Chinook salmon are able to move through water especially fast by jumping out of

Question

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 6.89 m/s at 46.9 degree above the horizontal, salls through the air a distance L before retuning to the water, and then swims the same distance L underwater in a straight, horizontal line with velocity 3.82 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 3.82 m/s?

Explanation / Answer

Get a sheet of paper and divide it into three columns. Go ahead, get some paper and do this. You will see that as we go through this problem using these three columns the problem will practically solve itself. This three-column method will work for all word problems.

In the first column write down all the information from the problem. Be sure to include the information that the problem is asking you to look for. Your first column should look like this from top to bottom: va = 6.89m/s; ta = ?s; dL = ?m; vw = 3.82m/s; tw = ?s; dt = ?m; tt = ?s; vav = ?m/s; Theta = 46.9 degrees

The ‘va’ means the velocity in the air and the ‘vw’ means the velocity in the water. I think you can figure the rest out!

Ok, now we know what the variables are. We need to find the equations that have these variables so that we can solve them to find the unknowns. You should have a list of equations that you have already studied. The equations will go in the middle column.

You may or may not have learned this equation: dh = (vR^2 x sin 2theta) / g. This tells you that when an object is thrown or kicked at an angle you can calculate its range, how far it went horizontally, by squaring the velocity at which it was kicked or thrown or shot, multiplied by the sine of twice the angle given. In this problem you were given an angle of 46.9 degrees, so we will use the sine of 93.8 degrees.

To know how long the fish is in the air we have to know how high it went and how long it took it to go that high while accelerating at 9.8m/s^2. Well, the vertical side of the triangle opposite the 46.9-degree angle will tell you the vertical velocity. Since the final vertical portion of the velocity is zero meters per second, the time it takes the fish to go up, or down is calculated using this equation: t1/2 = sin theta vR / g. This comes from this equation: a = (vf –vi) / t.

Then, of course since the horizontal velocities are constant we can find the time by rewriting the constant velocity equation v = d / t as t = d / v. You should already know that average velocity is calculated by dividing the total time into the total distance.

Now that we have thought about our equations we can write our middle column: From top to bottom it should look something like this: dL = (vR^2 x sin 2theta) / g (Note that I changed the distance horizontal the terms given in the problem); t1/2 = sin theta vR / g; v = d / t; t = d / v; vav = dt / tt; tt = ta + tw; dt = 2dL. This looks like a lot of equations but that is OK because we only have to solve them one at a time!

In the third column all you have to do is put the numbers and the units from the first column into the equations from the middle column and do the math with both the numbers and the units. What could be easier?

Here we go. From top to bottom your third column might look like this: dL = (6.89m/s)^2 x sin 93.8degrees / 9.8m/s^2; dL = 4.83 m

Now we have our dL. Let’s get the time in the air. Remember that if you throw something up and it takes it 1.5s to get as high as it will go, it will take 1.5s to come back down again. t1/2 = (sin 46.9degrees x 6.89m/s) / 9.8m/s^2; t1/2 = 0.51 s. Remember that this it the time it takes to go up, so the total time in the air is 1.02s.

The fish swims in the water for 4.83m at 3.82m/s. This takes t = 4.83m / 3.82m/s. t = 1.26s. We add this to the time in the air and get tt = 1.02s + 1.26s. tt = 2.28s.

The average velocity is vav = 9.66m / 2.28s; vav = 4.24m/s.

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