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

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The discussion focuses on determining the optimal angle for firing a projectile up a hill to maximize range, given the projectile's speed and the hill's slope. The equation for range on a slope is presented, indicating that maximizing the range involves finding a relationship between the angles of elevation and slope. A successful approach to solving the problem involves taking the derivative of the range equation with respect to the elevation angle and applying trigonometric identities. Additionally, the conversation touches on potential errors in elevation when shooting on steep slopes, questioning why these errors might be equivalent. Understanding these relationships is crucial for accurate projectile motion analysis in varying terrains.
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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?
 
The book claims the answer is that all the magnitudes are the same because "the gravitational force on the penguin is the same". I'm having trouble understanding this. I thought the buoyant force was equal to the weight of the fluid displaced. Weight depends on mass which depends on density. Therefore, due to the differing densities the buoyant force will be different in each case? Is this incorrect?

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