How to Choose a Camera Position for Projectile Motion

For a projectile moving in a vertical plane, position the camera to the side of the flight path, with its optical axis perpendicular to the plane of motion. Keep the camera level and stationary, and place the calibration ruler in the same plane as the projectile. This gives you the most reliable measurements of horizontal and vertical position.

A camera pointed diagonally across the experiment might capture the entire flight, but that doesn’t mean it records accurate distances. It can compress the apparent horizontal motion, introduce perspective errors, and make a perfectly reasonable projectile behave rather strangely in your data.

The aim is to record the projectile’s actual two-dimensional motion as faithfully as possible. That starts with choosing the right camera position.

01Why camera position matters

Imagine launching a small ball horizontally across a laboratory and recording its flight.

You want to extract its position at different times, then investigate whether:

• Its horizontal velocity remains approximately constant.
• Its vertical acceleration is approximately constant.
• Its trajectory agrees with the predictions of the projectile-motion model.

The problem is that your camera doesn’t directly measure position in metres. It records a two-dimensional image made from light reaching its lens.

The position of the ball on that image depends on the ball’s actual position and the direction from which the camera observes it.

A poorly positioned camera can therefore make the measurements inaccurate, even if the projectile behaves exactly as predicted.

Understand the plane of motion

For a standard projectile-motion experiment, define three perpendicular directions:

• \(x\): horizontal, in the direction of the projectile’s travel.
• \(y\): vertical, positive upwards.
• \(z\): perpendicular to the vertical plane containing the projectile’s motion.

The ideal projectile travels in the \(x\)-\(y\) plane, meaning its \(z\)-coordinate remains constant.

Your camera should look directly towards this plane, along the \(z\)-direction.

This means its optical axis (the direction in which its lens points) should be perpendicular to the plane of motion.

The camera’s image sensor should consequently be parallel to the projectile’s plane of motion.

Why does this arrangement work?

When the camera is perpendicular to the motion plane, points within that plane are approximately the same distance from the camera. Their positions can therefore be converted using one consistent pixel-to-metre scale.

It also avoids the horizontal foreshortening associated with an oblique viewing angle.

Where should the camera actually go?

Use these guidelines:

The camera should also be:

• Level, with its optical axis approximately horizontal.
• Positioned at a height that places the middle of the trajectory near the centre of the frame.
• Far enough away to reduce the effects of small depth variations.
• Close enough that the projectile remains large and clear enough to track.
• Mounted on a stable tripod.

There is no universal distance that works for every experiment. Choose the largest practical distance that still gives you enough image resolution to identify the projectile’s centre accurately.

One important detail: increasing the camera distance does not fix an incorrect viewing angle. It reduces perspective changes caused by depth variations, but an oblique camera still produces foreshortening.

02Set up the experiment for reliable data collection

Choosing the camera position is only the first part. You also need a reference scale, a consistent coordinate system, and a reliable method of recording time.

Consider a hypothetical experiment in which a small ball is launched from a horizontal projectile launcher.

Step 1: Establish the plane of motion

Position the launcher so that the projectile travels through a clearly defined vertical plane.

Choose the launch point as your coordinate origin:

\[ x_0=0,\qquad y_0=0 \]

Take horizontal displacement to the right as positive and vertical displacement upwards as positive.

The origin should correspond to the centre of the projectile when it first leaves the launcher.

This is important because the launcher may exert a force on the projectile while it is still in contact with it. Your projectile-motion analysis should begin after release.

Step 2: Position a calibration ruler

Place a metre ruler or another known-length reference in the same plane as the projectile’s trajectory.

For example, a ruler with a known length of \(1.00\text{ m}\) can be used to establish the relationship between pixels and metres.

Do not place the ruler several centimetres in front of or behind the trajectory plane.

Objects at different distances from the camera appear at different sizes. A ruler positioned at the wrong depth can therefore introduce a systematic measurement error.

Step 3: Secure the camera

Mount the camera on a tripod.

Keep it stationary throughout the experiment. Avoid changing the zoom, orientation, or camera position between recordings.

A fixed camera gives you a consistent reference frame. If the camera moves, the apparent movement of the projectile includes both the projectile’s motion and the movement of the camera.

Use a plain background where possible. A projectile that contrasts clearly with its surroundings is easier to track.

Step 4: Choose an appropriate frame rate

A video is made up of individual frames recorded at particular time intervals.

