Why Parallax Distorts a Recorded Projectile Trajectory

Your video can show a ball following a smooth projectile path, yet the coordinates you extract might suggest that it is slowing down horizontally and experiencing an acceleration smaller than gravity. The ball might be behaving perfectly normally. The problem could be parallax.

Parallax distorts a recorded projectile trajectory because a camera converts three-dimensional positions into a two-dimensional image. When the projectile’s distance from the camera changes during flight, its apparent size and position change as well. If you use one pixel-to-metre conversion for every frame, you can mistake this changing image scale for a change in the projectile’s actual motion.

This matters when analysing projectile videos. An apparent departure from the expected parabolic trajectory does not automatically mean that air resistance is significant or that the projectile model has failed. First, you need to establish whether the camera has recorded the motion accurately.

01Why does parallax affect a projectile video?

Imagine launching a small ball horizontally across a laboratory. You place a camera on a tripod, record the flight, and use a ruler to convert pixel positions into metres.

Your aim is to investigate two familiar predictions:

• Horizontal velocity should remain constant if air resistance is negligible.
• Vertical acceleration should remain approximately constant at \(9.8\text{ m s}^{-2}\) downwards.

Now consider the camera position.

The projectile travels through a vertical plane, but the camera is positioned slightly to one side rather than directly perpendicular to that plane. As the ball moves horizontally, it also moves further away from the camera.

Before reading on, predict what should happen to the apparent spacing between the ball’s positions in consecutive frames.

If the ball travels equal horizontal distances in equal time intervals, you might expect equal pixel spacing. That prediction is correct only if the image scale remains constant.

As the ball moves further away, it occupies fewer pixels. The same real-world distance therefore produces a smaller distance on the image.

What does the camera actually measure?

A camera does not directly measure metres, horizontal displacement, or vertical displacement.

It records the position of light on an image sensor. That position is expressed in pixels.

To convert pixels into metres, you need a calibration reference, such as a ruler of known length. If a \(1.00\text{ m}\) ruler occupies 500 pixels, you can establish an image scale of:

\[ S=\frac{500}{1.00} =500\text{ pixels m}^{-1} \]

You might then assume that every 500 pixels represents one metre.

But there is a catch.

That conversion is only valid for objects at the same camera depth as the calibration reference, assuming the same viewing geometry.

Think about watching a football match from the grandstand. Players close to you appear larger than players at the opposite end of the field, even if they are physically the same height.

Your eyes are receiving light from objects at different distances, and their apparent sizes change.

A camera produces the same basic perspective effect. The difference is that your video analysis software does not automatically compensate for it.

The analogy is useful for understanding apparent size, but it is not a complete model of camera projection. Actual image coordinates depend on both an object’s position and its depth relative to the camera.

02How perspective changes the measured coordinates

To understand the distortion properly, establish three coordinate directions:

• \(x\): horizontal position, positive to the right.
• \(y\): vertical position, positive upwards.
• \(z\): depth, positive away from the camera.

Let \(D\) represent the camera depth of the projectile, measured along the camera’s optical axis.

If the projectile stays in a plane perpendicular to the optical axis, its depth remains approximately constant. However, if the camera views the projectile’s plane obliquely, the projectile’s depth changes as it travels along the plane.

This is where the problem begins.

The relationship between distance and apparent size

A simplified pinhole camera model describes how an object’s position appears on an image.

For a point with horizontal and vertical coordinates \(X\) and \(Y\), its image coordinates are proportional to:

\[ u=\frac{fX}{D} \]

\[ v=\frac{fY}{D} \]

Here:

• \(u\) is the horizontal image coordinate, in pixels.
• \(v\) is the vertical image coordinate, in pixels.
• \(f\) is the effective focal length expressed in pixel units.
• \(X\) and \(Y\) are the object’s transverse coordinates relative to the optical axis, in metres.
• \(D\) is its camera depth, in metres.

The important feature is the denominator.

As \(D\) increases, the image coordinates become smaller for the same physical displacement from the optical axis.

Suppose your calibration ruler is positioned at a camera depth of \(D_0\).

