Warp wrote:
>   How about this classical physics problem:
> 
>   You drop a rock into a water well. Then you hear the splash. Exactly
> 10 seconds pass between you letting go of the rock and when you hear the
> splash. How deep is the well?
> 
>   We can assume that air friction is negligible, that g = 9.81 m/s^2,
> and that the speed of sound is 340 m/s.
> 
>   Don't just give the answer, also write your whole reasoning.

Hmm, takes me back to school.

10 = t(sound) + t(rockfall)

t(sound) = x/340 where x is the well depth

x = 1/2 * g * t(rockfall)^2
so
t(rockfall) = sqrt(2x/g)

substituting back in,

10 = x/340 + sqrt(2x/g)

sub y = sqrt(x), rearrange

(1/340)*y^2 + sqrt(2/g)*y - 10 = 0

solve quadratic for positive y, square y to get x

I get x = 393.25m, sounds about right but my arithmetic might be wonky.

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