Question

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 0.800m 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° 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 0.600m away from the fish. The archerfish successfully shoots down the resting insect by spitting water drops at the same angle 60° above the surface and with the same initial speed v as before. At what height h above the surface was the insect?

 

Answer 1

Part A Answer

We know that the drop needs to fall into the water at x = 0.800m. So it needs to travel 0.800m horizontally at the same time it takes to go up and come down (y-direction). Solving this problem can be done in several ways, I did it by setting up a system of equations in order to solve for the flight time t, like this:

Step 1: Find the equation for the time it takes the drop to land in the water. Since the drop starts and ends in the water (y and y₀ are the same), the time it takes to the peak is exactly 1/2 of the total flight time:

v_fy = v_0y + a_yt_peak
t_peak = 0.5t

Therefore:

0 = sin(θ)v + a_y0.5t
0 = sin(θ)v + (-9.8)(0.5)t
0 = sin(60)v + (-4.9)t
4.9t = 0.866v

t = 0.1767v

Step 2: The other half of our system of equations will be the time it takes for the drop to go 0.800m in the x-direction. We’re just solving for the flight time again using a different equation (that we can set equal to the first one):

x-x₀ = cos(θ)vt + 1/2a_xt²
0.800 = 0.5vt
1.6 = vt
t = 1.6/v

Step 3: Now that we have our 2 equations, both in terms of t, they are equal to each other and we can solve the system:

(0.1767v) = (1.6 / v)
v = (9.05285 / v)
v² = 9.05285

v = 3.01m/s

Part B Anawer

We start by solving for t, using the velocity from Part A. Remember that the angle is 60 degrees:

x-x₀ = v_xt + 1/2a_xt²
0.600 = cos(60)(3.01)t
0.600 = 1.505t
t = 0.3987

Now solve for y, using t:

y-y₀ = v_yt + 1/2at²
y – 0 = sin(60)(3.01)(0.3987) + 1/2(-9.8)(0.3987)²
y = 1.0393 – 0.7789

y = 0.260m