Its pixel-to-metre scale is:

\[ S_0=\frac{f}{D_0} \]

At another depth \(D\), the scale becomes:

\[ S(D)=\frac{f}{D} \]

Combining these relationships gives:

\[ S(D)=S_0\frac{D_0}{D} \]

This is the key relationship.

If you continue using the original calibration scale \(S_0\), the position you record is:

\[ x_{\text{app}}=x_{\text{true}}\frac{D_0}{D} \]

Similarly, for the vertical coordinate:

\[ y_{\text{app}}=y_{\text{true}}\frac{D_0}{D} \]

These equations assume a fixed camera, negligible lens distortion, and that any constant angular foreshortening has already been accounted for.

The subscript \(\text{app}\) means apparent: the position calculated from the image using the original calibration.

The consequences are straightforward:

Notice that both horizontal and vertical coordinates can be affected.

That is why parallax can distort the apparent shape of the entire trajectory rather than simply making the projectile’s range look slightly shorter.

Why the trajectory can appear to have acceleration

Suppose a projectile travels horizontally at a constant velocity.

Its actual horizontal position follows:

\[ x_{\text{true}}=v_xt \]

where \(v_x\) is its constant horizontal velocity and \(t\) is elapsed time.

If the camera depth increases during flight, the recorded coordinate becomes:

\[ x_{\text{app}}= v_xt\frac{D_0}{D(t)} \]

The horizontal position is now affected by two changing quantities:

• The projectile’s actual horizontal position increases with time.
• The apparent image scale decreases as the projectile moves further away.

Consequently, the measured horizontal displacement in each successive time interval can become smaller.

The camera may therefore suggest horizontal deceleration even though the actual horizontal acceleration is zero.

The same issue affects vertical measurements. A changing image scale can make the vertical displacement appear smaller or larger than it really is, affecting an experimental estimate of gravitational acceleration.

05An important distinction: Parallax versus ordinary foreshortening

Not every incorrectly positioned camera produces changing apparent acceleration.

Consider a camera viewing a projectile’s motion plane at an angle. If the camera is sufficiently far away, the depth variation across the plane may be negligible.

In that situation, the horizontal displacement can be approximately described by:

\[ x_{\text{image}}=x_{\text{true}}\cos\phi \]

where \(\phi\) is the angle between the camera’s viewing direction and the perpendicular to the projectile’s motion plane.

For example, if:

\[ \phi=30^\circ \]

and the actual horizontal displacement is \(2.00\text{ m}\), then:

\[ x_{\text{image}}=2.00\cos30^\circ \]

\[ x_{\text{image}}=1.73\text{ m} \]

The recorded distance is approximately \(13.4\%\) smaller.

However, if the foreshortening factor remains constant, equal actual displacements still produce equal apparent displacements.

The camera has compressed the measurement, but it has not introduced changing scale throughout the flight.

This is different from parallax caused by changing depth.

The important distinction is whether the scale factor changes as the projectile moves.

A constant scale error can make a trajectory too large or too small. A changing scale can alter the apparent curvature and make the projectile appear to accelerate differently.

06How parallax affects experimental uncertainty

When analysing projectile videos, you need to distinguish between random uncertainty and systematic measurement error.

They are not interchangeable.

Random uncertainty

Random uncertainty occurs when repeated position measurements vary slightly.

For example, the projectile might be blurred, making its exact centre difficult to identify. One measurement might place the centre at 240 pixels, while another places it at 242 pixels.

If your image scale is:

\[ S=200\text{ pixels m}^{-1} \]

and your position uncertainty is approximately \(\pm2\) pixels, the corresponding positional uncertainty is:

\[ \delta x=\frac{2}{200} \]

\[ \boxed{\delta x=\pm0.010\text{ m}} \]

This uncertainty describes how precisely you can identify the projectile’s position, assuming the calibration scale is correct.

Systematic error

Parallax creates a different problem.

Suppose the projectile is actually \(1.20\text{ m}\) from its horizontal reference position, but the camera records \(1.091\text{ m}\).

The difference is:

\[ 1.20-1.091=0.109\text{ m} \]

That is a substantial error compared with the assumed random positional uncertainty of \(0.010\text{ m}\).

Repeating the same measurement will not automatically remove this bias.

