Archer fish shoots a bug (similiar to another question)

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An archer fish must launch water drops at a specific angle to hit an insect on a twig, which requires calculating the correct launch angle θ0 based on the observed angle φ and distance d. The initial calculations provided a height of 0.55 m and a horizontal distance of 0.709 m, leading to an initial velocity squared (V02) of 3.727. However, the correct launch angle was determined to be approximately 57.38 degrees, prompting further discussion on the calculations and the interpretation of variables. Participants debated the correct use of equations related to projectile motion, particularly focusing on the relationship between vertical and horizontal components of the launch. Clarification on the variables and equations used is necessary to resolve discrepancies in 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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