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Carnegie Mellon University

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Problem 27 Hard Difficulty

Suppose a chinook salmon needs to jump a waterfall that is 1.50 $\mathrm{m}$ high. If the fish starts from a distance 1.00 $\mathrm{m}$ from the base of the ledge over which the waterfall flows, (a) find the $x$ - and $y$ -components of the initial velocity the salmon would need to just reach the ledge at the top of its trajectory. (b) Can the fish make this jump? (Note that a chinook salmon can jump out of the water with an initial speed of 6.26 $\mathrm{m} / \mathrm{s} .$ )

Answer

a. $\quad 20 x = 1.81 \mathrm { m } / \mathrm { s } \quad 4 \mathrm { g } _ { \mathrm { y } } = 5.42 \mathrm { m } / \mathrm { s }$
b. $y e s ,$ it can

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Video Transcript

so for per day Ah, we're going to vision vision the salmon as a projectile And so we can say that then v y final squared equals V Y initial squared plus two times the acceleration in the UAE direction times delta y and we can then save me. Why initial is gonna be equaling the square root V? Why final squared minus two times the acceleration of the Y direction times delta y We know that the final velocity would be essentially zero, so this would be equal in the square root of negative two multiplied by negative 9.80 meters per second squared multiplied by 1.50 meters. And we find that then V why initial is equaling 5.42 meters per second Now the elapsed time for the upward flight, we can say is delta T and Delta T would be vey wife final essentially minus V. Why initial divided by the acceleration of the wind direction, the wife final is again zero. So this would be equaling negative 5.42 meters per second, divided by 9.80 meters per second squared And this is giving us 0.553 seconds. No if the horizontal displacement were to be one meter. This means that the initial velocity in the ex direction needs to be equaling. Delta X Divided by Delta T. This is equaling 1.0 meters, divided by a 0.553 seconds. This is giving us 1.81 meters per second. This would be our initial ex velocity or initial velocity in the extraction. For part A. Now for part B. The speed with which the salmon must leave the water we can save the initial must be equal in the square root of these of X squared plus B sub y squared, so this would be equal in the square root of 1.81 meters per second quantity squared plus 5.42 meters per second quantity squared. Stand the square root. This is giving us 5.71 meters per second. We know that the initial equaling 5.71 meters per second is of course, less than 6.26 meters per second, which is the capability essentially of a salmon. That's the fastest speed with which the salmon can exit the water. And so, yes, we can say the salmon is capable of making the jump. That is the end of the solution. Thank you for watching.

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