Ravine Measurement by Echo and Falling Stone
While preparing a crossing task, a section commander must estimate the width and depth of a ravine without survey equipment. Standing at the near edge, a soldier strikes two stones together and hears the echo from the far wall 1.8 s later. The air temperature is 25 °C. At the same spot a second soldier drops a stone over the edge and times 0.8 s until it hits the bottom. Take g = 9.8 m/s² and ignore air resistance for the falling stone. Use v = 331 + 0.6T m/s for the speed of sound in air, where T is the air temperature in °C. The section holds a single reel of 350 m of cable, and standing orders require 10 per cent slack beyond the measured span for sag and tying off. The far wall is near-vertical, but the route the cable will actually follow is not a straight line across the gap.
1. Calculate the width of the ravine from the echo. Show the formula and each step, and state the most common error.
Read the worked answer
Step 1 - speed of sound at 25 °C. v = 331 + 0.6 × 25 = 331 + 15 = 346 m/s. Using a book value of 340 m/s is acceptable only if the paper does not supply the temperature formula; here it does, so use it.
Step 2 - total distance travelled by the sound. d = v × t = 346 × 1.8 = 622.8 m.
Step 3 - one-way distance. The sound went to the far wall and came back, so the gap is half the path: 622.8 ÷ 2 = 311.4 m, i.e. about 311 m.
Most common error: reporting 622.8 m as the width by forgetting to halve the distance. A second common error is using 340 m/s and ignoring the stated temperature, which gives 306 m and understates the gap by about 5 m.
2. Calculate the depth of the ravine, then decide whether the single 350 m reel is enough. Give a recommendation.
Read the worked answer
Depth: h = ½gt² = 0.5 × 9.8 × (0.8)² = 0.5 × 9.8 × 0.64 = 3.14 m, about 3.1 m. (Ignoring air resistance slightly overstates the depth; reaction time in starting and stopping the watch also adds error.)
Cable check: required length = measured span × 1.10 = 311.4 × 1.10 = 342.5 m. This is within 350 m, so on paper the reel suffices - but only 7.5 m (about 2 per cent) is left over.
Recommendation: do not commit the reel on this margin. The echo gives a straight-line distance, while the cable must follow the actual route and be tied off at both ends, so real length will exceed 342.5 m. Timing error alone matters: a 0.1 s error in the echo time shifts the answer by roughly 17 m. Repeat the echo measurement two or three times and average, or hold the reel in reserve and request a second one before starting. The correct exam answer states both the arithmetic and the judgement, not just "342.5 m, so yes".
3. If the same 1.8 s echo had been measured at 5 °C, what width would be calculated, and has the ravine actually changed? What lesson follows?
Read the worked answer
Recalculate: v = 331 + 0.6 × 5 = 331 + 3 = 334 m/s. Path = 334 × 1.8 = 601.2 m, so width = 300.6 m, about 301 m.
Has the ravine changed? No. The physical gap is fixed; only the speed of sound changed with air temperature, so the calculated width fell by 311.4 − 300.6 = 10.8 m, roughly 3.5 per cent.
Lesson: an acoustic range estimate is only as good as the assumed speed of sound, so the air temperature must be recorded with the measurement and the same formula used throughout. It also shows that the method carries a few per cent of uncertainty from temperature alone - before counting timing reaction error - which is exactly why the cable decision in the previous task must not rely on a 2 per cent spare.