# Waterfall Freefall Time and Velocity

As of April 29, 2009, the world record was set by Tyler Bradt when he ran 186-foot-tall Palouse Falls in Washington.

When dropping over a waterfall, how long does it take a kayaker to fall a given distance and how fast is he/she going when they hit the pool below?

Someone photographing a kayaker running a waterfall may want to know ahead of time how many seconds the freefall event will last. If we ignore the air's frictional resistance force, the equation for calculating how far you fall in a given number of seconds can be simplified from formulas for acceleration to:

t squared = (2d/32) = d/16, or in terms of the distance fallen, d = 16 x (t squared). Where "t" is time in seconds and "d" is the height fallen in feet and 32 is the acceleration of gravity in ft/s/s.

For example:

d = 16', t = 1 sec.

d = 64', t = 2 sec.

d = 144', t = 3 sec.

d = 256', t = 4 sec.

However, in the real world the higher the velocity the greater the wind resistance. So for big drops the actual distance fallen would be less than this, or looked at the other way, the time it takes to fall a given distance would be slightly longer.

How fast is a kayaker going at the bottom of a waterfall (or seal launch)?

If we ignore the frictional resistance force from the air, the formula for calculating a kayaker's velocity at the bottom of a vertical drop is:

V = square root of (2 x 32 x d), where "V" is the Velocity in ft/s and "d" is the height of the drop in feet, and 32 is the acceleration of gravity in ft/s/s.

Or for V in miles/hour:

V = 0.68 x square root of (2 x 32 x d) = 0.68 x (square root of 64d). Where 0.68 is the ratio of (3600 seconds per hour/5280 feet per mile).

For example:

d = 10', V = 17mph

d = 16', V = 22mph (1sec.)

d = 32', V = 30mph

d = 64', V = 44mph (2sec.)

d = 100', V = 55mph

d = 144', V = 65mph (3sec.)

d = 200', V = 77mph

d = 256', V = 87mph

However, in the real world the higher the velocity the greater the wind resistance. So for big drops the actual velocity would be somewhat less than this.

What's the impact force when a kayaker hits the water at the bottom of a waterfall?

That's the $64 million question. The impact force when you hit the water depends on many variables such as the aeration of the water, shape and size of the kayak and kayaker, the angle of impact, the combined mass of the kayak and kayaker, and the velocity of the kayaker at impact (the kayaker's velocity squared to be more precise). So unfortunately the thing that matters most is too complicated to calculate; tests with accelerometers would yield interesting results.

Addendum

A Beginning River Kayaker's Take On This

Revised scale with the reasons for the "d"

d = 1'; V = 5.4 mph (I didn't see the drop)

d = 2'; V = 7.6 mph (this was a dare)

d = 3'; V = 9.4 mph (also a dare but after a lot of thought)

d = 4'; V = 10.9 mph (oh sh** -- rhymes with "it", a serious miscalculation in reading the river)

d = 5'; V = 12.2mph (after many beers and saying "hey bubba, watch this")

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