Archer fish shoots a bug (similiar to another question)

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Homework Help Overview

The problem involves an archer fish aiming to squirt water at an insect located above water. The fish must calculate the launch angle required for the water drop to reach the insect, given the angle of sight and distance to the insect.

Discussion Character

  • Exploratory, Assumption checking, Problem interpretation

Approaches and Questions Raised

  • Participants discuss the geometry of the problem, including the height and horizontal distance involved. There are attempts to calculate the necessary launch angle and initial velocity. Some participants question the definitions of variables used in the equations.

Discussion Status

There is ongoing exploration of the calculations and methods. One participant has provided a numerical answer but expressed uncertainty about how it was derived. Others are seeking clarification on the variables and the relationships between them, indicating a productive dialogue without a clear consensus.

Contextual Notes

Participants are working under the constraints of the problem as stated, including the specific angles and distances provided. There is a noted lack of clarity regarding the definitions of certain variables, which may affect the calculations.

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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\theta

max height =V02sin\theta2 / 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\theta = 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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anyone?
 
I haven't checked your calculations, but your method is certainly O.K.
 
the correct answer was

57.381943177852

Although I am not sure how they got this.
 
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.
 
I am still not piecing this together.
 
if anyone could tell me where my logic is going wrong it would be greatly appreciated! thanks!
 

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