+118609 Hi all, I was going to do this but it is a little harder than expected and I am a little short on time right now.
Would anyone else like to take a shot?
(This is Archimedeskay question, he is having trouble posting here )
+118609 mm
\(x=(v_0cos\alpha)t \qquad y=(v_0sin\alpha)t-16t^2\\ let\;\;a=v_0cos\alpha \qquad and \qquad b=v_0sin\alpha\\~\\ x=at \qquad y=bt-16t^2\\ t=\frac{x}{a} \quad so\\ y=b(\frac{x}{a})-16(\frac{x}{a})^2\\ y=-\frac{16}{a^2}\;x^2+b(\frac{x}{a})\\ -y=\frac{16}{a^2}\;x^2-\frac{b}{a}\;x\\ \frac{ -a^2y}{16}=\;x^2-\frac{a^2}{16}\frac{b}{a}\;x\\ \frac{ -a^2y}{16}=\;x^2-\frac{ab}{16}\;x\\ \frac{ -a^2y}{16}+(\frac{ab}{32})^2=\;x^2-\frac{ab}{16}\;x+(\frac{ab}{32})^2\\ \frac{ -a^2}{16}(y-(\frac{b^2}{64}))=\;(x-\frac{ab}{32})^2\\ \)
\(\;(x-\frac{ab}{32})^2=\frac{ -a^2}{16}(y-(\frac{b^2}{64}))\\ \;(x-\frac{(v_0)^2cos\alpha sin\alpha}{32})^2=\frac{ -(v_0cos\alpha)^2}{16}(y-\frac{(v_0sin\alpha)^2}{64})\\ \;(x-\frac{(v_0)^2 2cos\alpha sin\alpha}{64})^2=\frac{ -(v_0cos\alpha)^2}{16}(y-\frac{(v_0sin\alpha)^2}{64})\\ \;(x-\frac{(v_0)^2 sin(2\alpha)}{64})^2=\frac{ -4(v_0cos\alpha)^2}{64}(y-\frac{(v_0sin\alpha)^2}{64})\\\)
So assuming that I have not made a mistake - which is a big assumption.
This is a concave down parabola.
The vertex is \(\left(\frac{(v_0)^2 sin(2\alpha) }{64}, \frac{(v_0)^2 sin^2(\alpha) }{64} \right)\)
