Question

The archerfish uses a remarkable method for catching insects sitting on branches or leaves above the waterline. The fish rises to the surface and then shoots out a stream of water precisely aimed to knock the insect off its perch into the water, where the archerfish gobbles it up. Scientists have measured the speed of the water stream exiting the fish's mouth to be 3.7 m/s. An archerfish spots an insect sitting 15 cm above the waterline and a horizontal distance of 27 cm away. The fish aims its stream at an angle of 35 ∘ from the waterline. Determine the height above the waterline that the stream reaches at the horizontal position of the insect.

Answer 1

The height above the waterline that the stream reaches at the horizontal position of the insect is 15 cm.

Step-by-step explanation:

Please, see the attached figure for a description of the problem.

The motion is parabolic and this is the equation that describes the position of an object in such a motion:

r = (x0 + v0 · t · cos α, y0 + v0 · t · sin α + 1/2 · g · t²)

Where:

r = position vector

x0 = initial horizontal position

v0 = initial velocity

t = time

α = angle of the stream with the waterline

y0 = initial vertical position

g = acceleration due to gravity

First, let´s calculate how much time it takes the stream to reach the horizontal position of 0.27 m. For this, we will use the equation of the x-component of the vector position:

x = x0 + v0 · t · cos α

Since the origin of the reference system is located at the mouth of the fish, x0 = 0. Then:

0.27 m = 3.7 m/s · t · cos 35º

t = 0.27 m /(3.7 m/s · cos 35º)

t = 0.089 s

Now, with this time, we can calulate the vertical position (height) of the stream using the equation for the y-component of the vector "r":

y = y0 + v0 · t · sin α + 1/2 · g · t²

y = 0 m + 3.7 m/s · 0.089 s · sin 35º + 1/2 · (-9.8 m/s²) · (0.089s)²

y = 0.15 m

when the stream reaches 27 cm horizontally, it will reach 15 cm vertically and hit the insect!