If your camera records at 60 frames per second, the time between consecutive frames is:

\[ \Delta t=\frac{1}{60} \]

\[ \Delta t=0.0167\text{ s} \]

You can analyse every sixth frame to obtain measurements at intervals of approximately \(0.10\text{ s}\).

A higher frame rate can provide more detailed information about rapid motion. However, a high frame rate alone does not guarantee accurate measurements. The image must also be sufficiently sharp to identify the projectile’s position.

If using a phone’s slow-motion recording, use the actual capture frame rate when calculating time intervals, not simply the playback frame rate.

Step 5: Check the recording before analysing it

Before collecting your final dataset, record a short test video.

Check that:

• The entire trajectory remains inside the frame.
• The projectile is visible throughout its flight.
• The calibration ruler is clearly visible.
• The camera remains stationary.
• The camera is perpendicular to the motion plane.
• The projectile does not move significantly out of that plane.

A rectangular calibration grid placed in the motion plane can also help. If its geometry appears strongly distorted in the recording, reconsider the camera alignment before proceeding.

03Convert video frames into position measurements

Once you have a suitable recording, you can extract the projectile’s position at different times.

The important idea is to convert image coordinates into physical coordinates.

Calibrating the image

Suppose a ruler of known length \(L\) metres occupies \(N\) pixels in the video.

The image scale is:

\[ S=\frac{N}{L} \]

where:

• \(S\) is the image scale in pixels per metre.
• \(N\) is the measured ruler length in pixels.
• \(L\) is the actual ruler length in metres.

For example, if a \(1.00\text{ m}\) ruler occupies 480 pixels:

\[ S=\frac{480}{1.00} \]

\[ S=480\text{ pixels m}^{-1} \]

Every 480 pixels therefore represents approximately one metre.

If the camera is correctly aligned and the projectile remains in the motion plane, this scale can be applied to both horizontal and vertical distances.

Converting image coordinates

Let:

• \(u\) be the horizontal pixel coordinate, increasing to the right.
• \(v\) be the vertical pixel coordinate, increasing downwards.
• \((u_0,v_0)\) be the pixel coordinates of the chosen origin.
• \(S\) be the calibration scale in pixels per metre.

The physical horizontal position is:

\[ x=\frac{u-u_0}{S} \]

The physical vertical position is:

\[ y=\frac{v_0-v}{S} \]

Notice the reversed order in the vertical equation.

Most image coordinate systems increase downwards, whereas we have defined positive vertical displacement as upwards.

This sign convention prevents an otherwise unnecessary sign error in your analysis.

Example dataset

Suppose a hypothetical 60 fps recording produces the following measurements.

The camera is correctly positioned, a \(1.00\text{ m}\) ruler spans 480 pixels, and the projectile is tracked every six frames.

All measurements below are illustrative data, not results from an actual experiment.

Look at the horizontal measurements first.

The projectile moves approximately \(0.300\text{ m}\) every \(0.10\text{ s}\). This suggests constant horizontal velocity.

Now look at the vertical measurements.

The projectile initially moves upwards, but its vertical displacement eventually starts decreasing. The changes between consecutive vertical positions are:

\[ 0.151,\quad 0.053,\quad -0.045,\quad -0.143 \]

The differences between these values are approximately equal:

\[ -0.098\text{ m} \]

This is consistent with constant downward vertical acceleration.

05Account for measurement uncertainty

Your camera cannot identify the exact centre of a projectile with perfect precision.

The image has a finite resolution, the projectile may be slightly blurred, and the boundary of the object may be difficult to distinguish.

Suppose you estimate the projectile’s centre to within approximately \(\pm2\) pixels.

With a scale of 480 pixels per metre, the corresponding positional uncertainty is:

\[ \delta x=\frac{2}{480} \]

\[ \delta x\approx0.0042\text{ m} \]

Each position therefore has an uncertainty of approximately \(4.2\text{ mm}\), assuming the calibration scale is exact.

When calculating a displacement from two independent position readings, a conservative worst-case uncertainty is:

\[ \delta(\Delta x)=\frac{4}{480} \]

\[ \delta(\Delta x)\approx0.0083\text{ m} \]

For the previous worked example, the uncertainty in average velocity due to pixel reading alone is:

\[ \delta v_x=\frac{0.0083}{0.300} \]

\[ \delta v_x\approx0.03\text{ m s}^{-1} \]

So the result can be reported as:

\[ v_x=3.00\pm0.03\text{ m s}^{-1} \]

This uncertainty only accounts for the assumed pixel-reading uncertainty. It does not include uncertainty in the ruler’s length, calibration, frame rate, or camera alignment.

