---
title: "Fitting a free fall to measure gravity | Sandbox Physics"
description: "Fit release height, initial velocity and gravity to a recorded fall, read the residuals, and tell a precise result from an accurate one."
canonical_url: "https://sandboxphysics.com/learn/mechanics/testing-constant-acceleration/"
language: "en"
page_type: "story"
alternate_languages:
  zh-CN: "https://sandboxphysics.com/zh/learn/mechanics/testing-constant-acceleration/"
last_modified: "2026-10-28"
---

[Volume 1 · Motion, Force and Energy](https://sandboxphysics.com/learn/mechanics/)

Volume 1 · Chapter 3

# How do we test a constant-acceleration model?

Fit release height, initial velocity and gravity to a recorded fall, read the residuals, and tell a precise result from an accurate one.

Before you start: Chapters 1 and 2. Fitting and residuals are introduced here; the matrix derivation is an optional section.

After this chapter you can: Derive the constant-acceleration model, fit it to a fall record, and use residuals and repeated drops to judge whether the model and the result can be trusted.

A camera record of one fall is enough to measure the acceleration of gravity. Assume the acceleration is constant and derive height as a quadratic function of time. Then find the curve that comes closest to the record; its quadratic term gives the acceleration of gravity. Whether to trust the result depends on two things: whether the deviations left after the fit show a pattern, and whether repeated drops give stable results close to an independently known value.

Picture a ball released from the top of a 30-metre drop tower. A camera at the side takes 20 frames per second, and each frame gives the ball's height with an error of about a centimetre; the camera has been calibrated as in Chapter 2. A photogate is mounted halfway down the tower.

The data in this chapter come from the [Free Fall and Gravity Measurement](https://sandboxphysics.com/labs/free-fall/) Lab. As in the last chapter, the Lab knows the true motion and can tell us how far an estimate is off; a real experiment has only the checks described here. We treat the acceleration of gravity as a given constant. Why it is the same for every object waits for forces in Chapter 6 and gravitation in Chapter 14.

## 3.1 Coordinates and what we know

Take upward as positive and the ground as the origin, and write the ball's height as $y$, in metres. The release height is $h$ and the initial velocity $v_0$, positive upward. The ball is treated as a point, and air resistance is ignored for now.

Under these conditions we assume the ball's acceleration stays constant throughout the fall, equal to $-g$. The minus sign means the acceleration points down. Near the ground $g$ is about $9.81\,\mathrm{m\,s^{-2}}$, which is also the Lab's default. The task of this chapter is to test that assumption against a record and to measure $g$.

## 3.2 From constant acceleration to velocity and position

Chapter 1 showed that the change in velocity is the area under the acceleration graph and the displacement is the area under the velocity graph. With a constant acceleration $a$, the acceleration graph is a horizontal line, and the area from zero to $t$ is $at$. So

$$
v(t)=v_0+at.
$$

The velocity graph is therefore a sloping line. The region under it from zero to $t$ is a trapezium whose parallel sides are $v_0$ and $v(t)$ and whose width is $t$:

$$
y(t)-h=\frac{v_0+v(t)}{2}\,t=v_0t+\tfrac12at^2.
$$

For free fall, take $a=-g$:

$$
y(t)=h+v_0t-\tfrac12gt^2,
\qquad
v(t)=v_0-gt.
$$

Eliminating $t$ gives one more useful relation. Solve the velocity equation for $t=(v-v_0)/a$ and substitute into the trapezium area to get $y-h=(v^2-v_0^2)/(2a)$, that is,

$$
v^2=v_0^2+2a(y-h).
$$

Compare these relations with [OpenStax University Physics Volume 1, Section 3.4](https://openstax.org/books/university-physics-volume-1/pages/3-4-motion-with-constant-acceleration) and [Section 3.5](https://openstax.org/books/university-physics-volume-1/pages/3-5-free-fall). As a check, every term has units of metres or metres per second, and setting $a=0$ brings back the uniform motion of Chapter 1.

