# COnstant acceleration - stone in a well

1. Sep 9, 2011

### mstud

1. The problem statement, all variables and given/known data

You drop a stone down into a well to find the depth of the well. You hear the sound of the splash 3.5 s after you let the stone go.

2. Relevant equations

$$s(displacement)=v_0t \cdot \frac 12 at^2$$

3. The attempt at a solution

I have
$t_{total}=3.5 s$

Stone: $v_0$=0, and a=9.81 m/s^2.

Sound: v=340 m/s, a=0.

If the sound had not used a small part of the total time, the answer would have been:

$$s(displacement)=v_0t \cdot \frac 12 at^2=0*3.5 s + \frac 12 * 9.81 m/s^2*(3.5s)^2=60 m$$

But the answer key of my book says 55 m. I see why, but I don't know how I can get that answer... Please help, anybody out there!

2. Sep 9, 2011

### wukunlin

basically in this question there is 3 unknowns and 3 constraints

3 unknowns
- time for rock to reach bottom of well
- time for sound to travel out of the well
- depth of well

3 constrants
- time takes to reach the bottom of the well and depth of the well is related to the equation you used
- time taken for sound to travel out of the well follows a simple s = vt relationship
- total time is 3.5s

can you see how to solve this now?

3. Sep 9, 2011

### mstud

Hmmm, not quite sure. Depth of well, s, is the same both when the stone and the sound travels it.

Sound: $s(displacement)= vt = 340m/s *t_{sound}$

Stone:

$$s(displacement)=v_0t\cdot \frac 12 at^2=0∗(3.5s-t_{sound}) +\frac 12 ∗ 9.81m/s^2 ∗ (3.5s-t_{sound})^2=0+12∗9.81m/s^2∗(3.5s-t_{sound})^2=12∗9.81m/s^2∗(3.5s-t_{sound})^2$$.

I can relate these two equations to each other: $340m/s *t_{sound}=12∗9.81m/s^2∗(3.5s-t_{sound})^2$ and solve this equation for $t_{sound}$. Am I on the right track now?

4. Sep 9, 2011

### wukunlin

looks like you are :)

5. Sep 9, 2011

### mstud

That was right :)

Thanks a lot !!!

6. Sep 9, 2011