The archerfish is a type of fish well known for its ability to catch resting insects by spitting a jet of water at them. This spitting ability is enabled by the presence of a groove in the roof of the mouth of the archerfish. The groove forms a long, narrow tube when the fish places its tongue against it and propels drops of water along the tube by compressing its gill covers.

When an archerfish is hunting, its body shape allows it to swim very close to the water surface and look upward without creating a disturbance. The fish can then bring the tip of its mouth close to the surface and shoot the drops of water at the insects resting on overhead vegetation or floating on the water surface.

Part A: At what speed v should an archerfish spit the water to shoot down a floating insect located at a distance 0.800 m from the fish? Assume that the fish is located very close to the surface of the pond and spits the water at an angle 60 degrees above the water surface.

Part B: Now assume that the insect, instead of floating on the surface, is resting on a leaf above the water surface at a horizontal distance 0.600 m away from the fish. The archerfish successfully shoots down the resting insect by spitting water drops at the same angle 60 degrees above the surface and with the same initial speed v as before. At what height h above the surface was the insect?

This situation is related to projectile motion or parabolic motion, in which the travel of the water shoot made by the archerfish has two components: x-component and y-component. Being their main equations as follows:

x-component:

[tex]x=V_{o}cos\theta t[/tex] (1)

Where:

[tex]V_{o}[/tex] is the water shoot initial speed

[tex]\theta=60\°[/tex] is the angle at which the water was shot

[tex]t[/tex] is the time since the water is shot until it hits the insect

y-component:

[tex]y=y_{o}+V_{o}sin\theta t-\frac{gt^{2}}{2}[/tex] (2)

Where:

[tex]y_{o}=0m[/tex] is the initial height of the water shoot (assuming that the fish is located very close to the surface of the water)

[tex]y=0 m[/tex] is the final height of the the water shoot (assuming that the insecto is also located very close to the surface of the water)

[tex]g=-9.8m/s^{2}[/tex] is the acceleration due gravity (always directed downwards)

With these given conditions, let's begin with the first answer:

A) Find water shoot initial speed

Firstly, we will have to isolate [tex]t[/tex] from both equations:

From (1):

[tex]t=\frac{x}{V_{o}cos \theta}[/tex] (3)

From (2):

[tex]t=\frac{-2V_{o}sin\theta}{g}[/tex] (4)

Making (3)=(4):

[tex]\frac{x}{V_{o}cos \theta}=\frac{-2V_{o}sin\theta}{g}[/tex] (5)

Isolating [tex]V_{o}[/tex]:

[tex]V_{o}=\sqrt{-\frac{x g}{2sin\theta cos \theta}}[/tex] (6)

Knowing that [tex]2sin\theta cos \theta=sin(2\theta)[/tex], we can rewrite (6) as:

[tex]V_{o}=\sqrt{-\frac{x g}{sin(2\theta)}}[/tex] (7)

Substituting the known values:

[tex]V_{o}=\sqrt{-\frac{(0.8 m)(-9.8 m/s^{2})}{sin(2(60\°))}}[/tex] (8)

[tex]V_{o}=3.0087 m/s \approx 3 m/s[/tex] (9) This is the initial velocity

B) Find the insects height

Now, let's change two of the conditions:

- [tex]x=0.6 m[/tex]

-[tex]y[/tex] is different from zero and we have to find its value.

So, lets's begin with (3), keeping the initial velocity we found in part (A) and find the time [tex]t[/tex]:

[tex]t=\frac{x}{V_{o}cos \theta}[/tex] (3)

[tex]t=\frac{0.6 m}{3 m/s}cos(60\°)}[/tex] (10)

[tex]t=0.4 s[/tex] (11)

On the other hand, we have:

[tex]y=0 m +V_{o}sin\theta t-\frac{gt^{2}}{2}[/tex] (12)

Substituting (11) in (12):

[tex]y=0 m +(3 m/s) sin(60\°) (0.4 s)-\frac{(9.8 m/s^{2})(0.4 s)^{2}}{2}[/tex] (13)

Finally:

[tex]y=0.255 m[/tex] This is the height of the insect above the water surface