## 3.3 Initial conditions pick the parabola

For a given $g$, every height–time curve is a downward-opening parabola of the same shape. $h$ sets the height the curve starts from and $v_0$ its slope at the start. Together they are the initial conditions. Knowing $g$ alone is not enough to predict a particular motion.

For example, throw a ball upward at $14\,\mathrm{m\,s^{-1}}$ from a height of 15 metres. It reaches the top when its velocity is zero:

$$
t_{\mathrm{top}}=\frac{v_0}{g}\approx1.43\,\mathrm s,
\qquad
y_{\max}=h+\frac{v_0^2}{2g}\approx24.99\,\mathrm m.
$$

It lands when $y=0$. Solving $15+14t-4.905t^2=0$ and taking the positive root gives $t\approx3.684\,\mathrm s$. From the velocity–height relation, the landing velocity is $-\sqrt{v_0^2+2gh}\approx-22.14\,\mathrm{m\,s^{-1}}$, the minus sign meaning downward.

The Lab's “Upward launch” preset uses exactly these initial conditions and gives a landing time of $3.68427\,\mathrm s$. The Lab solves the motion numerically in steps of $0.01\,\mathrm s$, and its result agrees with the formula to all these digits.

## 3.4 Worked example: predicting a drop

A ball is released from rest at the top of a 30-metre tower, with $g=9.81\,\mathrm{m\,s^{-2}}$. Find the landing time, the landing velocity, and the ball's velocity as it passes a photogate 15 metres up. The photogate measures speed by timing a 10-centimetre flag carried by the ball; find how long the flag blocks the beam.

Released from rest, $v_0=0$. Landing means $y=0$, so

$$
t_{\mathrm{impact}}=\sqrt{\frac{2h}{g}}=\sqrt{\frac{60\,\mathrm m}{9.81\,\mathrm{m\,s^{-2}}}}\approx2.473\,\mathrm s.
$$

The landing velocity is $-gt\approx-24.26\,\mathrm{m\,s^{-1}}$. An independent check with $v^2=2g(h-y)$ gives $\sqrt{2\times9.81\times30}\,\mathrm{m\,s^{-1}}\approx24.26\,\mathrm{m\,s^{-1}}$, in agreement.

At the photogate the ball has fallen 15 metres, so its speed is $\sqrt{2\times9.81\times15}\,\mathrm{m\,s^{-1}}\approx17.16\,\mathrm{m\,s^{-1}}$. The flag takes about

$$
\frac{0.1\,\mathrm m}{17.16\,\mathrm{m\,s^{-1}}}\approx5.83\,\mathrm{ms}
$$

to pass. Here the velocity is taken as constant during the passage. In that time it changes by only $g\times5.83\,\mathrm{ms}\approx0.057\,\mathrm{m\,s^{-1}}$, about 0.3% of the speed, so the approximation is reasonable.

The camera takes a frame every $0.05\,\mathrm s$, so 50 frames are recorded from 0 to $2.45\,\mathrm s$. The Lab's “Earth tower” preset gives a landing time of $2.4731\,\mathrm s$, a photogate speed of $17.1552\,\mathrm{m\,s^{-1}}$ and a flag transit time of $5.8292\,\mathrm{ms}$, as predicted.

## 3.5 From a record back to parameters: fitting

Now turn the problem round. We have 50 noisy heights $y_n$ with their times $t_n$, and we want $h$, $v_0$ and $g$.

For any trial set of values, each frame has a residual, the recorded value minus the model value:

$$
r_n=y_n-\Bigl(h+v_0t_n-\tfrac12gt_n^2\Bigr).
$$

A least-squares fit picks the parameters that make $S=\sum r_n^2$ as small as possible. Squaring has three advantages: positive and negative residuals cannot cancel; points further off count for more; and because the parameters appear only to the first power in the model, the minimum is fixed uniquely by three linear equations. The method is described in the [NIST Engineering Statistics Handbook, Section 4.1.4.1](https://www.itl.nist.gov/div898/handbook/pmd/section1/pmd141.htm).