The distinction matters. A measurement can have very little random uncertainty and still be systematically inaccurate because the camera was positioned incorrectly.

06What happens when the camera is positioned incorrectly?

The most tempting mistake is assuming that a ruler in the video automatically makes every distance accurate.

It doesn’t.

A ruler only provides a reliable scale when its orientation and position are appropriate for the measurements being made.

Case 1: The camera is positioned at an angle

Suppose the camera is positioned \(30^\circ\) away from the perpendicular direction to the plane of motion.

In a simplified far-field projection, the apparent horizontal displacement becomes:

\[ \Delta x_{\text{image}} \approx\Delta x\cos\phi \]

where \(\phi\) is the angle between the camera’s optical axis and the normal to the motion plane.

For a true horizontal displacement of \(1.00\text{ m}\):

\[ \Delta x_{\text{image}} =1.00\cos30^\circ \]

\[ \Delta x_{\text{image}}=0.866\text{ m} \]

The horizontal displacement appears approximately \(13.4\%\) shorter.

If you use a vertical ruler to calibrate the image, this horizontal foreshortening will not be corrected.

A horizontal ruler in the motion plane can provide a separate horizontal scale in an idealised far-field arrangement, but perspective distortion may still vary across the image.

For a straightforward school experiment, it is much better to position the camera correctly than attempt to correct a poorly aligned recording.

Case 2: The camera is behind the launcher

This is another tempting arrangement.

You can see the projectile leaving the launcher, rising, and falling. It might even look like a reasonable trajectory.

But the camera is looking along the direction of horizontal travel.

As a result, much of the projectile’s horizontal displacement occurs towards or away from the camera rather than across the image.

You may be able to observe the vertical motion, but you cannot reliably extract the horizontal range using a simple two-dimensional image scale.

Move the camera to the side of the projectile’s flight path instead.

Case 3: The calibration ruler is at a different depth

This is a particularly important source of systematic error.

Imagine a camera positioned \(4.00\text{ m}\) from the projectile’s plane of motion.

A \(0.50\text{ m}\) ruler is placed in that plane and occupies 250 pixels.

Now imagine that you accidentally position another identical ruler \(0.50\text{ m}\) closer to the camera.

It will appear larger, even though its actual length is unchanged.

Using the closer ruler to calibrate the projectile’s position would produce an incorrect pixel-to-metre conversion.

This is why the calibration reference must be in the actual plane of motion, not merely somewhere visible in the recording.

Case 4: The camera moves during the experiment

Suppose you pan the camera to keep the projectile near the centre of the frame.

The projectile’s position relative to the image is now affected by two things:

• The projectile’s actual movement.
• The camera’s changing orientation.

A stationary object in the background may appear to move across the video.

This means the image coordinates no longer directly represent positions relative to a fixed laboratory reference frame.

A fixed tripod is the simplest way to avoid this problem.

It is possible to correct some camera movement using stationary reference points, but this requires additional image analysis and appropriate geometric assumptions. It is unnecessary complication for a basic projectile-motion investigation.

07Understand the limits of your measurements

Even with a carefully positioned camera, video analysis has limitations.

It is useful to separate measurement errors from limitations of the physical model.

There is another limitation that is easy to overlook.

A single camera cannot establish that the projectile really stayed in the plane of motion.

It records a two-dimensional projection of the object. If the projectile moves towards or away from the camera, its apparent size and measured coordinates can change.

The recording may still look like a smooth curved trajectory, even though the actual motion is three-dimensional.

If out-of-plane motion is a concern, use a second synchronised camera positioned at a substantially different viewing angle. With appropriate calibration, the two recordings can provide information about the projectile’s depth as well as its horizontal and vertical positions.

Don’t confuse measurement errors with air resistance

Suppose your calculated horizontal velocity appears to decrease throughout the flight.

It is tempting to conclude that air resistance is responsible.

That may be correct, but first check the experimental arrangement.

Was the camera stationary? Was the calibration ruler in the correct plane? Did the projectile move towards or away from the camera? Were the frame intervals calculated using the actual recording frame rate?

Only after checking these possibilities should you attribute the changing horizontal velocity to the physical behaviour of the projectile.

A well-positioned camera helps you distinguish between an actual departure from the ideal projectile model and an apparent departure caused by the measurement process.

For a fuller explanation of the physical assumptions behind the model, see Projectile Motion Assumptions in HSC Physics Explained.

