A stone in the well

It was a deep well. I dropped a pebble and counted until I heard the splash: two point one seconds.

A deep well. I dropped a pebble and counted: the splash came after $T=2.1$ s.

Splash: $T=2.1\text{s}$.

A falling stone speeds up. In the first second it falls about five metres (four point nine); in two seconds about twenty metres (nineteen point six), four times as far, not twice. In three seconds about forty-four metres (forty-four point one), nine times as far. The distance grows like the square of the time.

Falling: $s=\frac{1}{2}g{t}^2=4.9t^2$ m. $t=1$: $4.9$ m. $t=2$: $19.6$ m, $4\times$ as far, not $2\times$. $t=3$: $44.1$ m, $9\times$. So $s\propto t^2$.

$s=\frac{1}{2}g{t}^2=4.9t^2$; $s(1)=4.9$, $s(2)=19.6$, $s(3)=44.1$; $s\propto t^2$.

If all two point one seconds were falling, the well would be about twenty-one point six metres deep. But that time also includes the sound coming back up, and sound is slow enough to matter. Counting the sound's trip at three hundred and forty-three metres per second, the fall itself lasted about two point zero four seconds.

Naively $4.9\times 2.1^2\approx 21.6$ m. But $T$ includes the sound coming up at $343$ m/s: $t+4.9t^2/343=2.1\Rightarrow t\approx 2.04$ s.

$4.9\times 2.1^2\approx 21.6$; $t+4.9t^2/343=2.1\Rightarrow t\approx 2.04$.

So the well is about twenty point four metres deep, not twenty-one point six: the sound's trip had misled me by more than a metre. And the surprise: a well four times as deep needs only twice the time to fall, about four seconds.

Depth: $4.9\times 2.04^2\approx 20.4$ m, not $21.6$ m; the sound misled me by $\approx 1.2$ m. A well $4\times$ as deep: the fall takes $2\times$ as long, about $4$ s.

$s=4.9\times 2.04^2\approx 20.4$; $21.6-20.4=1.2$; $4s\Rightarrow 2t$.