Recent content by Justinjah91

Ah yeah that makes sense. Sorry about that! Unfortunately, plugging that in makes it blow up into a crazy 4 order polynomial that I have no clue how to solve. I'm probably going to just use a numerical solver rather than trying for the analytical solution. I really would have expected it to be...

Yes I suppose it does. That would mean that ##\frac {dh} {dθ} = \frac {d(Δx)} {dθ} =0##. I've gone ahead and given that a try, but the displacement I get is dependent on that optimum launch angle, and I still don't know that. I'm fairly certain that I've done the derivatives correctly, but...

Here's a fully typed version of the problem with a diagramMy attempt: Given the angle of the hill, I know that the horizontal displacement of the arrow and my vertical height on the hill are related by ##Δx=d+\frac h {Tan(60)}## ...(1) where d is the distance of the enemy from the base of...