A stone is thrown down into a well with initial speed v_0 =…

A stone is thrown down into a well with initial speed $v_0 = 20\ \mathrm{m/s}$. The sound of it hitting the water is heard $t = 2\ \mathrm{s}$ after it was thrown. The speed of sound is $c = 340\ \mathrm{m/s}$. How deep is the well?

Explanation of Method

Depth of a well from the sound of a thrown stone.

Step 1: Split the 2 s

Let the stone fall for $t_1$; then the sound needs $t - t_1$ to come back up the same depth $H$: $H = v_0 t_1 + \tfrac12 g t_1^2$.

\[ H = c\,(t - t_1). \]

Step 2: A quadratic for $t_1$

\[ \tfrac12 g t_1^2 + (v_0 + c)\,t_1 - ct = 0 \;\Longrightarrow\; 4.905\,t_1^2 + 360\,t_1 - 680 = 0. \]

\[ t_1 = \frac{-360 + \sqrt{360^2 + 4 \cdot 4.905 \cdot 680}}{2 \cdot 4.905} \approx 1.843\ \mathrm{s}. \]

Step 3: Depth

\[ H = c\,(t - t_1) = 340 \cdot 0.157. \]

\[H \approx 53.5\ \mathrm{m}\]

Check

Stone: $ 20 \cdot 1.843 + 4.905 \cdot 1.843^2 = 36.9 + 16.7 = 53.5\ \mathrm{m}\ \checkmark $; sound: $ 53.5/340 = 0.157\ \mathrm{s} $; total $ 2.00\ \mathrm{s}\ \checkmark $

$\text{time } (\mathrm{s})$ $H\ (\mathrm{m})$ $0$ $20$ $53.5$ $100$ $1$ $2$ $\text{fall + sound}$ $\text{fall only}$ $H \approx 53.5\ \mathrm{m}$

total time until the sound is heard
stone's fall time

Total time against depth for a stone thrown down at 20 m/s. It equals 2 s at a depth of about 53.5 m (the stone falls 1.84 s, the sound returns in 0.16 s).

Result

\[\boxed{\,H \approx 53.5\ \mathrm{m}\,}\]

Remark

About the original solution. Its quadratic was written incorrectly, and its answer is not consistent: with $t_1 = 1.8\ \mathrm{s}$ the stone would fall $20 \cdot 1.8 + 4.905 \cdot 1.8^2 \approx 51.9\ \mathrm{m}$, not the 68 m it reported. The correct depth is about 53.5 m. Compare with Gravitational Kinematics 003, the same idea with a dropped stone.