Why Projectile Experiments Can Produce Nonparabolic Tracks

A projectile experiment can produce a track that looks nonparabolic even when the theory predicts a parabola. The reason is that an experimental track is not the same thing as an ideal mathematical model. Your camera measures positions, your measurements contain uncertainty, and the projectile may experience forces that the simplified theory ignores.

The important question is whether the deviations are simply measurement noise or evidence that the projectile is not following the ideal model. Those are different conclusions, and your experimental analysis needs to distinguish between them.

## Why doesn't the experimental track look like a parabola?

Imagine launching a small ball horizontally from a table and recording its motion with a camera. Under the ideal projectile model, the ball has constant horizontal velocity and constant downward acceleration due to gravity.

The horizontal position changes uniformly with time, while the vertical displacement increases quadratically with time. If we ignore air resistance and take upward as positive, the equations are

\[
x = x_0 + u_x t
\]

\[
y = y_0 + u_y t - \frac{1}{2}gt^2
\]

Here:
- \(x\) and \(y\) are the horizontal and vertical positions, measured in metres.
- \(x_0\) and \(y_0\) are the initial coordinates, in metres.
- \(u_x\) and \(u_y\) are the initial horizontal and vertical velocity components, in metres per second.
- \(t\) is the elapsed time, in seconds.
- \(g\) is the magnitude of gravitational acceleration, in metres per second squared.

Because horizontal position changes linearly with time and vertical position changes quadratically, eliminating time produces a quadratic relationship between \(y\) and \(x\). That is the mathematical reason the ideal trajectory is parabolic.

[IMAGE: Side-on scientific illustration of a ball launched horizontally from the edge of a table. Show the table, launch point, curved downward trajectory, horizontal and vertical axes, and several equally spaced time positions along the ideal parabola. Clearly distinguish the physical ball positions from the smooth theoretical curve.]

But an experiment introduces several complications:

1. **Measurement uncertainty:** The camera cannot locate the ball's centre with perfect accuracy. Pixel resolution, blur, and difficulty identifying the centre all affect the recorded coordinates.
2. **Timing uncertainty:** The recorded frames may not correspond to perfectly known time intervals, or the chosen first frame may not represent the actual release.
3. **Experimental conditions:** Air resistance, wind, and spin can produce forces that are absent from the ideal model.
4. **Camera geometry:** Perspective distortion, a tilted camera, or an incorrectly positioned measurement scale can distort the apparent trajectory.
5. **Data processing:** Rounding coordinates, choosing inconsistent reference points, or fitting an inappropriate curve can make a good trajectory appear poor.

The key distinction is between a track that is physically nonparabolic and a track that only appears nonparabolic because of how it was measured.

## What should the experimental data look like?

Suppose a ball is launched horizontally, and a camera records its position at equal time intervals.

For an ideal horizontal launch:
- The horizontal displacement between successive frames should be approximately constant.
- The downward vertical displacement should increase as time passes.
- The vertical displacement should be proportional to the square of elapsed time.

That last point gives us a useful experimental test. If the ball starts at \(y=0\), its ideal vertical displacement is

\[
\Delta y = -\frac{1}{2}gt^2
\]

A graph of vertical displacement against time should be quadratic, while a graph of vertical displacement against the square of time should be linear.

[IMAGE: Three-panel educational diagram showing the same horizontal projectile experiment represented as (1) a curved vertical displacement versus horizontal position graph, (2) a straight-line horizontal position versus time graph, and (3) a downward-curving vertical displacement versus time graph. Include clearly labelled axes and distinguish ideal theoretical curves from hypothetical experimental points.]

This gives you more than one way to test the model. A track that looks slightly uneven on a photograph might still produce data consistent with constant gravitational acceleration.

Conversely, a smooth-looking curve does not automatically prove that the projectile followed ideal projectile motion. A plotted curve can look convincing even when the measurements have been distorted.

## How the experiment produces a set of measurements

A video does not give you a continuous record of the ball's position. It gives you individual frames.

