Problem 112
Medium Gravitational Kinematics

A stone is dropped into a chasm and falls freely. The sound…

A stone is dropped into a chasm and falls freely. The sound of it hitting the water is heard $t = 4.9\ \mathrm{s}$ after it was released. How deep is the chasm? The speed of sound is $c = 340\ \mathrm{m/s}$.

Explanation of Method

Depth of a chasm from the sound of a falling stone.

Step 1: Split the 4.9 s

The time consists of the stone's fall and the sound's trip back up the same depth $H$: \[t = t_f + t_s = \sqrt{\frac{2H}{g}} + \frac{H}{c} = 4.9\ \mathrm{s}.\]

Step 2: A quadratic equation

Let $x = t_f$ (the fall time). Then $H = \tfrac12 g x^2$ and the sound time is $\tfrac{g x^2}{2c}$: $x + \frac{g}{2c}\,x^2 = 4.9 \;\Longrightarrow\; 0.014426\,x^2 + x - 4.9 = 0$.

\[ x = \frac{-1 + \sqrt{1 + 4 \cdot 0.014426 \cdot 4.9}}{2 \cdot 0.014426} \approx 4.595\ \mathrm{s}. \]

Step 3: Depth

\[ H = \tfrac12 g x^2 = 4.905 \cdot 4.595^2. \]

\[H \approx 104\ \mathrm{m}\]

Check

Fall: $ \sqrt{2 \cdot 103.6 / 9.81} \approx 4.595\ \mathrm{s} $; sound: $ 103.6/340 \approx 0.305\ \mathrm{s} $; total $ 4.90\ \mathrm{s}\ \checkmark $

$\text{time } (\mathrm{s})$ $H\ (\mathrm{m})$ $0$ $50$ $103.6$ $150$ $200$ $250$ $2$ $4$ $4.9$ $6$ $\text{fall + sound}$ $\text{fall only}$ $H \approx 103.6\ \mathrm{m}$
  • total time until the sound is heard
  • falling time alone
Total time until the sound is heard, as a function of the depth. It equals 4.9 s at a depth of about 104 m. The gap between the two curves is the sound's travel time, which is small (0.3 s here).

Result

\[\boxed{\,H \approx 104\ \mathrm{m}\,}\]

Remark

About the original solution. Its equation was set up correctly, but the numerical answer, 316 m, is wrong: a stone would need almost 9 s just to fall 316 m. Solving the quadratic gives about 104 m. Most of the 4.9 s is falling; the sound takes only 0.3 s.