Trajectory problem - launching from origin.

[Eclair_de_XII]

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 (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?"

Homework Equations

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

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?

 

[haruspex]

$0.728m=\frac{v_0^2}{9.8\frac{m}{s^2}}sin72°$

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.

 

[Eclair_de_XII]

I guess... 2x = R?

 

[Eclair_de_XII]

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

 

[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...

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

 

[ehild]

,

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

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

 

[ehild]

I guess... 2x = R?

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

 

[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.

 

[ehild]

0.529 m...

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

Yes. So what is θ?

 

[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.

 

[haruspex]

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

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

 

[ehild]

Yes.

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

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

 

[Eclair_de_XII]

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

 

[Eclair_de_XII]

Like I said, I need the v_x and v_y to do the equation.

 

[ehild]

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

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

 

[ehild]

Like I said, I need the v_x and v_y to do the equation.

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.

 

[haruspex]

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

http://www.australasianscience.com.au/article/issue-december-2010/spitting-image.html
But I think the problem setter does not intend this to be taken into account.

 

[ehild]

http://www.australasianscience.com.au/article/issue-december-2010/spitting-image.html
But I think the problem setter does not intend this to be taken into account.

Well, the distance d is not really needed to solve the problem. The article is very interesting, but it is quite complicated how the fish sees that fly. So I think the angle given is the angle in air the direction to the fly makes with the horizontal.

 

[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.

 

[haruspex]

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.

Well done.