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anonymous

  • one year ago

Relative velocity: A chinook salmon can jump out of water with a speed of 6.26 m/s. If the salmon is in a stream with water speed equal to 1.5 m/s, how high in the air can the fish jump if it leaves the water travelling vertically upwards relative to the Earth?

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  1. anonymous
    • one year ago
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    I know that the basic relative velocity equation is \[v_{AB} = v_{AE} - v_{BE}\], but this problem is giving me more trouble than the others have been for some reason. I'm calling the water W, the salmon S, and Earth E, so it would become \[v_{SW} = v_{SE} - v_{WE}\], but the only known quantity seems to be \[v_{WE}\], which is <1.5 m/s, 0 m/s> if we use due east as postive x and up as positive y. I'm really clueless as to how to go from here.

  2. dan815
    • one year ago
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    hmm weird lol -.- why would the 1.5 m/s water speed even come into effect when that shud be off to the side, and the fish is jumping upward

  3. anonymous
    • one year ago
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    substituting v_0= 7.76 m/s and v=0 m/s for the velocity at the highest point, I get: (0 m/s)^2 = (7.76 m/s)^2 - 2(9.80 m/s^2)*y_f and then y_f comes out to equal 3.07 m... the book says it's 1.88 m. I think you're not taking into account the direction of the vectors. If I just ignore the water speed and use v_0=6.26 m/s, y_f is still wrong, coming out to 2.00 m.

  4. dan815
    • one year ago
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    unless its jumping up or down a waterfall

  5. anonymous
    • one year ago
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    yea so the stream is in horizontal direction

  6. anonymous
    • one year ago
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    @dan815 it depends on the scenario. but not affecting is what he meant

  7. anonymous
    • one year ago
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    I think the problem is that the direction of the fish's jumping velocity is unknown... It swims straight up relative to Earth, which would mean it's swimming upstream as well, and the magnitude of the jumping velocity is 6.26 m/s but the upward component would be less than that.

  8. anonymous
    • one year ago
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    I figured it out... First, the velocity of the fish relative to the water has to be calculated. Since it's swimming vertically relative to Earth, it has to swim opposite how the water is moving in the x-direction (so that they balance out to 0 m/s relative to Earth). This makes the x-component of v_SE_x = -1.50 m/s. Then, we know that the magnitude of v_SE = 6.26 m/s, but the y-component is unknown (though we know it has to be positive, because the fish is swimming upward). So use the pythagorean theorem since component vectors form a right triangle: v_SE_y = sqrt(6.26^2 + (-1.50)^2) m/s = 6.08 m/s. Now, the fish jumps out of the water, so the velocity relative to the water now becomes relative to the Earth, and we can say that v_SE_0 = < -1.50 m/s, 6.08 m/s>. Apply the time-independent free-fall equation v_y^2 = v_0y^2 - 2g*y. The y-velocity at the peak will be 0 m/s: (0 m/s)^2 = (6.08 m/s)^2 - 2(9.08 m/s^2)*y -36.9664 m^2/s^2 = -19.16 m/s^2 * y 1.89 m = y ~ 1.88 m (the textbook answer) So, that was that... Thanks for trying to help, at least.

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