If you need to revisit how positions and components are described in two dimensions, see Relative Position and Motion in Two Dimensions for HSC Physics.

08Questions and solutions

The following questions build from basic camera calibration to more challenging situations involving perspective, moving reference frames, and three-dimensional motion.

Question 1

A student records a projectile using a fixed side-on camera operating at 120 frames per second.

A \(1.50\text{ m}\) calibration ruler, positioned in the plane of motion, occupies 750 pixels.

The projectile’s centre is recorded at horizontal pixel coordinates of 210 pixels and 390 pixels in frames 8 and 20, respectively. Both coordinates are measured relative to the same origin.

Calculate:

(a) The time interval between the measurements.

(b) The projectile’s horizontal displacement.

(c) Its average horizontal velocity.

Solution 1

The average horizontal velocity is \(3.60\text{ m s}^{-1}\) to the right.

(a) Calculate the time interval.

There are:

\[ 20-8=12\text{ frames} \]

At 120 frames per second:

\[ \Delta t=\frac{12}{120} \]

\[ \boxed{\Delta t=0.100\text{ s}} \]

(b) Calculate horizontal displacement.

First, calculate the image scale:

\[ S=\frac{750}{1.50} \]

\[ S=500\text{ pixels m}^{-1} \]

The horizontal pixel displacement is:

\[ \Delta u=390-210=180\text{ pixels} \]

Therefore:

\[ \Delta x=\frac{180}{500} \]

\[ \boxed{\Delta x=0.360\text{ m}} \]

(c) Calculate average horizontal velocity.

\[ v_x=\frac{\Delta x}{\Delta t} \]

\[ v_x=\frac{0.360}{0.100} \]

\[ \boxed{v_x=3.60\text{ m s}^{-1}} \]

The result represents the average horizontal velocity over the measured interval. The use of a fixed, correctly positioned camera and a calibration ruler in the motion plane makes this calculation appropriate.

Question 2

A student positions a camera at an angle of \(25^\circ\) away from the perpendicular direction to a projectile’s vertical plane of motion.

The camera is sufficiently far away that an orthographic projection is a reasonable approximation.

The student uses a vertical \(1.00\text{ m}\) ruler to calibrate the vertical scale of the image.

The projectile’s actual horizontal displacement is \(2.00\text{ m}\).

(a) Calculate the apparent horizontal displacement recorded by the camera when the student applies the vertical calibration scale.

(b) Calculate the percentage error in the measured horizontal displacement.

(c) Explain why calibrating the vertical direction does not automatically correct the horizontal measurement.

Solution 2

The apparent horizontal displacement is approximately \(1.81\text{ m}\), representing an underestimate of approximately \(9.4\%\).

(a) Calculate the apparent horizontal displacement.

The camera is angled relative to the normal of the motion plane.

The horizontal displacement is therefore foreshortened:

\[ \Delta x_{\text{image}} =\Delta x\cos\phi \]

Substitute:

\[ \Delta x_{\text{image}} =2.00\cos25^\circ \]

\[ \boxed{\Delta x_{\text{image}}=1.81\text{ m}} \]

(b) Calculate the percentage error.

The absolute error is:

\[ \text{Error}=2.00-1.81 \]

\[ \text{Error}=0.19\text{ m} \]

The percentage error is:

\[ \%\text{ error} =\frac{0.19}{2.00}\times100 \]

\[ \boxed{\%\text{ error}\approx9.4\%} \]

The measurement underestimates the actual horizontal displacement.

(c) Explain the calibration problem.

The vertical ruler provides a scale for the vertical direction.

However, the camera’s oblique position changes the apparent horizontal scale through foreshortening.

The student has incorrectly assumed that the same image scale can be applied to both directions.

A better experimental design is to reposition the camera so that its optical axis is perpendicular to the motion plane. This allows a common scale to be used for both directions, assuming the projectile remains in that plane and lens distortion is negligible.

Question 3

A camera is positioned \(4.00\text{ m}\) from a projectile’s plane of motion.

A \(0.50\text{ m}\) calibration ruler placed in the motion plane occupies 250 pixels.

However, a student accidentally uses an identical ruler positioned \(0.50\text{ m}\) closer to the camera. This ruler occupies 286 pixels.

The projectile moves horizontally through 600 pixels in the recording.

(a) Calculate the horizontal displacement reported when the student uses the incorrectly positioned ruler.

(b) Calculate the correct horizontal displacement.