The simplest case shows why it is sensible. If the model is just a constant $c$, then

$$
S(c)=\sum_{n=1}^N(y_n-c)^2=Nc^2-2c\sum y_n+\sum y_n^2.
$$

This is a quadratic in $c$ that opens upward. Its slope is zero at the bottom, where $c=\frac1N\sum y_n$. Fitting a constant by least squares gives exactly the mean. With three parameters the idea is the same, except that three slope-zero conditions must hold at once; the derivation is in Section 3.9.

For the Earth tower's default record, the fit gives

$$
\hat g=(9.8182\pm0.0055)\,\mathrm{m\,s^{-2}},
\qquad
\hat h=29.9975\,\mathrm m,
\qquad
\hat v_0=0.0106\,\mathrm{m\,s^{-1}}.
$$

The number after the $\pm$ is the standard error. It is computed from the size of the residuals and the spread of the recording times, and says roughly how far results would scatter if the same experiment were repeated. Section 3.7 checks it with repeated drops.

## 3.6 Residuals: what is the model missing?

A fit always returns some parameters, even when the model is wrong. To judge whether the model is adequate, look at the residuals left after the fit.

If the model is complete and the noise random, the residuals scatter above and below zero with no trend, and their RMS is close to the size of the noise. The default record's residual RMS is $8.7\,\mathrm{mm}$, comparable with the $1\,\mathrm{cm}$ noise that was set. The three fitted parameters have absorbed some of the noise, so residuals a little smaller than the noise are normal.

The “Drag biases the fit” preset adds linear air damping at a rate of $0.5\,\mathrm{s^{-1}}$, with everything else as in the Earth tower. The drag grows with speed, so the acceleration is no longer constant. Fitting the same quadratic model gives

$$
\hat g=(4.73\pm0.08)\,\mathrm{m\,s^{-2}},
\qquad
\hat h=30.61\,\mathrm m,
\qquad
\hat v_0=-2.68\,\mathrm{m\,s^{-1}}.
$$

$\hat g$ is less than half the true value, and although the ball was released from rest the fit reports a downward initial velocity. All of these parameters are patching over the drag the model leaves out.

**Figure 3.1** Fit residuals for two 30-metre drops, both fitted with the same constant-acceleration model. Top: no air resistance; the residuals scatter randomly and stay within $22\,\mathrm{mm}$. Bottom: linear air damping added; the residuals go negative, positive, negative and positive again, reaching half a metre. Both vertical axes are in millimetres, on different scales. The data come from the Free Fall Lab's model code with the same noise realization.

The shape of the residuals gives the problem away. In the bottom panel the residual RMS is $232\,\mathrm{mm}$, more than twenty times the noise, and the residuals form a wave: about $-618\,\mathrm{mm}$ at the start, positive at frames 10 and 20, negative from frame 40 to 50, and about $+499\,\mathrm{mm}$ at the last frame. Random noise does not line up like that. A patterned residual means the model has left out some physical process.

The standard error here is only $0.08\,\mathrm{m\,s^{-2}}$, which looks very precise, yet the result is wrong by half. The standard error describes only the scatter of the data and does not check whether the model is right. Checking the model means looking at the residuals; see the [NIST Handbook, Section 4.4.4](https://www.itl.nist.gov/div898/handbook/pmd/section4/pmd44.htm), for how to read them.

## 3.7 Precision and accuracy

Metrology separates two words that are often used interchangeably. Precision is how closely repeated measurements agree with one another; accuracy is how closely a measured value agrees with the true value. See the International Vocabulary of Metrology, [VIM 2.15](https://jcgm.bipm.org/vim/en/2.15.html) and [VIM 2.13](https://jcgm.bipm.org/vim/en/2.13.html). A result can be very precise and still inaccurate.