To turn those frames into useful data, you need three things:

- A known time interval between frames.
- A consistent conversion from pixels to metres.
- A consistent method of identifying the ball's position.

For example, a camera recording at 100 frames per second captures one frame every \(0.010\text{ s}\). If you analyse every tenth frame, your measurements are separated by \(0.100\text{ s}\).

For each selected frame, record the centre of the ball. Use the same coordinate system throughout the experiment, with the release point as the origin, rightwards as positive \(x\), and upwards as positive \(y\).

[IMAGE: Annotated side-on projectile experiment showing a stationary camera on a tripod facing the vertical plane of motion, a calibration ruler in the same plane as the ball, and the ball at several recorded positions. Include the release-point origin, positive horizontal and vertical axes, and a clear indication that the camera records discrete frames rather than a continuous path.]

There is an important limitation here. The camera measures pixels, not metres. You convert pixels into distances using a known reference length, such as a metre ruler.

If the ruler is not positioned in the same plane as the projectile, the conversion can be inaccurate. If the ball moves towards or away from the camera, its apparent size and position can also change.

For more detail on this measurement problem, see [How to Choose a Camera Position for Projectile Motion](https://tutorgum.com/learn/resources/hsc-physics-how-to-choose-a-camera-position-for-a-projectile-motion-experiment/).

### A hypothetical dataset

Consider a ball launched horizontally with an initial horizontal velocity of \(2.00\text{ m s}^{-1}\).

Assume:
- The release point is the origin.
- The initial vertical velocity is zero.
- Gravitational acceleration is \(9.80\text{ m s}^{-2}\) downwards.
- Air resistance is negligible.
- The camera records positions at equal time intervals.

The theoretical vertical position is

\[
y=-4.90t^2
\]

The theoretical horizontal position is

\[
x=2.00t
\]

The following measurements are illustrative, not results from a real experiment.

| Time (s) | Horizontal position (m) | Measured vertical position (m) | Predicted vertical position (m) |
|---:|---:|---:|---:|
| 0.10 | 0.20 | -0.050 | -0.049 |
| 0.20 | 0.40 | -0.180 | -0.196 |
| 0.30 | 0.60 | -0.430 | -0.441 |
| 0.40 | 0.80 | -0.800 | -0.784 |
| 0.50 | 1.00 | -1.210 | -1.225 |

Suppose the uncertainty in each vertical position measurement is approximately \(\pm0.020\text{ m}\).

Notice that the measured values do not exactly match the theoretical predictions. That is normal. An experiment is not expected to reproduce every theoretical value perfectly.

The more useful question is how large the differences are compared with the measurement uncertainty.

### Worked example: Is the measurement consistent with the theoretical prediction?

At \(t=0.40\text{ s}\), the measured vertical position is \(-0.800\text{ m}\). Determine whether this measurement is consistent with the ideal projectile model.

**Step 1**: Calculate the predicted vertical position.

Using upward as positive:

\[
y=-\frac{1}{2}gt^2
\]

Substitute the known values:

\[
\begin{aligned}
y&=-\frac{1}{2}(9.80)(0.40)^2\\
&=-0.784\text{ m}
\end{aligned}
\]

**Step 2**: Calculate the difference between the measured and predicted positions.

Define the residual as

\[
r=y_{\text{measured}}-y_{\text{predicted}}
\]

Therefore,

\[
r=-0.800-(-0.784)
\]

\[
r=-0.016\text{ m}
\]

The measured ball position is \(0.016\text{ m}\) below the predicted position.

**Step 3**: Compare the difference with the uncertainty.

The measurement uncertainty is \(\pm0.020\text{ m}\).

Since

\[
|r|=0.016\text{ m}<0.020\text{ m}
\]

the measurement is consistent with the theoretical prediction within the stated uncertainty.

This does not prove that the projectile followed a perfect parabola. It means the difference between this measurement and the model is small enough to be explained by the stated measurement uncertainty.

