Physics problem - Salmon jumping waterfall

[FlowiwGhar]

Homework Statement

Salmon often jump waterfalls to reach their
breeding grounds.
Starting downstream, 2.73 m away from a
waterfall 0.614 m in height, at what minimum
speed must a salmon jumping at an angle of
26.7◦
leave the water to continue upstream?
The acceleration due to gravity is 9.81 m/s^2.
Answer in units of m/s.

Relevant Equations

N/A

m * g * h + (1/2) * m * v² = m * g * y

Simplifying the equation:

g * h + (1/2) * v² = g * y

Substituting the values:

g * 0.614 + (1/2) * v² = g * 2.73 * sin(26.7°)

Now, let's solve for v:

(1/2) * v² = g * 2.73 * sin(26.7°) - g * 0.614

v² = 2 * (g * 2.73 * sin(26.7°) - g * 0.614)

v = √(2 * (g * 2.73 * sin(26.7°) - g * 0.614))

Substituting the value of g = 9.81 m/s² and performing the calculations:

v ≈ √(2 * (9.81 * 2.73 * sin(26.7°) - 9.81 * 0.614))

v ≈ √(2 * (53.803 - 6.018))

v ≈ √(2 * 47.785)

v ≈ √95.57

v ≈ 9.78 m/s

 

[haruspex]

Homework Statement: Salmon often jump waterfalls to reach their
breeding grounds.
Starting downstream, 2.73 m away from a
waterfall 0.614 m in height, at what minimum
speed must a salmon jumping at an angle of
26.7◦
leave the water to continue upstream?
The acceleration due to gravity is 9.81 m/s^2.
Answer in units of m/s.
Relevant Equations: N/A

m * g * h + (1/2) * m * v² = m * g * y

Simplifying the equation:

g * h + (1/2) * v² = g * y

Substituting the values:

g * 0.614 + (1/2) * v² = g * 2.73 * sin(26.7°)

Now, let's solve for v:

(1/2) * v² = g * 2.73 * sin(26.7°) - g * 0.614

v² = 2 * (g * 2.73 * sin(26.7°) - g * 0.614)

v = √(2 * (g * 2.73 * sin(26.7°) - g * 0.614))

Substituting the value of g = 9.81 m/s² and performing the calculations:

v ≈ √(2 * (9.81 * 2.73 * sin(26.7°) - 9.81 * 0.614))

v ≈ √(2 * (53.803 - 6.018))

v ≈ √(2 * 47.785)

v ≈ √95.57

v ≈ 9.78 m/s

Have you posted this because you want someone to check it or because you have some reason to think your answer is wrong?

 

[FlowiwGhar]

Have you posted this because you want someone to check it or because you have some reason to think your answer is wrong?

The answer is incorrect, so I'd like someone to check it please

 

[nasu]

There is nothing to check. Your formulas and the values you plugged in looks like some random manipulation of symbols. For example., what is the meaning of the variable labeled "y" in your first equation? What is the meaning of the first equation and how is relevant to the problem?

 

[haruspex]

Judging from your equations, y-h is the height the salmon would clear the dam by were it not for gravity. How that leads to your equation is mysterious. E.g. you add the KE to the PE needed to reach the top of the dam (mgh), but the PE is partly spent on that.
Start with sound physical principles and write equations from those.

 

[kuruman]

From your energy conservation equation it appears that $y-h$ is the height which the salmon reaches with zero speed. Is that what is happening here? As you know from projectile motion problems, the horizontal component of the velocity is constant. That means that the speed is never zero unless the fish jumps straight up which is not the case here.

Furthermore, it seems that you have misunderstood how to relate the given quantities to each other. In your equation you substitute $y=2.73\sin(26.7^{\circ}).$ That is incorrect and makes no sense. Draw a diagram showing $y$, $h$, the 2.73 m horizontal distance and the roughly parabolic trajectory of the fish to guide your thinking.

I would carry @haruspex 's suggestion one step further and recommend that you write an equation for the parabolic trajectory that relates the vertical displacement to the horizontal displacement, the initial speed and the angle of projection.

I note that since the angle of projection and the displacement of the salmon are all specified, there is no real minimization to be done.