In the Lab, “Capture a new drop” keeps the physical parameters and swaps in a new set of noise. We repeated the Earth tower 500 times, with noise realizations numbered 431 to 930:

**Table 3.1** Fit results for 500 repeated Earth tower drops.

| Quantity | Value ($\mathrm{m\,s^{-2}}$) |
| --- | --- |
| Mean of the 500 values of $\hat g$ | 9.80988 |
| Standard deviation of the 500 values of $\hat g$ | 0.00621 |
| Mean of the reported standard errors | 0.00606 |

A single drop reports a standard error of about 0.0061, and the actual standard deviation of 500 results is about 0.0062. They agree. When the model is right and the noise independent, the standard error of a single fit reliably predicts the scatter of repeated experiments, that is, the precision.

The mean of the 500 results differs from the true 9.81 by 0.00012. The uncertainty of the mean itself is about $0.00621/\sqrt{500}\approx0.00028$, larger than the difference, so no systematic bias shows. Under these conditions the result is both precise and accurate.

With drag the picture reverses: a standard error of 0.08 but a bias of $-5.08\,\mathrm{m\,s^{-2}}$, more than sixty standard errors. The result is precise but inaccurate. The “Imperfect camera” preset is a third case: $\hat g=9.29\pm0.43$, off by about 1.2 standard errors. This single result is imprecise, and its offset lies inside that imprecision.

## 3.8 Photogates and clock resolution

A photogate takes no pictures. It records how long a flag blocks its light beam, $\Delta t_{\mathrm{gate}}$, and the speed estimate is the flag length divided by that time. It measures the velocity around a single moment, which makes it an independent check on the camera fit: from the fitted $\hat g$ and $v^2=2g(h-y)$ you can predict the speed at the gate.

A timer can read time only to some smallest unit, its clock resolution. Readings are rounded to a whole number of units, so the rounding error is at most half a unit. The shorter the interval being timed, the larger the share of that error.

The “Clock misses the gate” preset sets the clock resolution to $0.05\,\mathrm s$. The flag blocks the beam for only $5.83\,\mathrm{ms}$, which rounds to zero, so the photogate cannot give a speed. The camera fit is untouched: the camera takes a frame every $0.05\,\mathrm s$ anyway, so the frame times are already whole multiples of the clock unit and rounding leaves them unchanged.

The “Imperfect camera” preset has a clock resolution of $0.02\,\mathrm s$, a release height of 8 metres and the gate at 4 metres. The true speed at the gate is about $8.86\,\mathrm{m\,s^{-1}}$ and the transit time about $11.3\,\mathrm{ms}$, which rounds to $20\,\mathrm{ms}$, so the photogate reports $5.00\,\mathrm{m\,s^{-1}}$, 44% low. The same clock also rounds each camera frame time by up to $0.01\,\mathrm s$, and that enters the fit error too.

The rule of thumb is that the clock resolution must be much smaller than the interval being timed. For a rounding error below 1%, the resolution has to be below 2% of the interval.

## 3.9 Optional: least squares with matrices, and the standard error

This section needs matrix multiplication, transposes and inverses, plus the idea of variance from Section 2.4. Skipping it does not affect later chapters.

Write the $N$ recorded heights as a column vector $\mathbf y$ and the parameters as $\boldsymbol\beta=(h,v_0,g)^{\mathsf T}$. The design matrix $X$ has $N$ rows; row $n$ is $(1,\ t_n,\ -t_n^2/2)$. The model and the sum of squared residuals are

$$
\mathbf y=X\boldsymbol\beta+\mathbf e,
\qquad
S(\boldsymbol\beta)=(\mathbf y-X\boldsymbol\beta)^{\mathsf T}(\mathbf y-X\boldsymbol\beta).
$$

Changing only the $j$th parameter while holding the others fixed, the slope of $S$ is called a partial derivative. At the minimum all three partial derivatives vanish. The one for parameter $j$ is $-2\sum_n X_{nj}\bigl(y_n-(X\boldsymbol\beta)_n\bigr)=0$, and together the three give

$$
X^{\mathsf T}X\,\hat{\boldsymbol\beta}=X^{\mathsf T}\mathbf y,
\qquad
\hat{\boldsymbol\beta}=\bigl(X^{\mathsf T}X\bigr)^{-1}X^{\mathsf T}\mathbf y.
$$

These are the normal equations. They say that the residuals of the best fit are "orthogonal" to every column of $X$: the sum of their products is zero.

