I dont feel like enough info is given.
Upon spotting an insect on a twig overhanging water, an archer fish squirts water drops at the insect to knock it into the water. Although the fish sees the insect along a straight-line path at angle φ and distance d, a drop must be launched at a different angle θ0 if its parabolic path is to intersect the insect. If φ = 38.0° and d = 0.900 m what θ0 is required for the drop to be at the top of the parabolic path when it reaches the insect?
Just poking at the problem I have found that the height of the insect is .55m.
If i plug the height into the max height formula, i get velocity at time 0 = 5.33m/s
I am not sure if this is even right, or if I am on the right track.
Now substitute a = -9.8 m/s^2, final velocity = 0, x = 0.55 m and initial velocity into an appropriate uniform acceleration formula. Call this equation (1).
In the horizontal direction you know that a = 0, x = ...., t = (calculated above) and initial velocity . Substitute into an appropriate uniform acceleration formula. Call this equation (2).
Now solve equations (1) and (2) simultaneously for .