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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?
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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anonymous
 one year ago
Best ResponseYou've already chosen the best response.0I 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.

dan815
 one year ago
Best ResponseYou've already chosen the best response.0hmm 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

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0substituting 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.

dan815
 one year ago
Best ResponseYou've already chosen the best response.0unless its jumping up or down a waterfall

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0yea so the stream is in horizontal direction

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0@dan815 it depends on the scenario. but not affecting is what he meant

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0I 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.

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0I 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 xdirection (so that they balance out to 0 m/s relative to Earth). This makes the xcomponent of v_SE_x = 1.50 m/s. Then, we know that the magnitude of v_SE = 6.26 m/s, but the ycomponent 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 timeindependent freefall equation v_y^2 = v_0y^2  2g*y. The yvelocity 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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