The standard error comes from how the noise passes into the parameters. Substituting $\mathbf y=X\boldsymbol\beta+\mathbf e$ gives $\hat{\boldsymbol\beta}-\boldsymbol\beta=(X^{\mathsf T}X)^{-1}X^{\mathsf T}\mathbf e$, so each parameter error is a weighted sum of the frame noises. If the frame noises are unrelated, average zero and share the same standard deviation $\sigma$, the add-the-squares rule of Section 2.4 gives

$$
\operatorname{Var}\bigl(\hat{\boldsymbol\beta}\bigr)=\sigma^2\bigl(X^{\mathsf T}X\bigr)^{-1}.
$$

The true $\sigma$ is unknown, so it is estimated from the residuals as $s^2=S_{\min}/(N-3)$. We divide by $N-3$ rather than $N$ because the three parameters have already fitted away some of the noise, leaving the residuals systematically small. The standard error of $\hat g$ is then $s$ times the square root of the diagonal element of $(X^{\mathsf T}X)^{-1}$ that belongs to $g$.

An independent check: the Earth tower record has 50 frames at $t_n=n/(20\,\mathrm{Hz})$ with $\sigma=0.01\,\mathrm m$. Computing the matrix directly gives an expected standard deviation of $\hat g$ of $0.00608\,\mathrm{m\,s^{-2}}$, in line with the 0.00621 from 500 repeats in Table 3.1.

This result has conditions. It needs the model to be right and the frame errors to be unrelated and equal in size. If the model leaves out drag, the formula still gives the scatter of repeated experiments but cannot see the bias at all. If the errors are related, for example when clock rounding pushes neighbouring frames off together, the formula can understate the real uncertainty. That is why Section 3.6 looks at the residuals before trusting the standard error.

## 3.10 Checking in the Free Fall Lab

Open the [Free Fall and Gravity Measurement](https://sandboxphysics.com/labs/free-fall/) Lab. A complete record is ready when the page opens, and the data row below shows the camera record and the “Fit residuals · bias leaves a pattern” plot.

**Predict, then compare.** Choose “Earth tower”. Use Section 3.4 to predict the landing time, the photogate speed and the flag transit time, then compare with the “Complete record duration”, “Photogate speed” and “Flag transit time” readouts.

**Repeat the drop.** Press “Capture a new drop” five times and note “Gravity fit and standard error” each time. Work out the mean and standard deviation of your five values of $\hat g$ and compare them with the single-drop standard error. The standard deviation of five values itself fluctuates a lot, which is exactly why more repeats are needed.

**Make the model fail.** Choose “Drag biases the fit”. Before looking, predict whether $\hat g$ will come out high or low. Look at the shape of the residual plot, compare the “Photogate speed” readout with the drag-free prediction of Section 3.4, and say which assumption has broken down.

**Check the clock.** Choose “Clock misses the gate” and explain why the photogate gives no speed while the camera fit is unchanged. In the advanced controls, reduce “Clock resolution” step by step until the photogate gives a speed, then see how far off it is at $1\,\mathrm{ms}$.

**Transfer to the Moon.** Choose Moon tower, where $g=1.62\,\mathrm{m\,s^{-2}}$ and the position noise is $2\,\mathrm{mm}$. Predict the landing time and the number of camera frames first, then explain why the standard error is so much smaller than on Earth.

Table 3.2 gives each preset's results with the default noise realization, so a reader without the Lab can follow the argument.

