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## Homework Statement

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.