That is a much more defensible conclusion than claiming the experiment has failed because two numbers do not match exactly.

## How to tell measurement noise from a genuine departure

A useful way to investigate a projectile track is to examine its residuals.

A residual is the difference between an experimental measurement and the corresponding theoretical prediction.

\[
r=y_{\text{measured}}-y_{\text{predicted}}
\]

A positive residual means the measured position is above the predicted position. A negative residual means it is below.

The pattern of these residuals can tell you much more than simply looking at the plotted trajectory.

### Case 1: Random measurement uncertainty

Imagine that you repeat the measurements and obtain small differences from the predicted curve.

Some points sit slightly above the curve. Others sit slightly below it.

This is what you might expect from random measurement uncertainty.

Possible causes include:
- Difficulty identifying the exact centre of the ball.
- Slight motion blur.
- Limited pixel resolution.
- Small variations in selecting the video frame.
- Minor rounding during calculations.

The differences do not follow a clear pattern.

[IMAGE: Scientific graph comparing a smooth ideal parabolic trajectory with five hypothetical experimental points scattered slightly above and below it. Include vertical uncertainty bars and a residual plot underneath showing small positive and negative residuals distributed randomly around zero.]

If the residuals are small compared with the uncertainties, the experimental data are consistent with the ideal model.

They do not establish that the model is exactly true. They establish that the experiment has not detected a significant disagreement.

### Case 2: Systematic deviation from the predicted curve

Now imagine that almost every measured point falls below the theoretical curve, and the difference becomes larger as the projectile travels further.

That is a different result.

A consistent pattern suggests that something other than random measurement uncertainty may be affecting the experiment.

Consider this second hypothetical dataset.

The predicted positions are unchanged, but the measured positions are now:

| Time (s) | Predicted \(y\) (m) | Measured \(y\) (m) | Residual (m) |
|---:|---:|---:|---:|
| 0.10 | -0.049 | -0.060 | -0.011 |
| 0.20 | -0.196 | -0.222 | -0.026 |
| 0.30 | -0.441 | -0.500 | -0.059 |
| 0.40 | -0.784 | -0.900 | -0.116 |
| 0.50 | -1.225 | -1.400 | -0.175 |

Again, assume a vertical measurement uncertainty of \(\pm0.020\text{ m}\).

The residuals are now consistently negative, and their magnitudes increase with time.

[IMAGE: Comparison graph with two side-by-side residual plots. The first shows random points scattered around zero within uncertainty bars. The second shows residuals becoming progressively more negative, with several points clearly outside their uncertainty bars. Label the horizontal axis time and the vertical axis residual.]

### Worked example: Does the second dataset support the ideal model?

Use the final measurement to determine whether random positional uncertainty can reasonably explain the observed difference.

**Step 1**: Calculate the residual at \(t=0.50\text{ s}\).

The predicted position is \(-1.225\text{ m}\), while the measured position is \(-1.400\text{ m}\).

\[
\begin{aligned}
r&=-1.400-(-1.225)\\
&=-0.175\text{ m}
\end{aligned}
\]

**Step 2**: Compare the residual with the measurement uncertainty.

The magnitude of the residual is \(0.175\text{ m}\), while the stated uncertainty is \(0.020\text{ m}\).

\[
\frac{|r|}{\text{uncertainty}}
=\frac{0.175}{0.020}
=8.75
\]

The discrepancy is 8.75 times the stated positional uncertainty.

**Step 3**: Examine the pattern across the dataset.

The residuals become increasingly negative rather than fluctuating randomly around zero.

This is evidence that random positional uncertainty alone is not an adequate explanation for the disagreement.

The ideal model may be incomplete, or a systematic experimental error may be affecting the measurements.

You cannot yet conclude that air resistance is responsible. You need to investigate other possible causes before making that claim.

## When the projectile really does follow a different path

Measurement problems are only one possibility. Sometimes the projectile genuinely does not follow the ideal parabolic trajectory.

