# Trajectory problem - launching from origin.

1. Mar 9, 2016

### Eclair_de_XII

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 (Fig. 4-38). 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 = 36.0° and d = 0.900 m, what launch angle θ0 is required for the drop to be at the top of the parabolic path when it reaches the insect?"

2. Relevant equations
$y=(tanθ_0)x-\frac{gx^2}{2(v_0cosθ_0)^2}$
$R = \frac{v_0^2}{g}sin2θ_0$
$θ=36°$
$|d|=0.9m$

3. The attempt at a solution
$\vec d = (0.728m)i+(0.529m)j$
$0.728m=\frac{v_0^2}{9.8\frac{m}{s^2}}sin72°$
$7.1344=v_0^2(sin72°)$
$v_0=2.335\frac{m}{s}$

$0.529m=(0.72812)(tanθ)-\frac{(9.8\frac{m}{s^2})(0.728m)^2}{2[(2.335\frac{m}{s})(cosθ°)]^2}sin72°$
$(-0.729)(sec^2θ)+(0.72812)tanθ-0.529=0$
$(-0.729)(tan^2θ+1)+(0.72812)tanθ-0.529=0$
$(-0.729)tan^2θ+(0.72812)tanθ-1.257=0$

Let:

$q=tanθ$
so:
$q^2-0.999q+1.724=0$

And then I come to an impasse at the 4ac>b2 block. I'm thinking I'm mistaking my R for my x... So how would I find R for v0?

Last edited: Mar 9, 2016
2. Mar 9, 2016

### haruspex

36 degrees is Φ, not θ0.
And on what basis are you using the range equation here? (You can make use of it, but not in the way you have applied it.) In the range equation, how exactly is the range defined?

Incidentally, in reality, the archer fish also has to allow for refraction.

3. Mar 9, 2016

### Eclair_de_XII

I guess... 2x = R?

4. Mar 9, 2016

### Eclair_de_XII

If 36° = Φ, then that leaves two unknowns in the equation...

5. Mar 9, 2016

### Eclair_de_XII

$0.728m=\frac{v_0^2}{9.8\frac{m}{s^2}}sinθcosθ$

Yeah, two unknowns... I can't really solve for anything. The least I can do is group them...

$7.1344\frac{m^2}{s^2}=(v_0sinθ)(v_0cosθ)$

6. Mar 9, 2016

### ehild

No, the water drop is launched from the surface of the water.

7. Mar 9, 2016

### ehild

Yes. A figure helps. What is y of the fly?

8. Mar 9, 2016

### Eclair_de_XII

0.529 m...

I should really start drawing these pictures on my own, instead of trying to do these problems without one.

9. Mar 9, 2016

### ehild

Yes. So what is θ?

10. Mar 9, 2016

### Eclair_de_XII

I'm still missing v0, though. Without it, I can't complete the operation. And it doesn't give me a time.

11. Mar 9, 2016

### haruspex

Sure, but are the fish's eyes above water?

12. Mar 9, 2016

### ehild

You can get the expression for x and the maximum y in therms of θ. What is their ratio?

13. Mar 9, 2016

### Eclair_de_XII

No, I derived x and y from Φ, so it'll just give me 36° again.

14. Mar 9, 2016

### Eclair_de_XII

Like I said, I need the vx and vy to do the equation.

15. Mar 9, 2016

### ehild

It can be . Otherwise the problem would indicate to take refraction into account, and both the angle and the distance would be different.

16. Mar 9, 2016

### ehild

you have x= $\frac{v_0^2}{2g}sin2θ_0$. What is the maximum of the parabola? Substitute for x into the formula for y. the speed cancels if you take the ratio y/x.

17. Mar 9, 2016

### haruspex

18. Mar 10, 2016

### ehild

19. Mar 10, 2016

### Eclair_de_XII

Okay, I got it.

$\frac{y}{x}=tanθ-(\frac{g\frac{v_0^2}{g}sinθcosθ}{2v_0^2cos^2θ})=tanθ-\frac{sinθcosθ}{2cos^2θ}=tanθ-\frac{1}{2}tanθ$
$0.727=\frac{1}{2}tanθ$
$1.454=tanθ$
$θ=55.48°$

Geez, this was much harder than it looked.

20. Mar 10, 2016

Well done.