In your proof you have the following:
cot(2π−ϕ)=tan(ϕ)Surely this would mean
cot(ϕ−2π)=tan(−ϕ)Which would leave you with
ϕ−2π=cot−1(tan(−ϕ))Linking this to the equation
2(ϕ+θ)=cot−1(tan(−ϕ))⇒2(ϕ+θ)=ϕ−2π+nπ,n∈ZSeeing as this would yield a negative angle, taking the initial condition that
n=1 would result in your equation, and the same answer. Really a pedantic thing though.
The co-ordinate system picked is beautiful for the problem however.
Edit: When you subbed your t into your
vy equation, how did you get that expression?
I get
vy=−v0(sinθ+2cosθtanϕ)And just from inspection for your equation for
vy, certain values roughly above
ϕ=21 and
ϕ=23 would yield a positive solution, which seeing as you took upwards as positive, would be illogical. A value such as
ϕ=6π would yield
vy=0.