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The beautiful Multnomah Falls in Oregon are approximately 206m high. If the Columbia river is flowing horizontally at 2.90m/s just before going over the falls, what is the overall velocity of the water when it hits the bottom?

Of course, I can help you with that!

To find the overall velocity of the water when it hits the bottom of Multnomah Falls, we need to consider two components: the horizontal velocity of the river and the vertical velocity due to the height of the falls.

Let's break down the problem:

1. Horizontal velocity: Given that the Columbia river is flowing horizontally at 2.90 m/s, this remains the same throughout the fall. So, the horizontal velocity remains at 2.90 m/s.

2. Vertical velocity: To determine the vertical velocity component, we can use the equations of motion. The height of the falls is given as 206 m.

Using the equation vf^2 = vi^2 + 2gh, where vf is the final velocity, vi is the initial velocity (which is 0 since the water starts from rest at the top), g is the acceleration due to gravity (approximately 9.8 m/s^2), and h is the height.

Plugging in the values, we have:
vf^2 = 0 + 2(9.8 m/s^2)(206 m)
vf^2 = 2(9.8)(206)
vf^2 = 4031.2
vf ≈ 63.5 m/s

Now, we can find the magnitude of the overall velocity by using the Pythagorean theorem.

Overall velocity = sqrt(horizontal velocity^2 + vertical velocity^2)
Overall velocity = sqrt(2.9^2 + 63.5^2)
Overall velocity ≈ sqrt(8.41 + 4032.25)
Overall velocity ≈ sqrt(4040.66)
Overall velocity ≈ 63.6 m/s

Therefore, the overall velocity of the water when it hits the bottom is approximately 63.6 m/s.

figure the time to fall 206m: 4.9t^2 = 206

The vertical speed v = 9.8t
so the final speed is √(v^2+2.9^2)