The most familiar cause is air resistance.

### Air resistance changes the acceleration

In the ideal model, gravity is the only significant force acting on the projectile after release.

Real projectiles also experience air resistance, or drag.

Drag acts opposite to the projectile's velocity relative to the surrounding air.

For example, when a ball travels downwards and to the right, the drag force acts upwards and to the left.

[IMAGE: Free-body diagram of a ball travelling diagonally downwards and to the right. Show the velocity direction, gravitational force vertically downwards, and air resistance acting diagonally upwards and to the left. Draw the forces from the ball's centre and distinguish the velocity arrow from the force arrows.]

This changes both components of the motion.

- The horizontal velocity is no longer necessarily constant.
- The vertical acceleration is no longer necessarily equal to \(9.80\text{ m s}^{-2}\) downwards.

The projectile's acceleration now depends on the additional force from the air.

As the velocity changes, the direction and magnitude of drag can change too.

The equations for ideal projectile motion therefore no longer describe the complete motion.

The trajectory may depart from a parabola.

Other possible causes include:

- **Wind:** The projectile experiences an aerodynamic force due to its motion relative to the surrounding air.
- **Spin:** A spinning projectile can experience a sideways or vertical aerodynamic force, changing its trajectory.
- **An unintended sideways launch:** The projectile may move outside the assumed vertical plane of motion.
- **Changing experimental conditions:** A moving launcher or an unstable release can make different trials produce different trajectories.

An important distinction: air resistance does not automatically make every measured track visibly nonparabolic. For a dense, compact ball moving relatively slowly, its effect may be too small to distinguish from experimental uncertainty.

You need evidence that the effect is significant.

### What about changes in gravitational acceleration?

The ideal model also assumes constant gravitational acceleration.

In reality, gravitational acceleration changes slightly with distance from Earth's centre. Its direction also changes across sufficiently large distances.

However, for an ordinary school projectile experiment, the projectile travels a tiny distance compared with Earth's radius.

The variation in gravitational acceleration is negligible for the measurements involved.

It is therefore unlikely to explain an obviously distorted track in a typical HSC experiment.

## A diagnostic guide: What does the pattern actually tell you?

When an experimental track does not look parabolic, avoid jumping straight to an explanation.

Use the measurements to narrow down the possibilities.

| What you observe | Possible explanation | What to investigate |
|---|---|---|
| Small deviations above and below the curve | Random measurement uncertainty | Position resolution and repeated measurements |
| Nearly constant vertical offset | Incorrect choice of vertical origin | Coordinate reference point |
| Increasing deviation across the flight | Air resistance or a position-dependent measurement error | Drag, camera geometry, and calibration |
| One isolated point far from the curve | Tracking error or an obstructed frame | Recheck the original video frame |
| Different trajectories between repeated launches | Inconsistent launch conditions | Launch speed, angle, and release mechanism |
| Different apparent tracks from different camera positions | Perspective or projection effects | Camera alignment and three-dimensional motion |

There is a subtle but useful distinction between a constant error and an error that changes across the experiment.

Suppose you accidentally place the vertical origin \(0.05\text{ m}\) below the actual release point.

Every measured vertical coordinate will be displaced by the same amount.

Your measured equation might become

\[
y=0.05-\frac{1}{2}gt^2
\]

rather than

\[
y=-\frac{1}{2}gt^2
\]

The graph has moved upwards, but it is still parabolic. The curvature has not changed.

Similarly, a constant scale error changes the numerical values of the coordinates but does not necessarily destroy the parabolic shape.

A position-dependent perspective error is more troublesome because it can distort the shape itself.

This is why you should not treat every difference between an experimental graph and a theoretical graph as evidence of a failure in the physics model.

## A practical decision process

When analysing a projectile experiment, use this sequence.

**1. Check the coordinate system.**

Confirm that the origin, positive directions, and release point are consistent.

**2. Check the measurements.**

Look for unclear ball positions, incorrect pixel-to-metre conversions, and inconsistent frame selection.