**Table 3.2** Representative results for each Free Fall Lab preset (noise realization 431).

| Preset | Camera frames | $\hat g$ ($\mathrm{m\,s^{-2}}$) | Bias against truth | Residual RMS (mm) | Photogate speed ($\mathrm{m\,s^{-1}}$) |
| --- | --- | --- | --- | --- | --- |
| Earth tower | 50 | $9.8182\pm0.0055$ | +0.08% | 8.7 | 17.16 |
| Moon tower | 122 | $1.61978\pm0.00011$ | −0.01% | 1.7 | 6.97 |
| Upward launch | 74 | $9.8121\pm0.0020$ | +0.02% | 8.8 | 18.52 |
| Imperfect camera | 11 | $9.29\pm0.43$ | −5.3% | 84 | 5.00 |
| Drag biases the fit | 63 | $4.73\pm0.08$ | −51.8% | 232 | 12.56 |
| Clock misses the gate | 50 | $9.8182\pm0.0055$ | +0.08% | 8.7 | unresolved |

## 3.11 Exercises

1. In the Free Fall Lab, which quantities are recorded directly and which are estimated? Sort the camera frame times, heights, flag transit time, $\hat g$, $\hat h$ and photogate speed.
2. A ball is thrown straight up at $8\,\mathrm{m\,s^{-1}}$ from a height of 2 metres, with $g=9.81\,\mathrm{m\,s^{-2}}$. Find the time to the top, the maximum height, the landing time and the landing velocity.
3. A photogate is 1.2 metres below the release point, the ball is released from rest, and the flag is 2 centimetres long. Find the transit time. With a clock resolution of $1\,\mathrm{ms}$, what is the largest possible relative error in the speed?
4. Someone fits a straight line (a uniform-motion model) to the height record of a fall. What shape will the residuals have? Which term does that shape say is missing, and with what sign?
5. Two groups of students measure $g$. Group A repeats five times and gets $9.62\pm0.02\,\mathrm{m\,s^{-2}}$; group B gets $9.9\pm0.3\,\mathrm{m\,s^{-2}}$. Which group is more precise? Which agrees with 9.81? What might cause group A's difference?
6. Continuing: if group A assumed a camera scale 2% larger than the true one, how far off would the fitted $g$ be? Compare with group A's result.
7. A single drop has a standard error of $0.0055\,\mathrm{m\,s^{-2}}$. Roughly how many independent drops are needed to bring the uncertainty of the mean down to $0.001\,\mathrm{m\,s^{-2}}$? What does this estimate assume?
8. Transfer. A cart is released from rest on a slope and its position is recorded every second: 0.00, 0.31, 1.18, 2.72 and 4.80 metres. Use the differences of Chapter 2 to test whether it is roughly in constant acceleration, and estimate the acceleration. If the second differences kept getting smaller, what would you suspect?
9. Optional, needs Section 3.9. Use the matrix formula to show that when a constant is fitted to $N$ data points with noise standard deviation $\sigma$, the mean has a standard deviation of $\sigma/\sqrt N$.

## 3.12 Answers and explanations

Exercise 1

Recorded directly: the camera frame times, the height in each frame and the flag transit time. $\hat g$ and $\hat h$ are estimates from fitting a model. The photogate speed is an estimate too, the flag length divided by the transit time, assuming the speed stays constant during the passage. Note that the frame times have been rounded by the clock and the heights depend on the camera calibration, so even the recorded quantities carry errors.

Exercise 2

The top comes at $t=v_0/g\approx0.815\,\mathrm s$, and the maximum height is $2\,\mathrm m+v_0^2/(2g)\approx5.26\,\mathrm m$.

For the landing, solve $2+8t-4.905t^2=0$; the positive root is $t\approx1.85\,\mathrm s$. The landing speed is $\sqrt{v_0^2+2gh}=\sqrt{64+39.24}\,\mathrm{m\,s^{-1}}\approx10.16\,\mathrm{m\,s^{-1}}$, downward. Checking with $v=v_0-gt$ gives $8-9.81\times1.85\approx-10.15$, the same within rounding.

Exercise 3

The speed at the gate is $\sqrt{2\times9.81\times1.2}\,\mathrm{m\,s^{-1}}\approx4.85\,\mathrm{m\,s^{-1}}$, so the transit time is about $0.02/4.85\,\mathrm s\approx4.12\,\mathrm{ms}$.

