Jet Ski Mishap

[BMAD]

Danny, who wasn't quite awake when he took his jet ski out on a lake, forgot to untie his jet ski from its anchor.  As soon as he got to the end of the rope, the ski stopped sending Danny flying.  The jet ski, which thankfully was in the middle of the lake, continued to go in circles from this point at a speed of 20 miles per hour. After six hours, the jet ski made 12 laps.  How deep is the anchor?

[bonanova]

If we knew that, we could tell you the length of the rope.

Or, the answer could be it's on the bottom.

[BMAD]

Could we write the depth as a function of the rope and given information?

[The_unbeliever]

I might not be a jetski expert, but are Jetskis NOT anchored, but tied to the pier instead?

If you define "anchor" by where the other end of the rope is tied, it's at pier's height. Definitely not underwater.

[The_unbeliever]

Could we write the depth as a function of the rope and given information?

Was squareroot(R-5/Pi), in miles, R as the rope's length, the answer you expected otherwise?
Sry, can't edit previous post :/

[BMAD]

Hmmm i have a different answer. Can you explain yours so i can see my mistake. 

 

I might not be a jetski expert, but are Jetskis NOT anchored, but tied to the pier instead?

If you define "anchor" by where the other end of the rope is tied, it's at pier's height. Definitely not underwater.

the owner of the jet ski can anchor it how they like

[The_unbeliever]

Hmmm i have a different answer. Can you explain yours so i can see my mistake. 

I might not be a jetski expert, but are Jetskis NOT anchored, but tied to the pier instead?

If you define "anchor" by where the other end of the rope is tied, it's at pier's height. Definitely not underwater.

the owner of the jet ski can anchor it how they like

Well, I usually set an anchor so my boat doesn't go adrift... Huh, I'm nitpicking, ain't I? >_>
In fact, I was wondering whether you meant a logical problem in the guise of a simple geometrical one (with the rope length missing on purpose).

And yes, I think I should keep the maths to pen & paper - I'm obviously missing ^2s here :/ My bad.

I've considered an ABC triangle with A as the center of the lap, C the anchor point :), B any point in the Jetski trajectory.
(AB) perpendicular to (AC), with AC = h = anchor depth, BC = r = rope length, AB = lap radius = 10 miles / 2Pi
1 line later,    h = sqroot(r^2 - (5/Pi)^2)