Timing a Drop to Certify a 12 m Wall
A cadet team must confirm the height of a training wall before a rope-descent exercise, because the drawing has been mislaid. They drop a stone from the top edge and time the fall with a phone stopwatch. Six readings are recorded: 1.52 s, 1.48 s, 1.61 s, 1.50 s, 1.55 s and 2.10 s. One cadet squares the largest reading and obtains about 21.6 m, concluding the wall is far higher than believed. A second cadet averages all six readings, gets 1.63 s and reports 13.0 m. The safety officer's rule is that the exercise may proceed only if the height is confirmed to within ±0.5 m of 12 m. The stopwatch operator is timing by eye and hand, and the stone is a flat, light chip that tumbles as it falls. Nothing has been done to estimate the operator's reaction time.
1. Decide which reading, if any, should be discarded, and justify the decision with numbers rather than intuition.
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Discard 2.10 s. Justification across three independent checks:
- Internal spread. The five accepted readings span 1.48–1.61 s, a range of 0.13 s. Their mean is 1.532 s. The 2.10 s reading sits 0.57 s above that mean - more than four times the entire spread of the others. That is not scatter; it is a different event.
- Physical plausibility. From a 12 m wall, t = √(2h/g) = √(24/9.81) ≈ 1.56 s. The five readings cluster within 0.08 s of that. To take 2.10 s the wall would have to be ½ × 9.81 × 2.10² ≈ 21.6 m - visibly inconsistent with a wall that cadets descend by rope in a single pitch.
- Likely cause. A late stop press, a fumbled catch, or the stone striking the wall on the way down. All three lengthen the time; none shortens it.
Discarding the largest number because it is inconvenient is bad practice. Discarding it because it is an outlier on a quantified test is correct - and the reason must be stated in the report.
2. Compute the best estimate of the wall's height from the accepted readings, and explain whether the ±0.5 m tolerance can be met by this method at all.
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Best estimate. t = 1.532 s; h = ½ × 9.81 × 1.532² = 4.905 × 2.347 ≈ 11.5 m.
So the wall is about 0.5 m below the drawing value - exactly at the edge of the tolerance, not comfortably inside it.
The method cannot meet the tolerance. Differentiate h = ½gt²: dh/dt = gt ≈ 9.81 × 1.53 ≈ 15 m per second of timing error. Therefore:
- a 0.1 s error ⇒ about 1.5 m of height error;
- a 0.05 s error ⇒ about 0.75 m - already larger than the ±0.5 m allowed.
Hand-timed stopwatch reaction alone is typically 0.2–0.3 s, and it biases in one direction: the operator stops late, so measured time is too long and the computed height is too high. Air drag and the tumbling of a flat, light chip lengthen the fall as well, pushing the estimate up again. The 11.5 m figure is therefore an upper bound, not a centre value; the true wall is probably slightly lower.
3. The safety officer asks for your recommendation. State it and give the alternative measurement that would actually settle the question.
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Recommendation: do not authorise on this evidence. Three reasons:
- The estimate, 11.5 m, differs from the drawing by about 0.5 m - precisely the tolerance boundary, so any further error decides the question by accident.
- Every known bias in the method (reaction lag, drag, tumbling) inflates the result. Correcting for them moves the answer away from 12 m, not towards it, so '11.5 m' should be read as '11.5 m or less'.
- The decision is a safety control, not a physics exercise; the cost of being wrong is a fall from height.
What would settle it: a direct length measure - a tape or pre-marked rope lowered from the top edge, or a laser rangefinder held at the base. Both give an uncertainty of a few centimetres, which is two orders of magnitude better than the stopwatch. If only a timing method is available, time 20 drops with an electronic gate or video frame-stepping, fit a mean with a stated standard error, and calibrate the operator on a known 12 m drop first. Until then the exercise waits.