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, the object has an initial velocity v0 = 0 m/s.

Assuming zero air resistence, the object as infinite terminal velocity 
and accelerates uniformly at 9.81 m/s^2. This yields

   vt = 9.81 t

The integral of this would be the object's position:

   pt = 9.81/2 t^2 = 4.905 t^2

We hear the splash at t=10. Assuming that sound is initiated at time t=K 
and it propogates towards the observer at 340 m/s, we have

   K = 10 - d/340

where d is the distance to the bottom of the well.

So, we know that d must obey

   K = 10 - d/340   and   4.905 K^2 = d

Rearranging, we have

   K - 10 = -d/340
   340 (K - 10) = -d
   340 K - 3,400 = -d
   -340 K + 3,400 = d

So d = -340 K + 3,400 = 4.905 K^2.

At this point, I have insufficient knowledge of algebra to deduce a 
solution, although obviously one exists.

-- 
http://blog.orphi.me.uk/
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