The rounding error is at most $0.5\,\mathrm{ms}$, a relative error of at most about $0.5/4.12\approx12\%$. In fact 4.12 rounds to 4, so the speed comes out about 3% high. For better accuracy, use a finer clock or a longer flag.

Exercise 4

The height curve of a fall bends downward, and a straight line can only cut through its middle. The best line lies above the curve at both ends of the record and below it in the middle, so the residuals are negative at the ends and positive in the middle, forming an arch.

The arch says a term in $t^2$ is missing, and since it bends the curve downward its coefficient is negative: the $-\tfrac12gt^2$ term. With that term added, the residuals should lose their pattern.

Exercise 5

Group A's standard error is much smaller, so group A is more precise. Group B's result differs from 9.81 by 0.09, less than one standard error, so it agrees. Group A differs by 0.19, more than nine standard errors, which random error cannot explain.

Group A probably has a systematic error, such as a wrong calibration scale or air resistance that cannot be ignored. Repeating the same experiment cannot reveal such errors. Look at the residuals, measure another way, or check the calibration.

Exercise 6

With the scale assumed 2% too large, every height is divided by 1.02. The model is linear in the heights, so the fitted parameters are divided by 1.02 as well: $\hat g\approx9.81/1.02\,\mathrm{m\,s^{-2}}\approx9.62\,\mathrm{m\,s^{-2}}$, exactly group A's result. That makes it a reasonable suspect, but only filming a ruler to recalibrate can confirm it.

Exercise 7

The uncertainty of the mean falls like $1/\sqrt N$, so $N\approx(0.0055/0.001)^2\approx30$ drops.

This assumes the errors of different drops are unrelated and the model has no systematic bias. Averaging does not shrink a systematic bias: with drag present, more repeats only give a wrong number more precisely.

Exercise 8

The first differences are 0.31, 0.87, 1.54 and 2.08 metres, and the second differences 0.56, 0.67 and 0.54 metres. For constant acceleration every second difference should equal $a(\Delta t)^2$. These three scatter around $0.59\,\mathrm m$ with no trend, so the motion is roughly constant acceleration with $a\approx0.59\,\mathrm{m\,s^{-2}}$.

The second differences scatter by about $\pm0.06\,\mathrm m$. If the position error is about $2\,\mathrm{cm}$, the rule from Chapter 2 gives second-difference noise of about $\sqrt6\times0.02\,\mathrm m\approx0.05\,\mathrm m$, the same size. If the second differences kept shrinking, the acceleration would be falling with time or speed. Suspect friction or air resistance growing with speed; the constant-acceleration model would then not be enough.

Exercise 9

When only a constant is fitted, $X$ is a column of $N$ ones and $X^{\mathsf T}X=N$. The normal equations give $\hat c=\frac1N\sum y_n$, the mean. The variance formula gives $\operatorname{Var}(\hat c)=\sigma^2/N$, a standard deviation of $\sigma/\sqrt N$. That is the $1/\sqrt N$ rule used in Exercise 7.

## Next: a different observer

This ends the first unit. We have learned to describe a motion, estimate it from a camera record and test a model against the record. Every description so far came from one camera standing still on the ground. Chapter 4 (in preparation) asks: for an observer who is moving, which of position, velocity and acceleration change, and which stay the same?