(c) Determine the percentage error and explain why the two rulers produce different calibration scales.

Solution 3

The student reports approximately \(1.05\text{ m}\), whereas the correct horizontal displacement is \(1.20\text{ m}\). The result is an underestimate of approximately \(12.6\%\).

(a) Calculate the incorrect displacement.

The student uses the nearer ruler to calculate the image scale:

\[ S_{\text{near}}=\frac{286}{0.50} \]

\[ S_{\text{near}}=572\text{ pixels m}^{-1} \]

The reported displacement is:

\[ \Delta x_{\text{reported}}=\frac{600}{572} \]

\[ \boxed{\Delta x_{\text{reported}}=1.05\text{ m}} \]

(b) Calculate the correct displacement.

The ruler in the actual motion plane occupies 250 pixels.

Its calibration scale is:

\[ S_{\text{correct}}=\frac{250}{0.50} \]

\[ S_{\text{correct}}=500\text{ pixels m}^{-1} \]

Therefore:

\[ \Delta x_{\text{correct}}=\frac{600}{500} \]

\[ \boxed{\Delta x_{\text{correct}}=1.20\text{ m}} \]

(c) Calculate the percentage error.

The absolute error is:

\[ 1.20-1.05=0.15\text{ m} \]

Therefore:

\[ \%\text{ error} =\frac{0.15}{1.20}\times100 \]

\[ \boxed{\%\text{ error}\approx12.5\%} \]

Using the rounded measured pixel values gives approximately \(12.6\%\).

The nearer ruler appears larger because it is closer to the camera.

The camera therefore assigns it more pixels per metre than it assigns to an object in the actual projectile plane.

When the student uses this larger scale to interpret the projectile’s position, the calculated displacement is too small.

The important principle is that a calibration object must be at the same depth as the object being measured.

Question 4

A student records a projectile using a camera that slowly pans horizontally to follow its movement.

The camera records four frames separated by \(0.10\text{ s}\).

The horizontal pixel coordinates of the projectile and a stationary background marker are:

Assume the projectile and marker are in the same plane, and the image scale is \(500\text{ pixels m}^{-1}\).

(a) Calculate the projectile’s average horizontal velocity relative to the laboratory.

(b) Calculate the velocity a student would incorrectly obtain by tracking only the projectile’s pixel coordinates.

(c) Explain why the fixed background marker is useful and why a stationary camera would be preferable.

Solution 4

The projectile’s actual average horizontal velocity is \(1.20\text{ m s}^{-1}\). Tracking only the projectile’s image coordinates gives an incorrect result of \(1.12\text{ m s}^{-1}\).

(a) Calculate the projectile’s displacement relative to the laboratory.

Because the camera pans, we must account for the movement of the stationary background marker.

At each time, subtract the marker’s pixel coordinate from the projectile’s pixel coordinate.

The relative horizontal displacement is:

\[ \Delta u=480-300 \]

\[ \Delta u=180\text{ pixels} \]

Convert to metres:

\[ \Delta x=\frac{180}{500} \]

\[ \Delta x=0.360\text{ m} \]

The time interval is:

\[ \Delta t=0.30\text{ s} \]

Therefore:

\[ v_x=\frac{0.360}{0.30} \]

\[ \boxed{v_x=1.20\text{ m s}^{-1}} \]

(b) Calculate the incorrect velocity.

If the student tracks only the projectile’s image coordinates:

\[ \Delta u=568-400 \]

\[ \Delta u=168\text{ pixels} \]

The apparent displacement is:

\[ \Delta x=\frac{168}{500} \]

\[ \Delta x=0.336\text{ m} \]

The incorrectly calculated velocity is:

\[ v_x=\frac{0.336}{0.30} \]

\[ \boxed{v_x=1.12\text{ m s}^{-1}} \]

(c) Explain the role of the background marker.

The marker is stationary relative to the laboratory, but it appears to move because the camera is panning.

By measuring the projectile’s position relative to the marker, we can account for the camera’s movement in this simplified arrangement.

This correction assumes the marker and projectile lie in the same plane and that the camera movement can be treated as a simple horizontal pan.

A fixed tripod would be preferable because it removes the additional movement of the camera and gives a much simpler reference frame.

Question 5

A student performs a projectile-motion experiment using a camera positioned perpendicular to the intended vertical plane of motion.

The camera is \(4.00\text{ m}\) from this plane. A \(1.00\text{ m}\) calibration ruler in the plane occupies 500 pixels.

