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  1. #1
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    physics trajectory problem

    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.
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    I dont feel like enough info is given.

    Any ideas?
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  3. #3
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    Quote Originally Posted by silencecloak View Post
    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.
    Quote Originally Posted by silencecloak View Post
    I dont feel like enough info is given.

    Any ideas?
    In the vertical direction you know that a = -9.8 m/s^2, final velocity = 0 (because you're at the top of the parabolic path) and x = 0.55 m (I didn't check this value by the way). Calculate t using an appropriate uniform acceleration formula.

    Now substitute a = -9.8 m/s^2, final velocity = 0, x = 0.55 m and initial velocity = u \sin \theta 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 = u \cos \theta. Substitute into an appropriate uniform acceleration formula. Call this equation (2).

    Now solve equations (1) and (2) simultaneously for \theta.
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