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Assume no sliding resistance. But include wind resistance. Angle from 0 to 90. A vertical slope is fair. What angle would produce the maximum speed. I know 90 vertical seems the obvious but less than vertical would contribute speed in the horizontal direction.

This could be a bicycle with no rolling resistance.

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closed as off-topic by Kyle Kanos, Jim, Jon Custer, ZeroTheHero, Bill N Jul 28 '17 at 22:51

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    $\begingroup$ Can't get faster than just falling. Obvious answer is obvious $\endgroup$ – Jim Jul 28 '17 at 14:03
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    $\begingroup$ @Jim I think a reasonable person knows what that means. $\endgroup$ – paparazzo Jul 28 '17 at 14:08
  • $\begingroup$ If the skier is rolling, they have a problem... $\endgroup$ – Jon Custer Jul 28 '17 at 15:58
  • $\begingroup$ @JonCuster Rolling resistance is a pretty common term. Do you have a question? $\endgroup$ – paparazzo Jul 28 '17 at 16:01
  • $\begingroup$ @Paparazzi - while 'rolling resistance' is a common term, it does not apply to skis slipping on a surface. The one you are looking for is 'sliding resistance'. Sorry, my tribologist friends would insist... $\endgroup$ – Jon Custer Jul 28 '17 at 16:03
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Without wind resistance, the speed you each at a given moment is a function of the potential energy you lost (which is equal to the kinetic energy you gained).

When you add wind friction, then the terminal velocity will be reached when the force pushing you forward is equal to the wind resistance pushing back. Now wind resistance goes as the square of velocity (approximately), and the force of gravity along the slope (the direction of motion) is proportional to $\sin\theta$ where $\theta$ is the angle of the slope ($\theta=0$ is horizontal).

It follows that the terminal velocity will be greatest in free fall ($\sin\theta=1$).

The point you made "horizontal velocity adds a component" is a red herring as the total kinetic energy can be no greater than the potential energy lost (and the work done against drag, for a given drop in height, will be greater if you have covered a greater distance to get there).

I could add more detailed equations but I don't think that's needed for the argument to hold.

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