Question

BIO EST The archerfish (genus Toxotes) preys on land-based insects inhabiting tree limbs above the water. It has the ability to shoot a jet of water droplets from its specialized mouth, which can knock an insect down and onto the water surface. The archerfish typically shoots the water jet at an angle of 70‚àò above the horizontal. These droplets can reach a height of 10 m! Taking the angle to be 70‚àò and the maximum height to be 10 m, estimate the initial speed of the water droplets.

Answer 1

Answer:

The initial speed of the water droplets is approximately 14.904 meters per second.

Explanation:

We can describe the water jet from archerfish as part of parabolic motion, which consists of the superposition of two different motions. First, an horizontal motion at constant velocity and, second, a free fall motion. The maximum height is reached when vertical component of speed is zero. The equation of motion is described below:

$v^{2} = v_{o}^{2}\cdot \sin^{2} \theta + 2\cdot g \cdot (y-y_{o})$

Where:

$v_{o}$ - Initial speed of the water jet, measured in meters per second.

$v$ - Current speed of the water jet, measured in meters per second.

$g$ - Gravitational acceleration, measured in meters per square second.

$y_{o}$ - Initial height, measured in meters.

$y$ - Current height, measured in meters.

Now we clear the initial speed within equation:

$v_{o}^{2} =\frac{v^{2}-2\cdot g\cdot (y-y_{o})}{\sin^{2}\theta}$

$v_{o} = \sqrt{\frac{v^{2}-2\cdot g\cdot (y-y_{o})}{\sin^{2}\theta} }$

If we know that $v = 0\,\frac{m}{s}$, $g = -9.807\,\frac{m}{s^{2}}$, $y = 10\,m$, $y=0\,m$ and $\theta = 70^{\circ}$, the initial speed of the water droplets is:

$v_{o} = \sqrt{\frac{\left(0\,\frac{m}{s} \right)^{2}-2\cdot \left(-9.807\,\frac{m}{s^{2}} \right)\cdot (10\,m-0\,m)}{\sin^{2}70^{\circ}} }$

$v_{o} \approx 14.904\,\frac{m}{s}$

The initial speed of the water droplets is approximately 14.904 meters per second.