Question

3) 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?

Answer 1

Concepts and reason

The concepts used to solve this problem are kinematic equation and time of flight.

To calculate the speed of the archer fish $\left( v \right)$, first, calculate the time of flight of spitted water and then calculate the height$\left( s \right)$ of insect above the surface by using kinematic equation.

Fundamentals

The equation of kinematics which shows the relation among initial velocity, acceleration, distance and time is as follows;

$s = {v_0}t + \frac{1}{2}a{t^2}$

Here, ${v_0}$ is the initial velocity, $t$ is the time and $a$is the acceleration.

Similarly, the equation which shows the relation between final velocity and initial velocity and acceleration is as follows;

$v = u + at$

Here,$v$ is final velocity, $u$is initial velocity,$t$ is time and $a$ is acceleration.

The expression for time of the flight is,

$t = \frac{{2v\sin \theta }}{g}$

(A)

Calculate the speed of the archer fish.

The time of the flight of spitted water is,

$t = \frac{{2v\sin \theta }}{g}$

Substitute $9.8{\rm{ m}} \cdot {{\rm{s}}^{ - 2}}$ for $g$ and $60^\circ$ for $\theta$in above equation.

$\begin{array}{c}\\t = \frac{{2v\sin 60^\circ }}{{9.8{\rm{ m}} \cdot {{\rm{s}}^{ - 2}}}}\\\\ = \left( {0.1767\;v} \right){{\rm{m}}^{ - 1}} \cdot {{\rm{s}}^2}\\\end{array}$

Spitted water will travel$0.80{\rm{ m}}$ horizontally.

Displacement of water in this time period is,

$x = vt\cos \theta$

Substitute $\left( {0.1767\;v} \right){{\rm{m}}^{ - 1}} \cdot {{\rm{s}}^2}$ for $t$ ,$60^\circ$ for $\theta$ and $0.80{\rm{ m}}$for $x$in above equation.

$\begin{array}{c}\\0.80{\rm{ m}} = v\left( {0.1767\;v} \right){{\rm{m}}^{ - 1}} \cdot {{\rm{s}}^2}\left( {\cos 60^\circ } \right)\\\\0.80{\rm{ m}} = {v^2}\left( {0.1767{\rm{ }}} \right)\frac{1}{2}{{\rm{m}}^{ - 1}} \cdot {{\rm{s}}^2}\\\\v = \sqrt {\frac{{2\left( {0.80{\rm{ m}}} \right)}}{{0.1767\;{{\rm{m}}^{ - 1}} \cdot {{\rm{s}}^2}}}} \\\\ = 3.01{\rm{ m}} \cdot {{\rm{s}}^{ - 1}}\\\end{array}$

(b)

Calculate the height of insect above the surface.

Vertical component of the velocity is,

${v_v} = v\sin \theta$

Substitute $3.01\;{\rm{m}} \cdot {{\rm{s}}^{ - 1}}$for $v$and $60^\circ$ for $\theta$in above equation.

$\begin{array}{c}\\{v_v} = \left( {3.01\;{\rm{m}} \cdot {{\rm{s}}^{ - 1}}} \right)\sin 60^\circ \\\\ = 2.6067{\rm{ m}} \cdot {{\rm{s}}^{ - 1}}\\\end{array}$

Horizontal component of the velocity is,

${v_h} = v\cos \theta$

Substitute $3.01\;{\rm{m}} \cdot {{\rm{s}}^{ - 1}}$for $v$and $60^\circ$ for $\theta$in above equation.

$\begin{array}{c}\\{v_h} = \left( {3.01\;{\rm{m}} \cdot {{\rm{s}}^{ - 1}}} \right)\cos 60^\circ \\\\ = 1.505{\rm{ m}} \cdot {{\rm{s}}^{ - 1}}\\\end{array}$

When horizontal $\left( {0.60\;{\rm{m}}} \right)$distance away from the fish.

The time of flight for distance (d) is ,

$t = \frac{d}{{{v_h}}}$

Substitute $0.60\;{\rm{m}}$for $d$and $1.505{\rm{ m}} \cdot {{\rm{s}}^{ - 1}}$for ${v_h}$in equation $t = \frac{d}{{{v_h}}}$.

$\begin{array}{c}\\t = \frac{{0.60\;{\rm{m}}}}{{1.505{\rm{ m}} \cdot {{\rm{s}}^{ - 1}}}}\\\\ = 0.3987{\rm{ s}}\\\end{array}$

Distance of the insect above the surface is,

$s = {v_v}t + \frac{1}{2}g{t^2}$

Substitute $2.6067{\rm{ m}} \cdot {{\rm{s}}^{ - 1}}$for ${v_v}$and $0.3987{\rm{ s}}$for $t$and $- 9.8{\rm{ m}} \cdot {{\rm{s}}^{ - 2}}$for $g$in above equation.

$\begin{array}{c}\\s = \left( {2.6067{\rm{ m}} \cdot {{\rm{s}}^{ - 1}}} \right)\left( {0.3987{\rm{ s}}} \right) + \frac{1}{2}\left( { - 9.8{\rm{ m}} \cdot {{\rm{s}}^{ - 2}}} \right){\left( {0.3987{\rm{ s}}} \right)^2}\\\\ = 0.260{\rm{ m}}\\\end{array}$

Ans: Part A

The speed of the archer fish is $3.01\;{\rm{m}} \cdot {{\rm{s}}^{ - 1}}$.