Chinook salmon are able to move through water especially fast by jumping out of the water periodically. This behavior is called porpoising. Suppose a salmon swimming in still water jumps out of the water with velocity 6.41 m/s at 44.1° above the horizontal, sails through the air a distance L before returning to the water, and then swims the same distance L underwater in a straight, horizontal line with velocity 3.91 m/s before jumping out again.

(a) Determine the average velocity of the fish for the entire process of jumping and swimming underwater.

(b) Consider the time interval required to travel the entire distance of 2L. By what percentage is this time interval reduced by the jumping/swimming process compared with simply swimming underwater at 3.91 m/s?

(a) To determine the average velocity of the fish for the entire process, we need to consider the distance traveled and the time taken for each part (jumping and swimming).

Let's analyze the jumping part first. The salmon jumps out of the water with a velocity of 6.41 m/s at an angle of 44.1° above the horizontal. We can separate this velocity into horizontal and vertical components:

Vertical component: v_y = 6.41 m/s * sin(44.1°)
Horizontal component: v_x = 6.41 m/s * cos(44.1°)

During the jumping part, the salmon travels a distance L horizontally, so the time taken for this part is t_jump = L / v_x.

Now let's analyze the swimming part. The salmon swims underwater at a velocity of 3.91 m/s. During this part, the salmon also travels a distance L. The time taken for this part is t_swim = L / v_swim, where v_swim is the swimming velocity.

To calculate the average velocity for the entire process, we need to consider the total distance traveled (2L) and the total time taken:

Total distance traveled: 2L
Total time taken: t_total = t_jump + t_swim

The average velocity is then given by:

Average velocity = Total distance traveled / Total time taken

Substituting the values, we get:

Average velocity = (2L) / (t_jump + t_swim)

(b) To compare the time interval required to travel the entire distance of 2L with and without the jumping/swimming process, we need to calculate the time interval for swimming underwater at a constant velocity of 3.91 m/s.

The time taken to swim underwater for a distance of 2L is:

t_swim_only = (2L) / v_swim

To calculate the percentage reduction in time interval, we compare the time taken with the jumping/swimming process (t_total) to the time taken for swimming underwater only (t_swim_only):

Percentage reduction = [(t_swim_only - t_total) / t_swim_only] * 100%

To determine the average velocity of the fish for the entire process of jumping and swimming underwater, we need to calculate the total displacement and the total time taken.

(a) Total Displacement:

The fish jumps out of the water at an angle of 44.1° above the horizontal with a velocity of 6.41 m/s. The horizontal component of the initial velocity is given by:

Vx = V * cos(θ)
= 6.41 m/s * cos(44.1°)
= 4.545 m/s

The vertical component of the initial velocity is given by:

Vy = V * sin(θ)
= 6.41 m/s * sin(44.1°)
= 4.141 m/s

When the fish jumps, it will reach a maximum height and fall back down. The total horizontal displacement during this time is twice the horizontal component of the initial velocity, as it returns back to the starting point. Therefore, the total horizontal displacement is:

2L = 2 * (2 * Vx * t)
= 2 * (2 * 4.545 * t)
= 18.18 * t

Where t is the time taken for the fish to jump and fall back into the water.

When the fish is underwater, it is swimming in a straight, horizontal line with a velocity of 3.91 m/s. Therefore, the total horizontal displacement underwater is:

L = Vx_water * t_water
= 3.91 * t_water

where t_water is the time taken for the fish to swim underwater.

Since the total displacement is the same for both jumps and swimming, we equate the two displacements:

18.18 * t = 3.91 * t_water

We solve for t_water:

t_water = (18.18 * t) / 3.91

Total Time:

The total time taken for the entire process is the sum of the time taken for jumping (2t) and swimming underwater (t_water):

Total time = 2t + t_water
= 2t + (18.18 * t) / 3.91

Average Velocity:

The average velocity is given by the total displacement divided by the total time:

Average velocity = (Total displacement) / (Total time)
= (2L) / (2t + t_water)
= (2L) / (2t + (18.18 * t) / 3.91)

(b) To compute the percentage reduction in time compared to simply swimming underwater at 3.91 m/s, we need to compare the total time taken for the entire process with the time it would take to swim the same distance underwater at 3.91 m/s:

Time without jumping = L / Vx_water
= L / 3.91

Percentage reduction = 100 * (1 - (Total time) / (Time without jumping))

Now you can substitute the given values of L, Vx_water, and solve for the unknowns to get the final answers.

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