**3. Plot the experimental positions.**

Use a scatter plot rather than automatically joining every point with straight lines.

**4. Compare with the theoretical model.**

For an ideal projectile, the trajectory should follow a quadratic relationship between horizontal and vertical position.

**5. Examine the residuals.**

Check whether the differences are small and irregular or large and systematic.

**6. Investigate possible physical effects.**

If the deviations exceed the experimental uncertainty, consider air resistance, wind, spin, or motion outside the assumed plane.

**7. Repeat or improve the experiment.**

A better camera arrangement, clearer tracking, and repeated trials can help distinguish measurement problems from genuine physical effects.

For the underlying assumptions that make the ideal trajectory parabolic, see [Projectile Motion Assumptions in HSC Physics Explained](https://tutorgum.com/learn/resources/hsc-physics-projectile-motion-assumptions/).

## Questions and solutions

The following questions progress from checking an individual measurement to evaluating whether a two-dimensional experiment has captured the projectile's full motion.

### Question 1

A ball is launched horizontally with an initial horizontal velocity of \(1.80\text{ m s}^{-1}\).

Take the release point as the origin, with upwards positive, and assume gravitational acceleration is \(9.80\text{ m s}^{-2}\).

At \(t=0.25\text{ s}\), a video measurement gives a vertical position of \(-0.29\text{ m}\), with an uncertainty of \(\pm0.03\text{ m}\).

Determine whether this measurement is consistent with the ideal projectile model.

### Solution 1

The measurement is consistent with the ideal model because its difference from the predicted position is smaller than the stated uncertainty.

**Step 1**: Calculate the theoretical vertical position.

Since the ball is launched horizontally, its initial vertical velocity is zero.

\[
y=-\frac{1}{2}gt^2
\]

Substitute:

\[
\begin{aligned}
y&=-\frac{1}{2}(9.80)(0.25)^2\\
&=-0.30625\text{ m}
\end{aligned}
\]

**Step 2**: Calculate the residual.

\[
\begin{aligned}
r&=y_{\text{measured}}-y_{\text{predicted}}\\
&=-0.29-(-0.30625)\\
&=+0.01625\text{ m}
\end{aligned}
\]

**Step 3**: Compare with the uncertainty.

\[
0.01625\text{ m}<0.03\text{ m}
\]

The difference is within the stated measurement uncertainty.

Therefore, this measurement does not provide evidence that the projectile has departed from the ideal model.

One measurement cannot establish the shape of the entire trajectory, however. Several positions are needed to investigate whether the complete track is consistent with a parabola.

### Question 2

A student launches a ball horizontally and records its vertical position at equal time intervals.

The vertical positions are measured relative to a laboratory floor, rather than relative to the launch point.

The launch point is \(0.060\text{ m}\) above the floor.

The following illustrative data are obtained:

| Time (s) | Measured vertical position (m) |
|---:|---:|
| 0.00 | 0.060 |
| 0.10 | 0.011 |
| 0.20 | -0.136 |
| 0.30 | -0.381 |
| 0.40 | -0.724 |

Assume \(g=9.80\text{ m s}^{-2}\), negligible air resistance, and negligible measurement uncertainty for this question.

A student claims that the data are inconsistent with a parabolic trajectory because they do not fit the equation \(y=-4.90t^2\).

Evaluate the student's claim and determine the correct equation for the measured vertical position.

### Solution 2

The student's claim is incorrect. The data are consistent with a parabolic trajectory because the vertical coordinate has a non-zero initial value.

**Step 1**: Identify the coordinate origin.

The equation

\[
y=-4.90t^2
\]

assumes that the vertical origin is at the release point.

However, the experiment measures height relative to the laboratory floor.

