A Chinook salmon has a maximum underwater speed of 3.58 m/s, but it can jump out of water with a speed of 6.26 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 it can swim to the top if the water speed is 3.00 m/s.
(a) If water has a speed of 2.07 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. The water's velocity is initially horizontal as it passes over the ledge. Enter the magnitude of the distance only.)
_____ m
(b) 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?
_____ m
the water velocity increase is due to the change in gravitational potential energy
1/2 m 3.00^2 - 1/2 m 2.07^2 = m g h
3.00^2 - 2.07^2 = 2 g h
the kinetic energy of the jump is converted to gravitational potential energy
6.26^2 = 2 g h
the max waterfall is the sum of the two heights
(a) Well, it seems like the water is going to do some high-flying acrobatics! To find out how far below the ledge the water will be moving at a speed of 3.00 m/s, we'll have to set up a little equation.
Since the water is moving at 2.07 m/s as it passes over the ledge, and it needs to slow down to 3.00 m/s for our ambitious salmon, we can say that the change in velocity is 3.00 m/s - 2.07 m/s = 0.93 m/s.
Now, we'll need to summon our knowledge of projectiles. When we launch the water, the horizontal component of its velocity will remain the same, but the vertical component will change due to gravity. We can use the fact that the horizontal displacement is the same as the vertical displacement (because the water falls straight down), and set up an equation to solve for the vertical displacement.
Using the equation: v_f^2 = v_i^2 + 2aΔy, where v_f is the final velocity (0 m/s when it reaches the top), v_i is the initial velocity (0.93 m/s), a is the acceleration (-9.8 m/s² due to gravity), and Δy is the vertical displacement, we can plug in the values:
0 = (0.93 m/s)^2 + 2(-9.8 m/s²)Δy
Simplifying and solving for Δy, we get:
Δy = - (0.93 m/s)^2 / (2(-9.8 m/s²))
Δy ≈ 0.0465 m
So the water will be moving with a speed of 3.00 m/s about 0.0465 meters below the ledge.
(b) Ah, the brave salmon wants to test its leaping prowess! To find the maximum height of the waterfall that the salmon can clear, we'll use a similar approach.
Since the salmon can swim up the fall for the remaining distance once it reaches the point where the water speed is less than 3.58 m/s, we can say that the vertical displacement at that point is equal to the height of the waterfall.
Using the same equation we used before, v_f^2 = v_i^2 + 2aΔy, where v_f is the final vertical velocity (0 m/s when it reaches the top), v_i is the initial vertical velocity (6.26 m/s), a is the acceleration (-9.8 m/s² due to gravity), and Δy is the vertical displacement, we can solve for Δy:
0 = (6.26 m/s)^2 + 2(-9.8 m/s²)Δy
Simplifying and solving for Δy, we get:
Δy = - (6.26 m/s)^2 / (2(-9.8 m/s²))
Δy ≈ 1.947 m
So the maximum height of the waterfall that the salmon can clear is approximately 1.947 meters. That's one impressive salmon!
Remember to take these calculations with a grain of salmon seasoning, as they assume ideal conditions and neglect factors such as air resistance. Good luck to our ambitious little swimmer!
To solve this problem, we can use the principles of projectile motion.
(a) To find the distance below the ledge where the water is moving with a speed of 3.00 m/s, we can start by considering the horizontal motion of the water. Since the water velocity is initially horizontal as it passes over the ledge, the horizontal velocity remains constant.
Given:
Water speed as it passes over the ledge (horizontal velocity) = 2.07 m/s
Water speed required for the salmon to swim up (horizontal velocity) = 3.00 m/s
The horizontal motion of the water is governed by the equation:
horizontal distance = horizontal velocity × time
Since the horizontal velocity remains constant, we can equate the distances traveled by the water to find the time taken:
2.07 × time = 3.00 × time
Solving for time:
time = 0 (as the distance traveled by the water is zero)
Therefore, the water will be moving with a speed of 3.00 m/s immediately below the ledge. The distance below the ledge will be 0 meters.
(b) To find the maximum height of the waterfall that the salmon can clear, we can consider the vertical motion of the salmon when it jumps vertically upward.
Given:
Maximum jumping speed of the salmon = 6.26 m/s
Required water speed for the salmon to swim up (vertical velocity) = 3.00 m/s
The vertical motion of the salmon can be analyzed using the equation:
vertical displacement = initial vertical velocity × time + (1/2) × acceleration × time^2
In this case, the initial vertical velocity is 6.26 m/s (positive upward), the acceleration is the acceleration due to gravity (9.8 m/s^2), and the final vertical velocity is 3.00 m/s (in the opposite direction).
Plugging in the values:
0 = 6.26 × time + (1/2) × (-9.8) × time^2
Simplifying the equation, we get a quadratic equation:
-4.9 × time^2 + 6.26 × time = 0
Solving this equation, we find two possible solutions for time:
time = 0 (as the starting point)
time = 1.276 s (positive solution)
Since the salmon cannot jump backwards, we discard the solution with time = 0. Therefore, the salmon takes 1.276 seconds to reach its maximum height.
Using this time, we can find the maximum height using the equation:
vertical displacement = initial vertical velocity × time + (1/2) × acceleration × time^2
vertical displacement = 6.26 × 1.276 + (1/2) × (-9.8) × (1.276)^2
Calculating this, we find:
vertical displacement ≈ 4.796 m
Therefore, the maximum height of the waterfall that the salmon can clear is approximately 4.796 meters.
To solve this problem, we need to consider the principles of projectile motion.
(a) To find the distance below the ledge where the water is moving with a speed of 3.00 m/s, we can use the equation for projectile motion:
Vf^2 = Vi^2 + 2aΔx
In this case, the initial velocity (Vi) is 2.07 m/s, the final velocity (Vf) is 3.00 m/s, and the acceleration (a) is -9.8 m/s^2 (negative because it acts downward). We want to find Δx, the distance below the ledge.
Plugging in the values into the equation and isolating Δx, we get:
Δx = (Vf^2 - Vi^2) / (2a)
Δx = (3.00^2 - 2.07^2) / (2 * -9.8)
Calculating this gives Δx = 0.2125 m (rounded to four decimal places).
Therefore, the water will be moving with a speed of 3.00 m/s, approximately 0.2125 m below the ledge.
(b) To find the maximum height of the waterfall that the salmon can clear, we can use the concept of conservation of energy. At the base of the fall, the salmon's kinetic energy from swimming is converted solely into gravitational potential energy at the maximum height.
Kinetic energy (KE) = Gravitational potential energy (PE)
Since the kinetic energy is given by KE = (1/2)mv^2, where m is the mass and v is the velocity, and the gravitational potential energy is given by PE = mgh, where g is the acceleration due to gravity (9.8 m/s^2) and h is the height, we can equate the two:
(1/2)mv^2 = mgh
Canceling the mass (m) gives:
(1/2)v^2 = gh
Rearranging the equation to solve for h, we get:
h = (1/2)(v^2) / g
h = (1/2)(6.26^2) / 9.8
Calculating this gives h = 2.0002 m (rounded to four decimal places).
Therefore, the maximum height of the waterfall that the salmon can clear is approximately 2.0002 m.