Archer fish shoots a bug (similiar to another question)

  • Thread starter silencecloak
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In summary, an archer fish squirts water drops at an insect to knock it into the water, but must launch the drops at a different angle than the line of sight in order to hit the insect. Using the given values of φ = 38.0° and d = 0.900 m, a height of 0.55 m and length of 0.709 m were determined for the triangle formed by the insect, fish, and drop. Using these values and the equations Vy=V0sin\theta and max height =V02sin\theta2 / 2g, the initial upward speed of the water (V0sin(theta)) can be found. The water must reach across the 0.709 m
  • #1
silencecloak
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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. 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?


Homework Equations



Vy=V0sin[tex]\theta[/tex]

max height =V02sin[tex]\theta[/tex]2 / 2g

V2 = V02+2a(x-x0)

The Attempt at a Solution



Solving for the triangle gave me a height of .55m and a length of .709 m

Using these values to solve for V02 I ended up with V02 = 3.727.

Plugging these values into the max height formula and solving sin2[tex]\theta[/tex] = 2*g*h / 3.7272 gave me a theta value of 61.75 degrees

did i solve this correctly?

Thanks!






The Attempt at a Solution

 
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  • #2
anyone?
 
  • #3
I haven't checked your calculations, but your method is certainly O.K.
 
  • #4
the correct answer was

57.381943177852

Although I am not sure how they got this.
 
  • #5
silencecloak said:
Solving for the triangle gave me a height of .55m and a length of .709 m

Using these values to solve for V02 I ended up with V02 = 3.727.

Sorry, I should have checked your work more carefully. What is V0 supposed to represent? Is it the V0 in the second equation you gave, or the third? (Note that the second equation is derived from the third, since Vy=0 at the top of a freefall trajectory.)

You could use your second equation to get V0sin(theta), the initial upward speed of the water. The water has to reach across the 0.709 m of horizontal distance in time for it to reach its highest point, which gives you a second equation. Work it out from here.
 
  • #6
I am still not piecing this together.
 
  • #7
if anyone could tell me where my logic is going wrong it would be greatly appreciated! thanks!
 

1. How does the archer fish shoot a bug out of the water?

The archer fish has a specialized mouth that allows it to shoot a stream of water at its prey. It uses its tongue to create a suction and then rapidly closes its gills, forcing the water out in a focused jet.

2. Why does the archer fish shoot a bug out of the water?

The archer fish uses this unique hunting method to catch insects that are out of its reach on overhanging branches or leaves. It also allows the fish to conserve energy by not having to swim up to the surface to catch its prey.

3. How accurate is the archer fish's shooting ability?

The archer fish is incredibly accurate and can shoot its prey with precision from a distance of up to several feet away. This is due to its ability to adjust the velocity and direction of the water jet to compensate for the refraction of light in the water.

4. What types of bugs does the archer fish typically shoot?

The archer fish typically shoots small insects such as flies, moths, and beetles. However, it has also been observed shooting larger prey such as spiders and even small lizards.

5. Is the archer fish's shooting ability learned or innate?

The archer fish's shooting ability is a learned behavior. Juvenile archer fish observe and mimic the shooting behavior of older, more experienced fish in their group. However, the fish must also have the proper physical adaptations in order to successfully shoot its prey.

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