## Figure and data notes

- The constant-acceleration and free-fall formulas follow OpenStax University Physics Volume 1, Sections 3.4 and 3.5. Least squares and residual checks follow the NIST Engineering Statistics Handbook, Sections 4.1.4.1 and 4.4.4. Precision and accuracy use the definitions of VIM 2.15 and 2.13.
- Tables 3.1 and 3.2 and Figure 3.1 were computed with the Free Fall Lab's model code: a point mass, upward positive, fourth-order Runge–Kutta integration with a $0.01\,\mathrm s$ step, Gaussian camera noise, default noise realization 431. “Drag biases the fit” changes only the linear air damping, to $0.5\,\mathrm{s^{-1}}$, from “Earth tower”; “Clock misses the gate” changes only the clock resolution, to $0.05\,\mathrm s$.
- The drag fit was checked independently against the exact solution for linear drag: without noise, the fit gives $g\approx4.72\,\mathrm{m\,s^{-2}}$ and a residual RMS of about $231\,\mathrm{mm}$, consistent with the Lab's noisy result. The independent check of the standard error is in Section 3.9.
- In Table 3.2 the photogate speeds for Earth, Moon and Upward launch match the true speed at the gate to the digits shown. The 5.00 for Imperfect camera is the reading after clock rounding; the true value is about 8.86.

## Test it in the Lab

Change one setting at a time, then compare what you see with the tables and answers in this chapter.

[Free Fall & Measuring Gravity → Drag the release ring, replay a simulated drop and compare the camera fit with the photogate. Add drag to see a precise gravity estimate become biased.](https://sandboxphysics.com/labs/free-fall/)

## Sources and reading scope

1. [OpenStax, University Physics Volume 1, §3.4 Motion with Constant Acceleration](https://openstax.org/books/university-physics-volume-1/pages/3-4-motion-with-constant-acceleration)

   The constant-acceleration relations for velocity and position, with the initial time set to zero. Chapter 3 derives them from areas under graphs.

2. [OpenStax, University Physics Volume 1, §3.5 Free Fall](https://openstax.org/books/university-physics-volume-1/pages/3-5-free-fall)

   Up-positive sign convention with acceleration −g, neglect of air resistance, and the stated range of g at Earth’s surface, about 9.78 to 9.83 m/s².

3. [NIST/SEMATECH e-Handbook of Statistical Methods, §4.1.4.1 Linear Least Squares Regression](https://www.itl.nist.gov/div898/handbook/pmd/section1/pmd141.htm)

   Least squares chooses the parameters that minimize the sum of squared deviations between data and model. The chapter’s standard-error formula and its assumptions are stated in the chapter.

4. [NIST/SEMATECH e-Handbook of Statistical Methods, §4.4.4 How can I tell if a model fits my data?](https://www.itl.nist.gov/div898/handbook/pmd/section4/pmd44.htm)

   Residual plots as the main check of model adequacy: random scatter suggests an adequate model; structure left in the residuals signals a poor fit.

5. [JCGM 200:2012, International Vocabulary of Metrology (VIM), entries 2.13 measurement accuracy and 2.15 measurement precision](https://jcgm.bipm.org/vim/en/2.13.html)

   Definitions of accuracy (closeness to a true value) and precision (closeness of replicate measurements), including the note that precision should not be used to mean accuracy. Entry 2.15 is at https://jcgm.bipm.org/vim/en/2.15.html.

[← Chapter 2 · How does a camera measure motion?](https://sandboxphysics.com/learn/mechanics/measuring-motion-with-a-camera/) Chapter 4 · What changes for a different observer? (in preparation)

Chapters in this volume

- [1. How do we describe a motion?](https://sandboxphysics.com/learn/mechanics/describing-motion/)
- [2. How does a camera measure motion?](https://sandboxphysics.com/learn/mechanics/measuring-motion-with-a-camera/)
- [3. How do we test a constant-acceleration model?](https://sandboxphysics.com/learn/mechanics/testing-constant-acceleration/)
- 4\. What changes for a different observer? in preparation
- 5\. Why does a constant-speed turn still accelerate? in preparation
- 6\. How do forces change motion? in preparation
- 7\. When does a contact slip? in preparation
- 8\. How do ropes and pulleys link motions? in preparation
- 9\. How does a force along a path change kinetic energy? in preparation
- 10\. When can we compare only the start and the end? in preparation
- 11\. What is conserved in a collision? in preparation
- 12\. Why does it matter where a force acts? in preparation
- 13\. When does a rotation stay the same? in preparation
- 14\. How does one law of gravity explain falling and orbits? in preparation