91Ó°ÊÓ

Before we dive into the fish's incredible aerial adventure, let's talk about its "Initial Conditions." When the eagle releases the fish, the initial conditions play a key role in the fish's subsequent motion.
The fish is dropped from the eagle's claws, but not from rest in terms of horizontal motion. It starts with an initial horizontal velocity of \( v_{0x} = 6.0 \, \mathrm{m/s} \), thanks to the eagle's flight. In the vertical direction, when the fish begins its fall, it doesn't have any initial vertical velocity. This means \( v_{0y} = 0 \). The force of gravity will soon change this, though.

- **Horizontal Motion**: Constant at \( 6.0 \, \mathrm{m/s} \).
- **Vertical Motion**: Starts from rest and accelerates due to gravity \( g = 9.8 \, \mathrm{m/s^2} \).

So, these initial conditions set the stage for the trajectory of the fish—its path will be a combination of this steady horizontal motion and increasing vertical speed.

Now that the fish is in freefall, we need to calculate how its velocity changes over time. This involves understanding both its horizontal and vertical components of velocity.

**Horizontal velocity (\(v_x\)):** Remains constant throughout the motion since no horizontal forces are acting on the fish. Therefore, \( v_x = 6.0 \, \mathrm{m/s} \).

**Vertical velocity (\(v_y\)):** As the fish falls, it accelerates downward due to gravity. The vertical velocity at any time \( t \) can be calculated using:
\[v_y = g t = 9.8 t \]

To find the fish's total speed at any moment, we use the Pythagorean theorem, combining horizontal and vertical components:
\[v = \sqrt{v_x^2 + v_y^2} = \sqrt{6.0^2 + (9.8t)^2} \]

This equation allows us to find the speed when doubled to \( 12.0 \, \mathrm{m/s} \) and then \( 24.0 \, \mathrm{m/s} \) in the sequential steps of the given problem.

The "Time of Flight" is really about when key velocity milestones occur, specifically when the fish reaches designated speeds during its descent.

**Doubling the Speed (\(12.0 \mathrm{m/s}\)):**- We start by setting the calculated speed formula equal to \( 12.0 \, \mathrm{m/s} \).- Solving \( \sqrt{6.0^2 + (9.8t)^2} = 12.0 \) involves squaring both sides and isolating \( t \).
- This yields an approximate \( t \approx 1.06 \, \mathrm{s} \).

**Doubling Again (\(24.0 \mathrm{m/s}\)):**- The same process applies, but setting the speed to \( 24.0 \, \mathrm{m/s} \).- Solving gives us \( t' \approx 2.37 \, \mathrm{s} \).
- To find the additional time taken after the first doubling, subtract the first time interval: \( 2.37 - 1.06 = 1.31 \, \mathrm{s} \).

Understanding the "Time of Flight" helps us appreciate how an object, even as unpredictable as a falling fish, adheres to predictable physical laws.