Arturo O. answered • 06/25/18

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Steven,

I have not seen you post in a while. Welcome back!

Nechama Z.

asked • 06/24/18Suppose a Chinook salmon needs to jump a waterfall that is 1.50m high. If the fish starts from a distance of 1.00 m from the base of the ledge over which the waterfall flows, 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. Can the fish make the jump? (Remember that a Chinook salmon can jump out of the water with a speed of 6.26m/s^2.)

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Arturo O. answered • 06/25/18

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Experienced Physics Teacher for Physics Tutoring

Steven,

I have not seen you post in a while. Welcome back!

Steven W. answered • 06/25/18

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Hi Nechama!

If we treat the salmon as a projectile, we can break the problem down into horizontal and vertical motion, and look at the conditions on each.

HORIZONTAL

---------------

In the horizontal direction, by definition, there is no acceleration for a projectile (since gravity is defined to be the only allowed acceleration for a projectile, and it is in the vertical). Thus, its velocity is constant (whatever horizontal velocity it has at launch will be the same horizontal velocity it has until it hits something and stops being a projectile), and the only equation of motion that matters is the constant-velocity one:

Δx = v_{x}t

where:

Δx = horizontal displacement (in flight)

v_{x} = (the constant) horizontal velocity

t = time

To solve for v_{x}, we have to know the other two values. The minimum requirement is that the salmon jump forward horizontally by 1 m (so Δx = 1 m). But the time is unknown, so we are stuck here for the moment. But, fortunately, we can go to the vertical direction to solve for time (as well as vertical initial velocity, which the problem also requires).

VERTICAL

-----------

In this direction, there is a non-zero acceleration (gravitational acceleration), so we have to fall back to using the kinematic equations (of uniformly accelerated motion) in their full glory. That means, to solve for one value, we need to know at least three others. Since the problem asks to solve for initial vertical velocity (v_{oy}), let's look at that first:

to find: v_{oy}

What do we know? We always know the acceleration, which is g, downward (which I will call the negative direction). We are also told that the salmon must reach a height of 1.5 m, so we know the vertical displacement (Δy).

We have to know a third quantity, and that is a little tricky. But in the vertical direction, the salmon, as a minimum condition, only has to JUST make it to a height of 1.5 m. The minimum effort this would take is to jump so as to reach that height and *no higher.* This would mean that, at that height, the salmon must come to rest in the vertical direction. If it were still moving upward at that point (or coming down), then it would have arrived with more than the minimum effort necessary. For this reason, on the condition of minimum effort, we can take the salmon's final velocity (v_{y}) at 1.5 m in height to be zero.

So now we have what we need to solve for v_{oy}:

to find: v_{oy}

know: a, Δy, v_{y}

a = -9.8 m/s^{2} (or whatever your class uses at the value of g)

Δy = 1.5 m

v_{y} = 0

You can now go to your stable of kinematic equations, pick out the one involving these four quantities, and solve for v_{oy}.

Once that is done, you can solve for the time it takes for the salmon to make that jump, using a different kinematic equation.

BACK TO THE HORIZONTAL

-------------------------------

Once you have solved for the time using the vertical direction quantities, you can carry that back to the horizontal equation we left above:

1 m = v_{x}t

And now, knowing t, you can solve for v_{x}.

At this point, you have both the horizontal and vertical components of the salmon's initial velocity, so you can solve for the required magnitude of the salmon's overall initial velocity using the Pythagorean theorem.

I hope that helps! If you have any questions or would like to check and answer, just let me know.

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