Upon spotting an insect on a twig overhanging water, an archer fish squirts water drops at the insect to knock it into the water (Fig. 4 -38). Although the fish sees the insect along a straight-line path at angle f and distance d, a drop must be launched at a different angle θ if its parabolic path is to intersect the insect. If φ= 36.0° and d 0.900 m, what launch angle (theta) is required for the drop to be at the top of the parabolic path when it reaches the insect?
2. Relevant equation
So the two equations i have used so far are, y-yi=vi*sin(θ)-.5gt^2, x-xi=vi*cos(θ).
The Attempt at a Solution
Firstly, I found both xi and yi by using the trigonometric relationships. .9sin(φ)=.53 and .9cos(φ)=.73 for the vertical and horizontal distances. This yields the maximum height and distance from the initial position to the insect. I substituted .53 into the horizontal distance equation which gave me .73/(vi*cos(θ))= t, which i then substituted into the vertical distance equation. This is the part where I am stuck. After simplifying the equation, i am left with two unknown variables, vi and θ. 0= (-3.577*vi^2-.53)+tan(θ)-3.577vi^2*tan^2(θ). Should I solve the quadratic equation for θ and go from there?? p.s Sorry that I didn't use LaTeX, i'm not very comfortable with it, since i've never used it before, but i will try to use it in my following posts.