# Archer fish shoots a bug (similiar to another question)

1. Oct 1, 2009

### silencecloak

1. The problem statement, all variables and given/known data

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?

2. Relevant equations

Vy=V0sin$$\theta$$

max height =V02sin$$\theta$$2 / 2g

V2 = V02+2a(x-x0)

3. 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!

3. The attempt at a solution

2. Oct 1, 2009

anyone?

3. Oct 1, 2009

4. Oct 1, 2009

### silencecloak

57.381943177852

Although I am not sure how they got this.

5. Oct 2, 2009

### ideasrule

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. Oct 2, 2009

### silencecloak

I am still not piecing this together.

7. Oct 4, 2009

### silencecloak

if anyone could tell me where my logic is going wrong it would be greatly appreciated! thanks!