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Rifleman's Rule: Best angle for a projectile up a hill [urgent]

Homework Statement



A projectile is fired at speed Vo at an elevation angle alpha up a hill of slope beta (alpha>beta). At what angle will the range (L) be maximum?

Homework Equations



L=(2Vo^2)/g*(cos(a)/cos(b))*(sin(a)-cos(a)tan(b)) is the distance up the slope that

The Attempt at a Solution



From the above equation we can see that if b=zero, we can maximize the equation by making a=45. If a=b, L=0. The question is, how do we make an equation that maximizes L. I tried integrating and setting the LHS to zero, but that didn't work. I need a relationship between L, b, and a that maximizes.

I'd like no full answers here since it is a homework question, just help as to how to find the relationship.
 
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Nevermind! The problem is solved. For future reference, you merely take the derivative of the L equation with respect to alpha. Do a few trig identities in order to get one alpha in the equation and solve.
 
Can anyone generalize and explain why shooting on a steep up-slope or down-slope might result in an error in elevation? Why might the error be equivalent?
 

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