Physics question 'Clear the Runway?

PHYSICS--Inclined plane Question...We ALL missed it!!!! Where are the errors?

• http://answers.yahoo.com/question/index;_ylt=AsWet5YwK.F1OJoFrhGYtSzty6IX;_ylv=3?qid=20081016001518AA0AI3B&show=7#profile-info-AxV8NLuNaa This is the link. I had to pose a question, so I did. I misread the question. I thought that the object was fired up the plane ON the plane, but obviously, it is a tad different. But no one got the actual question right! The object is fired at an angle theta from the beginning of a ramp at angle phi, let's say. The question was "What angle of the launch yields the maximum distance up the plane. The solutions were FANTASTICALLY complex. The answer is very simple (although the calculus "proof" is tedious. My freshman physics class got it right in under a minute. Since they have taken calculus and have studied projectile motion, I unleashed them on it. They correctly reasoned that the maximum distance up the plane corresponds to the maximum height off the ground which corresponds to the maximum range which occurs at 45 degrees (provided that the angle of the plane is < 45 degrees) otherwise, one fires the object directly up the plane at theta. It is intuitive and "elegant" with no math. As I said, the math is quite straight forward. Write the parabola in terms of x instead of parametrically. Solve for the point of intersection of the plane and the trajectory. Once you have an expression for y in terms of phi, differentiate it wrt phi and set to 0 and solve...out "pops" 45 degrees. Since I asked the question, I'll give "best" to anyone who finds the most errors in the other solutions...starting with the "best" of the wrong answers. I already outlined my error, but you are free to "rub my nose in it" if you like. Happy hunting! Now you will learn about the worst aspect of my job..."...but professor, I'm sure my answer is right. Would you please show me where I went wrong?" -Fred

• Answer:

  With all due respect Herr Professor, I feel I'm being told that Θ = 45° will give Dmax regardless of φ. The problem with that is that "intuitively" I know that when φ = 45, D = 0 if Θ = 45. I did the math and agree with the Wikipedia article for the expression for Rs (slant range); http://en.wikipedia.org/wiki/Trajectory To maximize this range one needs to differentiate it with respect to Θ, not φ. This differential comes out dRs/dΘ = K[cos²Θ - sin²Θ + 2m*cosΘ*sinΘ] where K = (2Vi²/g)*√(1+m²) and m= tanφ Setting = 0 gives a quadratic in tanΘ: tan²Θ - 2mtanΘ - 1 = 0 → tanΘ = tanφ + secφ (for maximum distance up the incline.) I have checked this numerically and it's good for -90° <φ< 90°. Interestingly, starting at φ = 0, for which the best range Θ IS 45°, each incremental increase in φ results in Θ incrementing by ½ that amount: φ........Θ(best range) 0........45 10......50 20......55 30......60 40......65 45......67.5 50......70 60......75 70......80 80......85 90......90 Letting φ take on negative values brings the value of Θ down to zero when φ = -90°. This eq and its results seem very intuitive to me..........