A Chinook Salmon has a maximum underwater speed of 3.58 m/s, but it can jump out of water with a speed of 6.68 m/s. to move upstream past a waterfall, the salmon does not need to jump to the top of the fall, but only to a point in the fall where the water speed is less than 3.58 m/s; it can then swim up the fall for the remaining distance. Because the salmon must make forward progress in the water, let's assume that it can swim to the top if the water speed is 3.00 m/s. If water has a speed of 1.40 m/s as it passes over a ledge, how far below the ledge will the water be moving with a speed of 3.00 m/s? (Note that water undergoes projectile motion once it leaves the ledge.)
If the salmon is able to jump vertically upward from the base of the fall, what is the maximum height of waterfall that the salmon can clear?
Do this problem in two steps.
<< If water has a speed of 1.40 m/s as it passes over a ledge, how far below the ledge will the water be moving with a speed of 3.00 m/s?>>
Use conservation of energy to get he distance Y1 that the water must fall to have a velocity of 3.00 m/s.
M g Y1 = (1/2) M [(3 m/s)^2 - (1.4 m/s)^2]
The mass M cancels out. Solve for Y1.
<<If the salmon is able to jump vertically upward from the base of the fall, what is the maximum height of waterfall that the salmon can clear?
>>
Use conservation of energy again. The distance Y2 that the salmon can rise is
given by
M g Y2 = (1/2) M (6.58)^2
The unknown M (of the salmon) once again cancels out. Solve for Y2
Y1 + Y2 is the maximum waterfall height that can be climbed by swimming from Y2 to the top (rising out of the water a distance Y1, re-entering the water and swimming Y2 to the top).
2.45
To answer the first part of the question, we need to use the principle of conservation of energy. The potential energy gained by the water as it falls is equal to the difference in kinetic energy between the speeds of the water over the ledge and at a lower point.
Starting with the equation:
M g Y1 = (1/2) M [(3 m/s)^2 - (1.4 m/s)^2]
Here, M represents the mass of the water, g is the acceleration due to gravity, and Y1 is the distance below the ledge where the water is moving with a speed of 3.00 m/s.
By canceling out the mass term on both sides of the equation, we can solve for Y1. This will give us the distance below the ledge where the water will be moving with a speed of 3.00 m/s.
Now, let's move on to the second part of the question.
To determine the maximum height of the waterfall that the salmon can clear, we need to consider the maximum jump height the salmon can achieve.
Using conservation of energy once again, we have the equation:
M g Y2 = (1/2) M (6.68 m/s)^2
Here, M represents the mass of the salmon, g is the acceleration due to gravity, and Y2 is the maximum height the salmon can jump vertically upward from the base of the fall.
By canceling out the mass term on both sides of the equation, we can solve for Y2. This will give us the maximum height of the waterfall that the salmon can clear.
Finally, to calculate the maximum height of the waterfall that the salmon can climb from the bottom to the top, we add Y1 and Y2 together. This will give us the total height the salmon can overcome by swimming from Y2 to the top.