Water skiing is a sport where an individual is pulled behind a boat or a cable ski installation on a body of water, skimming the surface.

Consider an idealized case where the boat is moving at a constant velocity $\overrightarrow v_0=\text{const}$ (relative to the water), independently of the skier. The boat and skier are connected by a massless, unstretchable rope. Surface of the water is assumed to be smooth.

Then, what is the maximum possible (instantaneous)speed of the skier $ v_\text{max}$ (relative to the water)?

My first guess, based on intuition, is that $v_\text{max}=2v_0$.

But I'm not at all sure.


Skier's velocity may vary. Consider vector projections of velocity vectors $\overrightarrow v_1 $ and $\overrightarrow v_2 $ of the boat and skier in the direction of the rope. Because the rope is unstretchable, these projections must be equal. That means, must hold the equality: $$v_1\cos{\alpha}=v_2\cos{\beta}$$

$\alpha$ is an angle between $\overrightarrow v_1 $ and the direction of the rope
$\beta$ is an angle between $\overrightarrow v_2 $ and the direction of the rope

So for example, if $\alpha<\beta$ then $v_2>v_1$. I.e. the skier is moving faster than the boat relative to the water surface.

  • 1
    $\begingroup$ I guess I'm missing something! How's the $v_{max}$ double the velocity by which it's being pulled?? $\endgroup$ – Vineet Menon Apr 9 '12 at 7:09
  • $\begingroup$ @VineetMenon: Even I feel that I'm missing something. There may be something to do with the water having varying speeds (due to dragging/viscosity), which changes the relative velocities. Dunno. IMO, it's $v=v_0$, no "max" involved--regardless of the orientation. $\endgroup$ – Manishearth Apr 9 '12 at 7:35
  • $\begingroup$ but OP said, water surface to be perfectly smooth, which eliminates the friction, drag... $\endgroup$ – Vineet Menon Apr 9 '12 at 7:48

Viewed from a comoving frame, the boat is stationary. The skier is then simply constrained to move on a circle. In this frame, the skier's velocity is $\omega r$, but since nothing in the problem bounds $\omega$ the skier can go arbitrarily fast.

This has little to do with the physics of water skiing that I can tell.

  • $\begingroup$ It is natural to assume that the velocity vector of the skier and the direction of the skis are parallel. Otherwise, your opinion is correct of course, i agree. $\endgroup$ – Martin Gales Apr 10 '12 at 6:12

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