The projectile therefore begins at

\[
y_0=+0.060\text{ m}
\]

**Step 2**: Include the correct initial position in the equation.

For a horizontal launch:

\[
y=y_0-\frac{1}{2}gt^2
\]

Substitute:

\[
\boxed{y=0.060-4.90t^2}
\]

**Step 3**: Check one of the measurements.

At \(t=0.30\text{ s}\):

\[
\begin{aligned}
y&=0.060-4.90(0.30)^2\\
&=0.060-0.441\\
&=-0.381\text{ m}
\end{aligned}
\]

This exactly matches the supplied measurement.

**Step 4**: Interpret the result.

The non-zero intercept does not mean that the trajectory is nonparabolic.

It simply reflects the choice of coordinate origin.

The coefficient of the quadratic term remains \(-4.90\), corresponding to the expected gravitational acceleration.

This is an important experimental distinction: a misplaced origin changes the position of a graph, not its fundamental shape.

### Question 3

A student records a projectile experiment at a constant capture rate of 120 frames per second.

The actual time between two selected frames is \(0.10\text{ s}\).

However, the student uses the slow-motion playback timestamps and incorrectly assumes that the time interval is \(0.20\text{ s}\).

The measured horizontal and vertical pixel coordinates are otherwise correct.

**(a)** Explain whether this timing error changes the geometric shape of the plotted horizontal-versus-vertical trajectory.

**(b)** The student uses the vertical displacement over equal intervals to calculate gravitational acceleration. Determine the value the student would obtain if the assumed time intervals are twice their actual values.

**(c)** Explain why an incorrect value of gravitational acceleration does not necessarily mean that the spatial trajectory is nonparabolic.

### Solution 3

**(a) The timing error does not change the geometric shape of the spatial trajectory.**

The horizontal-versus-vertical graph is constructed using measured positions.

If the pixel coordinates are correct and the same physical frames are selected, the positions remain unchanged.

Incorrect timestamps affect the relationship between position and time, but they do not move the measured points on an \(x\)-\(y\) graph.

The spatial track can therefore remain parabolic even though the time-based analysis is incorrect.

**(b) The student calculates \(2.45\text{ m s}^{-2}\).**

For equal time intervals, the difference between successive vertical displacements is related to acceleration by

\[
D=-g(\Delta t)^2
\]

where \(D\) is the difference between successive vertical displacements.

The student uses an interval twice as large as the actual interval.

Let the actual interval be \(\Delta t\). The assumed interval is \(2\Delta t\).

The calculated gravitational acceleration is therefore

\[
g_{\text{calculated}}
=\frac{g(\Delta t)^2}{(2\Delta t)^2}
\]

Simplifying:

\[
g_{\text{calculated}}=\frac{g}{4}
\]

Substitute the accepted value:

\[
\begin{aligned}
g_{\text{calculated}}
&=\frac{9.80}{4}\\
&=2.45\text{ m s}^{-2}
\end{aligned}
\]

The student significantly underestimates gravitational acceleration.

**(c) The spatial trajectory and the time-based analysis test different things.**

The spatial trajectory describes the relationship between horizontal and vertical position.

The time-based graphs describe how each position changes over time.

Incorrect timing can distort calculated velocities and accelerations without changing the actual measured spatial positions.

This is why you should identify exactly which experimental quantity is affected before concluding that the projectile's physical motion disagrees with the theoretical model.

### Question 4

Two cameras simultaneously record the same projectile launch.

Camera A is positioned perpendicular to the plane of motion. Its calibration ruler is placed in the same plane as the projectile.

Camera B is positioned at an oblique angle. Its calibration ruler is placed on a wall behind the projectile's flight path.

Both recordings are converted into metres and compared with the ideal projectile model.

The results are:

| Measurement | Camera A | Camera B |
|---|---:|---:|
| Maximum absolute residual | 0.018 m | 0.110 m |
| Typical positional uncertainty | ±0.020 m | ±0.020 m |
| Residual pattern | Small, alternating deviations | Increasingly negative deviations |

A student concludes that the projectile must experience significant air resistance because Camera B records a nonparabolic track.