The projectile’s actual horizontal motion is constant, with a horizontal velocity of \(3.00\text{ m s}^{-1}\).

However, the projectile gradually moves away from the camera, perpendicular to its original plane of motion.

The following information is known:

Assume an ideal pinhole camera, negligible lens distortion, and a fixed calibration scale based on the original plane.

(a) Calculate the horizontal positions that the camera would report at \(0.20\text{ s}\) and \(0.40\text{ s}\).

(b) Use the three reported horizontal positions to calculate the apparent horizontal acceleration, assuming equal time intervals.

(c) Explain why the student might incorrectly conclude that the projectile is experiencing horizontal deceleration.

(d) Recommend an experimental modification that would allow the student to distinguish genuine horizontal acceleration from this perspective effect.

Solution 5

The camera reports horizontal positions of approximately \(0.578\text{ m}\) and \(1.116\text{ m}\). These measurements produce an apparent horizontal acceleration of approximately \(-1.01\text{ m s}^{-2}\), even though the projectile’s actual horizontal acceleration is zero.

This is a perspective effect caused by the projectile leaving the plane used for calibration.

(a) Calculate the apparent horizontal positions.

For an ideal pinhole camera, the apparent image scale is inversely proportional to the object’s distance from the camera.

The original calibration plane is \(4.00\text{ m}\) away.

At \(0.20\text{ s}\), the projectile is:

\[ D_1=4.00+0.15 \]

\[ D_1=4.15\text{ m} \]

The apparent horizontal position is:

\[ x_1=x_{\text{actual}}\frac{4.00}{4.15} \]

\[ x_1=0.60\frac{4.00}{4.15} \]

\[ \boxed{x_1=0.578\text{ m}} \]

At \(0.40\text{ s}\), the projectile is:

\[ D_2=4.00+0.30 \]

\[ D_2=4.30\text{ m} \]

Therefore:

\[ x_2=1.20\frac{4.00}{4.30} \]

\[ \boxed{x_2=1.116\text{ m}} \]

The three horizontal positions recorded by the camera are therefore:

(b) Calculate the apparent horizontal acceleration.

For equal time intervals and constant acceleration:

\[ a_x=\frac{x_2-2x_1+x_0}{(\Delta t)^2} \]

Substitute the apparent positions:

\[ a_x= \frac{1.116-2(0.578)+0} {(0.20)^2} \]

\[ a_x=\frac{-0.040}{0.040} \]

\[ \boxed{a_x\approx-1.01\text{ m s}^{-2}} \]

The camera suggests that the projectile is decelerating horizontally.

However, the question explicitly states that its actual horizontal velocity is constant.

The apparent acceleration is therefore not caused by a real horizontal force. It is caused by the changing distance between the projectile and the camera.

(c) Explain the misleading result.

The tempting conclusion is that the projectile experiences horizontal deceleration, perhaps because of air resistance.

That conclusion would be reasonable if the camera’s scale remained constant throughout the flight.

But that assumption is incorrect.

As the projectile moves farther away, its image becomes smaller. The original calibration scale therefore becomes increasingly inaccurate.

The measured horizontal positions grow more slowly than the actual horizontal positions.

This creates an apparent decrease in horizontal velocity.

The key issue is that the projectile no longer remains in the plane used to calibrate the camera.

(d) Recommend an experimental modification.

The experiment should be redesigned to control or measure the projectile’s depth.

One option is to align the launcher so that the projectile remains within the intended vertical plane.

A second option is to use two synchronised cameras at substantially different viewing angles, with both views calibrated. The additional camera provides information about the projectile’s movement perpendicular to the original plane.

This question demonstrates an important limitation of video analysis: a convincing two-dimensional trajectory does not necessarily represent the projectile’s complete three-dimensional motion.

A camera can produce misleading measurements even when its original position and calibration appear correct.

09What to do with your data next

Once your camera position, calibration, and measurement uncertainty are under control, you can use the recorded positions to test the projectile-motion model.

Plot horizontal position against time and investigate whether the relationship is linear. Then plot vertical position against time and investigate whether it follows a quadratic relationship.

From these graphs, you can estimate horizontal velocity and vertical acceleration, compare your results with the theoretical predictions, and decide whether any disagreement is caused by experimental limitations or the physical behaviour of the projectile.

That is the real value of choosing the camera position carefully: it allows you to investigate the physics rather than spend the entire experiment trying to work out whether your camera has distorted it.