Optimizing Projectile Distance on an Inclined Plane

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The discussion focuses on optimizing the angle for projectile motion on an inclined plane to maximize distance up the hill. Two formulas are presented: 1/2arctan(-cotϕ) = θ and θ = π/4 + ϕ/2. The first formula is confirmed to be valid, but it fails at φ=0, while the second formula correctly predicts a 45-degree angle when φ=0. At φ=45 degrees, the first formula suggests a 67.5-degree angle, contrasting with the second formula's -22.5 degrees. The second formula is widely recognized, and users are encouraged to seek its derivation online for better understanding.
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Homework Statement


"A person stands at the base of a hill that is a straight incline making an angle ϕ with the horizontal. For a given initial speed v₀, at what angle θ (to the horizontal) should objects be thrown so that the distance d they land up the hill is as large as possible?"

I've looked up the same question on the internet, but the answers are either the formula (1/2arctan(-cotϕ) = θ) or π/4+ ϕ/2. I already figured out how to get 1/2arctan(-cotϕ) = θ, but I do not know how they got they π/4+ ϕ/2. Also, which is the correct answer?

Homework Equations



1/2arctan(-cotϕ) = θ
 
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The π/4 + φ/2 formula predicts 45 degrees when φ=0, which is certainly correct. The arctan formula doesn't work at φ=0. At φ=45 degrees, the first formula predicts a reasonable 67.5 degrees, while the arctan formula seems to say -22.5 degrees. Better check my calc; I make mistakes.

The second formula is pretty well known. No doubt you could find a derivation of it on the web. Try searching for "Projectile motion on an incline". Of course it is always better to figure it out yourself if you have the time.
 
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