Four Hundred Metres in Nine Seconds, and the Answer That Is Exactly Half

With no air, v squared equals 2gD gives 89.4 metres per second and t equals root of 2D/g gives 8.94 seconds, both stable under g = 10 or g = 9.81. Dividing the height by the impact speed returns 4.47 seconds, wrong by exactly a factor of two at every drop height, because a body released from rest averages half its final speed. Real air reverses the picture: a coin-sized disc reaches terminal velocity near 11.9 metres per second and takes about 34.6 seconds, so nine seconds is a floor and 200 miles an hour a ceiling.

A coin is released from rest and falls 400 metres. Ignore the air. How fast is it going when it lands, and how long was it falling?

About 90 metres per second, which is about 200 miles an hour, and about nine seconds. Both numbers come out of the same line of algebra, and one of them has a wrong answer so clean that it survives every check except the right one.

Two formulas, one substitution

Constant acceleration from rest gives $v = gt$ and $D = \tfrac12 g t^2$ . Eliminating $t$ between them removes time from the picture entirely:

v^2 = 2gD = 2 \cdot 10 \cdot 400 = 8000, \qquad v = \sqrt{8000} = 89.44\ \text{m/s}

(1)

Taking $g = 10$ instead of 9.81 costs under one percent: the textbook value gives 88.59 metres per second. Converting, $89.44 \times 3600/1609 = 200.1$ miles an hour, so the round figure is right to better than a percent as well.

The time then follows from $t = v/g$ , or directly:

t = \sqrt{\frac{2D}{g}} = \sqrt{80} = 8.944\ \text{s}

(2)

With $g = 9.81$ equation (2) gives 9.03 seconds, so "nine seconds" is accurate to a third of a percent rather than being a lazy rounding of 8.94. That is a useful thing to know about an estimate: both plausible values of the constant land on the same answer.

The wrong answer is exactly half

Here is the mistake worth studying. The coin falls 400 metres and it lands at 89.44 metres per second, so divide:

\frac{D}{v} = \frac{400}{89.44} = 4.47\ \text{s}

(3)

Dimensionally correct. Uses only numbers already established. Off by exactly a factor of two, at every drop height, forever. Equation (3) divides the distance by the speed the coin reached at the very end of the fall, and the coin was slower than that for the entire journey.

The average speed of something released from rest under constant acceleration is exactly half its final speed, because $gt$ is linear and the mean of a line starting at zero is half its endpoint. So the correct division is $D/(v/2) = 2D/v$ , which is equation (2) again. The factor is structural: for any height $D$ ,

\frac{D/\sqrt{2gD}}{\sqrt{2D/g}} = \frac12

(4)

Fig. 1 — Distance is the area under the speed curve. The trap in equation (3) measures the rectangle instead of the triangle, so it accounts for twice the fall in the same time.

Mean speed over a fall from rest

Since $v(t) = gt$ , the average of $v$ over $[0, T]$ is $gT/2 = v(T)/2$ . Here that is 44.72 metres per second. Any calculation that divides a distance by the final speed has silently assumed the fall happened at its fastest throughout.

Three ways to the same nine seconds

Eliminate $v$ : substitute $v = \sqrt{2gD}$ into $t = v/g$ and get $\sqrt{2D/g}$ .

Use the distance law directly: solve $400 = 5t^2$ for $t = \sqrt{80}$ , which needs no speed at all.

Use the average speed: $t = D/(v/2) = 800/89.44 = 8.94$ . This is the version worth saying out loud, because the same sentence produces the answer and diagnoses equation (3). The other two are correct and silent about the mistake.

Why the answer had to look like this

Before any arithmetic, the shape of the answer is forced. The problem contains one length $D$ and one acceleration $g$ , with dimensions $\mathsf{L}$ and $\mathsf{L}\mathsf{T}^{-2}$ . The only combination with the dimension of time is $\sqrt{D/g}$ , and the only one with the dimension of speed is $\sqrt{gD}$ . So

t = c_1\sqrt{\frac{D}{g}}, \qquad v = c_2\sqrt{gD}

(5)

with dimensionless constants, and integrating the motion supplies $c_1 = c_2 = \sqrt2$ . One consequence is immediate and useful: quadrupling the drop height doubles both the time and the impact speed. Another is that the mass never appears, which is the part of this that took humanity two thousand years.

What the air does, and which way

Every number above describes a fall in vacuum, and the direction of the correction matters more than its size. Drag always opposes the motion, so with air the fall takes longer and the arrival is slower. Nine seconds is a floor and 90 metres per second is a ceiling.

For quadratic drag the speed obeys $\dot v = g - kv^2$ and settles at a terminal value

v_{\infty} = \sqrt{\frac{2mg}{\rho\,C_D A}}

(6)

A small coin is about 2.5 grams across 19 millimetres, and a tumbling disc has a drag coefficient near 1. Equation (6) then gives 11.9 metres per second, which is 26.5 miles an hour, and the solution $v(t) = v_\infty \tanh(gt/v_\infty)$ reaches within a percent of that in under four seconds and thirty metres. The remaining 370 metres are covered at a near-constant crawl, so the fall takes about 34.6 seconds.

Fig. 2 — The same 400 metres, integrated with and without air. Air changes the answer by a factor of four in time and a factor of seven and a half in speed, which is why the no-air assumption belongs in the premise rather than in a footnote.

So the vacuum answer is not a prediction about a real coin. Twenty-six miles an hour is a thrown pebble, and the popular story about coins dropped from great heights being lethal is refuted by equation (6) rather than supported by equation (1).

What survives contact with reality

Two smaller assumptions are safe. Gravity is treated as constant over the drop, and it varies by about $2D/R_{\oplus} = 1.3 \times 10^{-4}$ , or 0.013 percent, over 400 metres, which is far below the rounding in $g = 10$ . Rotation of the earth deflects the fall by a few centimetres and changes nothing about the timing.

The assumption that does all the damage is the one about air, and it fails differently for different objects. A dense compact object of the same size as the coin but a hundred times heavier has a terminal speed ten times higher by equation (6), so for a steel ball the vacuum numbers are close to right. The vacuum calculation is not wrong; it answers a question about a particular idealisation, and it is worth being explicit about which one.

Sources and further reading

Equations (1) and (2) — Equations of motion
The constant that was rounded to 10 — Standard gravity
Equation (6) and the tanh solution — Terminal velocity
The argument in equation (5) — Dimensional analysis

Both falls were integrated step by step rather than evaluated from a formula: the vacuum fall at intervals of $10^{-5}$ seconds returned 8.944 seconds and 89.44 metres per second, the drag fall under a fourth-order scheme returned 34.6 seconds and 11.85 metres per second, and the factor of two in equation (4) was reproduced numerically at drop heights of 25, 100, 400 and 1000 metres.