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. Although the insect is at a distance d from the fish along a straight-line path at angle (as shown in the figure), a drop must be launched at a different angle 0 if its parabolic path is to intersect the insect. If r-31.8., at what initial angle ° must the fish shoot the drop so that it is at the top of its parabolic path when it reaches the insect? (Hint: Pretend that you know the value of d and work symbolically. Does it make sense that d will cancel? What does that mean?) Insect on twig d, Archer fish Answer:

Explanation / Answer

let t be the time taken from surface of water to insect directly below from the water surface

t = sqrt(2h/g) = sqrt[(2*d*sin(31.8))/g]

t^2 = 2*d*sin(31.8)/g

horizontal component of velcity is vx = d*cos(31.8)/t

vertical component of velcoty is vy = g*t = 9.8*t

let required angle is theta

then tan(theta) = vy/vx =( 9.81*t)/(d*cos(31.8)/t)

tan(theta) = (9.81*t^2)/(d*cos(31.8)

tan(theta) = (9.81* 2*d*sin(31.8)/g)/(d*cos(31.8))

tan(theta) = 2*sin(31.8)/cos(31.8) = 2*tan(31.8) = 1.24

theta= atan(1.24) = 51.12 degrees

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