A small archerfish (length 20 to 25 cm) lives in brackish waters of Southeast Asia from India to the Philippines. This aptly named creature captures its prey by shooting a stream of water drops at an insect, either flying or at rest. The bug falls into the water and the fish gobbles it up. The archerfish has high accuracy at distances of 1.2 m to 1.5 m, and it sometimes makes hits at distances up to 3.5 m. A groove in the roof of its mouth, along with a curled tongue, forms a tube that enables the fish to impart high velocity to the water in its mouth when it suddenly closes its gill flaps. Suppose the archerfish shoots at a target 1.85 m away, at an angle of 30.0° above the horizontal. With what velocity must the water stream be launched if it is not to drop more than 1.70 cm vertically on its path to the target?

Question

A small archerfish (length 20 to 25 cm) lives in brackish waters of Southeast Asia from India to the Philippines. This aptly named creature captures its prey by shooting a stream of water drops at an insect, either flying or at rest. The bug falls into the water and the fish gobbles it up. The archerfish has high accuracy at distances of 1.2 m to 1.5 m, and it sometimes makes hits at distances up to 3.5 m. A groove in the roof of its mouth, along with a curled tongue, forms a tube that enables the fish to impart high velocity to the water in its mouth when it suddenly closes its gill flaps. Suppose the archerfish shoots at a target 1.85 m away, at an angle of 30.0° above the horizontal. With what velocity must the water stream be launched if it is not to drop more than 1.70 cm vertically on its path to the target?

Answer 1

To find the velocity at which the water stream must be launched, we can use the principles of projectile motion and determine the initial velocity required to hit the target without dropping more than 1.70 cm vertically on its path.

First, let's break down the given information:
- Initial vertical displacement (Δy) = 1.70 cm = 0.017 m
- Horizontal distance (d) to the target = 1.85 m
- Launch angle (θ) = 30.0° above the horizontal

Using the equations of projectile motion, we can find the initial velocity (v₀) required. The horizontal and vertical components of the initial velocity can be calculated using the following equations:

Horizontal component: v₀x = v₀ * cos(θ)
Vertical component: v₀y = v₀ * sin(θ)

Since there is no acceleration in the horizontal direction (assuming ideal conditions), the horizontal displacement (Δx) can be given by:

Δx = v₀x * t

We can rearrange this equation to solve for time (t):

t = Δx / v₀x

To find the time of flight, we can use the vertical displacement (Δy):

Δy = v₀y * t + (1/2) * a * t²

Since the water stream is launched vertically, the acceleration in the vertical direction (a) is equal to the acceleration due to gravity (-9.8 m/s²).

Plugging in the values and solving the equation:

0.017 m = (v₀ * sin(θ)) * t + (1/2) * (-9.8 m/s²) * t²

Simplifying the equation:

0.017 m = (v₀ * sin(30°)) * (Δx / (v₀ * cos(30°))) + (1/2) * (-9.8 m/s²) * (Δx / (v₀ * cos(30°)))²

Simplifying further:

0.017 m = (v₀ * 0.5) * (Δx / (v₀ * 0.866)) - 4.9 * (Δx² / v₀² * 0.75)

0.017 m = 0.5 * Δx * 0.866 - 4.9 * (Δx² / (v₀² * 0.75))

Now, we can substitute the value of Δx = 1.85 m and solve the equation for v₀.

0.017 m = 0.5 * (1.85 m) * 0.866 - 4.9 * (1.85 m)² / (v₀² * 0.75)

Solving this equation will give us the value of v₀, the initial velocity required for the water stream to hit the target without dropping more than 1.70 cm vertically.