Evaluate this conclusion. Explain what the two recordings suggest and describe one additional investigation that would strengthen the conclusion.

### Solution 4

The student's conclusion is not justified. The difference between the two recordings suggests that camera geometry or calibration is affecting the measurements from Camera B.

**Step 1**: Compare the two sets of measurements.

Camera A has a maximum residual of \(0.018\text{ m}\), which is within its stated uncertainty of \(\pm0.020\text{ m}\).

Its measurements are therefore consistent with the ideal model within experimental uncertainty.

Camera B has a maximum residual of \(0.110\text{ m}\), which is substantially larger than its stated uncertainty.

Its measurements show a systematic departure from the theoretical prediction.

**Step 2**: Consider the camera arrangements.

Camera A is positioned perpendicular to the plane of motion, and its calibration ruler is in the same plane as the projectile.

These conditions help ensure that the measured pixel distances correspond consistently to actual distances.

Camera B is positioned obliquely, with its calibration ruler behind the projectile.

The projectile and ruler may therefore have different apparent scales in the image. Perspective effects can also distort the apparent positions of the projectile throughout its flight.

This provides a plausible explanation for the increasing residuals.

**Step 3**: Consider the fact that both cameras record the same launch.

The projectile cannot simultaneously follow two different physical trajectories.

The disagreement between the recordings must therefore involve differences in how the two cameras represent or process the motion, rather than proving that the projectile experiences air resistance.

Camera A provides the more reliable evidence under the stated experimental conditions.

**Step 4**: Propose an additional investigation.

Repeat the analysis after repositioning Camera B perpendicular to the plane of motion and placing its calibration ruler in the same plane as the projectile.

Keep the physical launch conditions unchanged.

If the residual pattern becomes much smaller, this supports the explanation that Camera B's original geometry introduced systematic measurement errors.

If both calibrated recordings show similar systematic departures, investigate physical effects such as air resistance or wind.

The important principle is that a systematic residual pattern is evidence of a disagreement, not an automatic identification of its cause.

### Question 5

A projectile is launched horizontally through a vertical plane.

Two synchronised cameras record its motion. Both cameras have been calibrated, and their position measurements have an uncertainty of \(\pm0.001\text{ m}\).

Camera A records the horizontal and vertical positions. Its data are consistent with

\[
x=2.00t
\]

and

\[
y=-4.90t^2
\]

within experimental uncertainty.

Camera B views the projectile from a direction perpendicular to the original plane of motion and measures its sideways displacement, \(z\).

The following illustrative data are obtained:

| Time (s) | Sideways displacement, \(z\) (m) |
|---:|---:|
| 0.00 | 0.000 |
| 0.10 | 0.002 |
| 0.20 | 0.012 |
| 0.30 | 0.042 |
| 0.40 | 0.112 |

Take the initial sideways position as zero and assume that the time intervals are accurate and equal.

A student argues:

"The side-on camera shows a perfect parabola, so the projectile must be following ideal projectile motion. The sideways measurements are just a minor experimental distraction."

Evaluate this argument.

Your response must:

**(a)** Explain what the first camera can and cannot establish about the projectile's motion.

**(b)** Calculate the successive sideways displacements and their second differences.

**(c)** Use the measurement uncertainty to determine whether the sideways acceleration can reasonably be treated as constant.

**(d)** Explain what this means for the complete three-dimensional trajectory and identify a possible physical cause.

### Solution 5

The student's argument is incorrect. Camera A establishes that the motion is consistent with a parabolic trajectory when projected onto the original vertical plane, but it does not establish that the complete motion is ideal projectile motion.

The sideways measurements provide evidence of additional, changing acceleration.

**(a) The first camera only measures a two-dimensional projection.**

Camera A records the horizontal and vertical coordinates.

Its measurements are consistent with

\[
x=2.00t
\]

and

\[
y=-4.90t^2
\]

These equations describe ideal projectile motion in the measured vertical plane.

However, the camera does not measure sideways displacement.

