Recent content by DavidFrank

Looking at my teacher's notes, he did this: cos(θ+α)*cosθ - sinθ*sin(θ+α) = 0 Thank you, I will give this a shot when I return home.

I am sorry but I am stuck. Do you want to me to take a few steps back before I found "R" (the distance traveled on the incline) or should I keep on going by using Vfy2 = Viy2 + 2g(Yf - Yi)? Because if I were to continue with the substitution of Vi^2 into "R" I'd get: 1 = [4 * sinΘ * sinα *...

Thank you for your response haruspex, so far I have used the position formula and the Vf^2 = Vi^2 + 2a(Δx) formulas

That is so far all I have gotten. My attempt was to find "R" which is the incline distance of this projectile problem. I sort of ran into some issue when I tried finding Vi and substituting it into R = [2sin(θ)cos(θ+ α) * v.i^2] / [g * cos^2(α)] since the R cancels out.

Oh okay so Vfy = 0 and Vfx = Vfcos(0°) = Vf. Let me think about this some more, thank you.

I am sorry, I generally would show some work but I have no clue on how to start off to find angle theta. The answer is given below and I have to get it. I don't know at all how to approach this problem and unfortunately I am not too strong on my trig identities either but I will try. May some...