Calculating Archerfish's Line of Sight Angle in Relation to Squirt Angle

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The discussion centers on calculating the archerfish's line of sight angle in relation to its squirt angle while aiming at a beetle above the water. The archerfish squirts water at a speed of 2.00 m/s and an angle of 50 degrees, targeting a beetle 3.00 cm above the water's surface. The calculated time for the water to hit the beetle yields two possible answers due to the projectile's trajectory. The horizontal distance for optimal targeting was found to be 0.023 m, and the line of sight angle was determined to be 52 degrees, raising questions about whether this angle can exceed the squirt angle. The discussion emphasizes the importance of accuracy in calculations, suggesting that the line of sight can indeed be greater than the squirt angle.
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Homework Statement


An archerfish hunts by dislodging an unsuspecting insect from its resting place with a stream of water expelled from the fish's mouth.
Suppose the archerfish squirts water with a speed of 2.00 m/s at an angle of 50 degrees above the horizontal, and aims for a beetle on a leaf 3.00 cm above the water's surface.
a. Find the time it takes for the water to hit the beetle; why are there 2 answers?
b. At what horizontal distance from the beetle should the archerfish fire if it is to hit its target in the least time?
c. Find the angle (above the horizontal) of the archerfish's line of sight; explain why the line of sight angle is not 50 degrees!

What I want to know is if the line of sight angle is allowed to be greater than the angle of the squirt?

Homework Equations


3 DVAT formulas

The Attempt at a Solution


After I solved for t (using quadratic formula, vertical acceleration, initial vertical velocity(which I got using law of sines), and delta y), I solved for delta x using the smaller of the two values I got for t (this time is when it reaches .03m high before the peak of the curve). The value I got for delta x was .023 m. I am 95% certain that this a correct value. Then to solve for the angle of the line of sight of the archerfish I used delta x, delta y, and law of tangents and got an angle of 52 degrees above the horizontal. However, the angle of the squirt of water was 50 degrees above the horizontal, and I am unsure if the angle of the line of sight is allowed to be larger than the angle of the squirt of water?
 
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If the archerfish is looking at the beetle, at what angle is his line of sight?
What does a basketball player look at when shooting free throws?
 
J Hann said:
If the archerfish is looking at the beetle, at what angle is his line of sight?
What does a basketball player look at when shooting free throws?
A basketball player would be looking at the net when shooting a freethrow, and the angle of his shot would be above his line of sight, so does that mean that the archerfish's lines of sight can't be greater and the angle of the squirt of water?
 
How would you find the line of sight if you don't know what the fish is looking at?
 
My teacher said it was assumed in the problem that the fish was looking at its prey, sorry, I guess I should have included that
 
RaeZ said:
After I solved for t (using quadratic formula, vertical acceleration, initial vertical velocity(which I got using law of sines), and delta y), I solved for delta x using the smaller of the two values I got for t (this time is when it reaches .03m high before the peak of the curve). The value I got for delta x was .023 m. I am 95% certain that this a correct value. Then to solve for the angle of the line of sight of the archerfish I used delta x, delta y, and law of tangents and got an angle of 52 degrees above the horizontal. However, the angle of the squirt of water was 50 degrees above the horizontal, and I am unsure if the angle of the line of sight is allowed to be larger than the angle of the squirt of water?
I suspect you are just not being accurate enough in your calculations. Keep at least 4 significant figues at each step.
I get a slightly larger number for x.
What do you get for the time?
 

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