You might track the projectile ten times and obtain very similar coordinates. Your results could be highly consistent but still systematically inaccurate.

Why taking more measurements is not always enough

A common experimental response is to collect more data and calculate an average.

That is useful when random uncertainty is important.

However, averaging measurements taken using the same incorrect camera geometry does not remove parallax.

If the scale is systematically wrong, every measurement is affected by the same underlying problem.

You should therefore consider several sources of uncertainty:

• Uncertainty in identifying the projectile’s centre.
• Uncertainty in the ruler’s length and calibration scale.
• Uncertainty in the recorded frame times.
• Changes in the projectile’s depth relative to the camera.
• Camera alignment and lens distortion.

Your reported uncertainty should account for relevant measurement limitations. A small spread between repeated readings does not prove that the measurements are accurate.

07How can you identify and reduce parallax?

The aim is to make the camera’s measurements represent the projectile’s actual motion as closely as possible.

There are several useful improvements.

1. Position the camera perpendicular to the motion plane

The camera should face the projectile’s vertical plane directly.

Its optical axis should be perpendicular to that plane, with the image sensor approximately parallel to it.

This helps ensure that the projectile remains at approximately the same camera depth throughout its flight.

It also reduces horizontal foreshortening.

2. Place the calibration ruler in the projectile’s plane

The calibration reference should occupy the same plane as the projectile’s motion.

If the ruler is placed in front of or behind the projectile, its apparent scale will differ.

A calibration grid can provide an additional check. If its squares appear to change size or become noticeably distorted across the region of interest, the camera geometry may need attention.

3. Keep the camera stationary

Mount the camera securely on a tripod.

Avoid panning, zooming, or changing the camera position during recording.

Camera movement creates another source of apparent motion, making it harder to distinguish the projectile’s displacement from changes in the reference frame.

4. Use two cameras if depth movement is significant

A single camera records a two-dimensional projection of the projectile’s motion.

It cannot independently establish the projectile’s depth.

A second synchronised camera positioned at a different viewing angle provides additional information about the projectile’s three-dimensional position.

With appropriate calibration, the two recordings can be used to reconstruct the projectile’s spatial trajectory.

5. Check whether the projectile’s apparent size changes

If you are recording a spherical projectile with a known, constant diameter, its apparent image diameter can provide an additional indication of changing depth.

For a fixed camera with unchanged zoom, a sphere further away occupies fewer pixels.

This can help identify a perspective effect even when the camera position is not ideal.

However, this method relies on the projectile remaining spherical, having a clearly identifiable boundary, and being recorded without significant motion blur or changes in camera zoom.

It is a useful diagnostic, not a universal replacement for correct camera positioning.

For a more detailed explanation of the physical setup and calibration process, see How to Choose a Camera Position for Projectile Motion.

08Questions and solutions

The following questions apply parallax to projectile-video analysis, measurement uncertainty, and experimental design. The later questions require you to distinguish apparent motion from actual motion.

Question 1

A camera is calibrated using a \(1.00\text{ m}\) ruler positioned \(2.40\text{ m}\) from the camera.

A projectile is recorded at a depth of \(3.00\text{ m}\). Its actual horizontal displacement from the optical axis is \(0.75\text{ m}\).

Assume a pinhole camera model and negligible lens distortion.

Calculate:

(a) The apparent horizontal displacement when the original calibration is used.

(b) The percentage error in the recorded displacement.

Solution 1

The apparent horizontal displacement is \(0.600\text{ m}\), representing an underestimate of \(20.0\%\).

(a) Calculate the apparent displacement.

Use:

\[ x_{\text{app}}=x_{\text{true}}\frac{D_0}{D} \]

Substitute:

\[ x_{\text{app}}= 0.75\times\frac{2.40}{3.00} \]

\[ \boxed{x_{\text{app}}=0.600\text{ m}} \]

The projectile appears closer to the optical axis than it actually is because it is further from the camera than the calibration ruler.

(b) Calculate the percentage error.

\[ \text{Error}= \frac{x_{\text{app}}-x_{\text{true}}} {x_{\text{true}}}\times100 \]

\[ =\frac{0.600-0.750}{0.750}\times100 \]

\[ \boxed{\text{Error}=-20.0\%} \]

The negative value indicates an underestimate.

