Upon spotting an insect on a twig overhanging water, an archer fish squirts wate

Question

Upon spotting an insect on a twig overhanging water, an archer fish squirts water drops at the insect to knock it into the water (see figure below). Although the fish sees the insect along a straight-line path at angle and distance d, a drop must be launched at a different angle 0 if its parabolic path is to intersect the insect. If = 35.5° and d = 0.870 m, what launch angle 0 is required for the drop to be at the top of the parabolic path when it reaches the insect? (Consider up to be the positive y direction and to the right to be the positive x direction.)

counterclockwise from the +x-axis

Explanation / Answer

If the fish is at (0,0), then the insect is at (0.870m*cos35.5, 0.870m*sin35.5), or
(0.708m, 0.505m)

max height = 0.505m = (V*sin(theta))^2 / (2*g)

0.5 range = 0.708m = (V^2 sin(2 theta)) / 2*g

Divide first by second:

0.505 /0.708 = 0.7132 = (sin^2 theta) / sin(2 theta)

Now use sin(2 theta) = 2 * (sin theta) * (cos theta)

0.7132 = (sin theta) / 2(cos theta)

0.7132 = 0.5 * (tan theta).....................(tan = sin / cos)

theta = arctan1.4264 = 54.96 degree

launch angle 0 is required for the drop to be at the top of the parabolic path when it reaches the insect is 54.96 degree

Related Questions

Navigate

• Browse (All)
• Browse U
• Subjects

Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.