A projectile can have a perfectly parabolic side-on projection while also moving outside that plane.

Therefore, Camera A alone cannot establish that the complete motion is confined to a single vertical plane.

**(b) Calculate the sideways displacements and second differences.**

The time intervals are all \(0.10\text{ s}\).

The successive sideways displacements are:

\[
\begin{aligned}
\Delta z_1&=0.002-0.000=0.002\text{ m}\\
\Delta z_2&=0.012-0.002=0.010\text{ m}\\
\Delta z_3&=0.042-0.012=0.030\text{ m}\\
\Delta z_4&=0.112-0.042=0.070\text{ m}
\end{aligned}
\]

Now calculate the differences between successive displacements:

\[
\begin{aligned}
D_1&=0.010-0.002=0.008\text{ m}\\
D_2&=0.030-0.010=0.020\text{ m}\\
D_3&=0.070-0.030=0.040\text{ m}
\end{aligned}
\]

For constant sideways acceleration, these second differences should be approximately equal for equal time intervals.

They clearly increase across the dataset.

**(c) Test whether the sideways acceleration is constant.**

For equal time intervals, the second difference is related to acceleration by

\[
D=a_z(\Delta t)^2
\]

where \(a_z\) is the sideways acceleration.

Therefore,

\[
a_z=\frac{D}{(\Delta t)^2}
\]

Using the three second differences:

\[
\begin{aligned}
a_{z1}&=\frac{0.008}{(0.10)^2}=0.8\text{ m s}^{-2}\\
a_{z2}&=\frac{0.020}{(0.10)^2}=2.0\text{ m s}^{-2}\\
a_{z3}&=\frac{0.040}{(0.10)^2}=4.0\text{ m s}^{-2}
\end{aligned}
\]

These estimates are substantially different.

We must also consider measurement uncertainty before deciding whether the differences are significant.

Each sideways position has an uncertainty of \(\pm0.001\text{ m}\).

A second difference is calculated using

\[
D=z_3-2z_2+z_1
\]

If we use the conservative maximum-error approach, its uncertainty is

\[
\Delta D=0.001+2(0.001)+0.001
\]

\[
\Delta D=0.004\text{ m}
\]

The three second differences therefore have approximate uncertainty ranges:

| Second difference | Value (m) | Conservative range (m) |
|---|---:|---:|
| \(D_1\) | 0.008 | 0.004 to 0.012 |
| \(D_2\) | 0.020 | 0.016 to 0.024 |
| \(D_3\) | 0.040 | 0.036 to 0.044 |

The ranges do not overlap.

Even allowing for the stated measurement uncertainty, the sideways acceleration is not consistent with a constant value.

**(d) Interpret the complete motion.**

The projectile has a sideways component of motion that changes with time.

Its trajectory therefore cannot be described by the ideal two-dimensional projectile model, even though its side-on projection agrees with the expected parabola.

The increasing sideways acceleration suggests an additional force that changes during the flight.

A possible explanation is an aerodynamic force caused by wind or spin.

The important point is that the first camera's result was not necessarily wrong. It was incomplete.

A two-dimensional projection can hide motion in a third direction. The second camera reveals information that the first camera cannot provide.

The data therefore challenge the assumption that gravity is the only significant force acting on the projectile.

This is a stronger conclusion than simply saying that the track looks nonparabolic. It identifies which part of the motion disagrees with the ideal model and supports that conclusion using measurements and uncertainty.

## What to take into your next experiment

A nonparabolic-looking experimental track does not automatically mean that projectile motion theory is wrong.

It may reflect random measurement uncertainty, a systematic error, a poorly chosen coordinate system, or a genuine physical effect such as air resistance.

The distinction comes from examining the data rather than judging the appearance of the curve.

For your next projectile investigation, pay particular attention to whether the measured landing position agrees with the theoretical prediction. Once you have established whether the model is consistent with the recorded trajectory, comparing the predicted and experimental landing positions gives you another independent way to evaluate the experiment.