The camera has not changed the projectile’s actual position. It has changed how that position is represented in the recorded image.

Question 2

A projectile is filmed using a fixed camera calibrated at a depth of \(5.00\text{ m}\).

The projectile moves horizontally at a constant actual velocity. Its actual positions and camera depths are given below.

Assume any fixed angular foreshortening has already been corrected.

(a) Calculate the apparent horizontal position at each time.

(b) Calculate the average apparent horizontal velocity during each time interval.

(c) Calculate the apparent horizontal acceleration using the three equally spaced measurements.

(d) Explain why the calculated acceleration does not demonstrate that air resistance is acting.

Solution 2

The apparent horizontal acceleration is approximately \(-1.11\text{ m s}^{-2}\), even though the projectile’s actual horizontal acceleration is zero.

(a) Calculate the apparent horizontal positions.

Use:

\[ x_{\text{app}}=x_{\text{true}}\frac{D_0}{D} \]

At \(t=0.00\text{ s}\):

\[ x_{\text{app}}=0 \]

At \(t=0.25\text{ s}\):

\[ x_{\text{app}}= 0.800\times\frac{5.00}{5.25} \]

\[ x_{\text{app}}=0.762\text{ m} \]

At \(t=0.50\text{ s}\):

\[ x_{\text{app}}= 1.600\times\frac{5.00}{5.50} \]

\[ x_{\text{app}}=1.455\text{ m} \]

(b) Calculate the apparent horizontal velocities.

For the first interval:

\[ v_{x1}=\frac{0.762-0}{0.25} \]

\[ v_{x1}=3.05\text{ m s}^{-1} \]

For the second interval, using unrounded coordinates:

\[ v_{x2}= \frac{1.454545-0.761905}{0.25} \]

\[ v_{x2}=2.77\text{ m s}^{-1} \]

The measured horizontal velocity appears to decrease.

(c) Calculate the apparent acceleration.

For three equally spaced measurements:

\[ a_x=\frac{x_3-2x_2+x_1}{(\Delta t)^2} \]

Substitute:

\[ a_x= \frac{1.454545-2(0.761905)+0} {(0.25)^2} \]

\[ \boxed{a_x=-1.11\text{ m s}^{-2}} \]

(d) Explain the apparent acceleration.

The projectile’s actual horizontal velocity is constant because the actual positions increase by \(0.800\text{ m}\) during each \(0.25\text{ s}\) interval.

The camera, however, records progressively smaller apparent displacements because the projectile moves further away.

This produces an apparent negative acceleration.

Air resistance could cause actual horizontal deceleration, but these measurements do not establish that it is responsible. The changing camera depth provides another explanation for the observed pattern.

Question 3

A camera is calibrated using a reference plane \(3.00\text{ m}\) from its lens.

A projectile travels through three equally spaced positions during equal time intervals. Its actual horizontal positions are:

\[ 0.00\text{ m},\quad0.50\text{ m},\quad1.00\text{ m} \]

Consider two experimental arrangements.

Arrangement A: The projectile remains at a constant camera depth of \(3.60\text{ m}\) throughout its motion.

Arrangement B: The projectile’s camera depths at the three positions are \(3.60\text{ m}\), \(3.90\text{ m}\), and \(4.20\text{ m}\), respectively.

The time interval between measurements is \(0.20\text{ s}\).

(a) Calculate the three apparent horizontal positions for Arrangement A.

(b) Calculate the three apparent horizontal positions for Arrangement B.

(c) Determine the apparent horizontal acceleration in each arrangement.

(d) Explain why Arrangement A produces a scale error without producing parallax-induced horizontal acceleration, whereas Arrangement B produces both effects.

Solution 3

Arrangement A produces zero apparent horizontal acceleration, while Arrangement B produces an apparent acceleration of approximately \(-1.37\text{ m s}^{-2}\).

(a) Calculate the apparent positions for Arrangement A.

The depth remains constant at \(3.60\text{ m}\).

The scale factor is:

\[ \frac{D_0}{D}=\frac{3.00}{3.60} =0.8333 \]

Apply this factor to each actual position:

\[ x_1=0\times0.8333=0 \]

\[ x_2=0.50\times0.8333=0.4167\text{ m} \]

\[ x_3=1.00\times0.8333=0.8333\text{ m} \]

(b) Calculate the apparent positions for Arrangement B.

The camera depth changes at each position.

First position:

\[ x_1=0 \]

Second position:

\[ x_2=0.50\times\frac{3.00}{3.90} \]

\[ x_2=0.3846\text{ m} \]

Third position:

\[ x_3=1.00\times\frac{3.00}{4.20} \]

\[ x_3=0.7143\text{ m} \]

(c) Calculate the apparent acceleration.

For Arrangement A:

\[ a_x= \frac{0.8333-2(0.4167)+0}{(0.20)^2} \]

\[ \boxed{a_x=0\text{ m s}^{-2}} \]

For Arrangement B:

\[ a_x= \frac{0.7143-2(0.3846)+0}{(0.20)^2} \]

\[ \boxed{a_x\approx-1.37\text{ m s}^{-2}} \]

(d) Explain the difference.

In Arrangement A, the projectile remains at a constant depth. Its recorded positions are all multiplied by the same scale factor.

The measurements are too small, but the equal actual horizontal displacements remain equal in the recording.

In Arrangement B, the projectile moves progressively further away. Each position is multiplied by a different scale factor.

The recorded horizontal displacements therefore become progressively smaller, producing an apparent negative acceleration.

The important distinction is that a constant scale error changes the size of the measurements, whereas a changing scale can change the apparent motion itself.

Question 4

A student analyses the same video frame five times to determine the projectile’s horizontal position.

The measurements are:

The expected actual horizontal position from the independently established projectile model is \(1.200\text{ m}\).

(a) Calculate the mean measured position.

(b) Calculate the range of the measurements and use half the range as a rough estimate of the measurement spread.

(c) Explain whether the measurements demonstrate good precision, good accuracy, both, or neither.

(d) Explain why collecting more measurements using the same camera arrangement might not resolve the disagreement. Recommend two experimental improvements.

Solution 4

The mean measured position is \(1.089\text{ m}\), with a half-range of \(0.003\text{ m}\). The results are closely grouped but systematically underestimate the actual position.

(a) Calculate the mean.

\[ \bar{x}= \frac{1.087+1.091+1.089+1.092+1.086}{5} \]

\[ \boxed{\bar{x}=1.089\text{ m}} \]

(b) Calculate the measurement spread.

The range is:

\[ \text{Range}=1.092-1.086 \]

\[ \text{Range}=0.006\text{ m} \]

Half the range is:

\[ \frac{0.006}{2}=0.003\text{ m} \]

The approximate repeatability spread is therefore:

\[ \boxed{\pm0.003\text{ m}} \]

This is a simple estimate of the spread, not a complete uncertainty analysis.

(c) Assess precision and accuracy.

The measurements show good precision because they are closely grouped around their mean.

However, their mean differs from the expected position by:

\[ 1.200-1.089=0.111\text{ m} \]

The result is therefore not accurate relative to the stated reference value.

The student has obtained consistent measurements, but consistency does not guarantee accuracy.

(d) Explain why averaging is insufficient and suggest improvements.

Taking additional measurements may reduce random uncertainty, but it cannot automatically remove systematic errors caused by incorrect camera geometry.

If every measurement is affected by the same parallax error, the mean will remain biased.

Two suitable improvements are:

• Reposition the camera so that its optical axis is perpendicular to the projectile’s motion plane, and place the calibration ruler in that plane.
• Introduce a second synchronised camera at a different viewing angle to measure depth changes and identify out-of-plane motion.

The student should also check whether the projectile’s apparent size changes during flight.

These improvements address the underlying geometry rather than simply collecting more readings.

Question 5

A student records a spherical projectile using a fixed camera.

The camera is calibrated at the projectile’s initial depth. The ball has a constant physical diameter of \(40.0\text{ mm}\).

The student records the following hypothetical data:

Assume:

• The camera follows an ideal pinhole model.
• The camera is stationary, with fixed zoom and negligible lens distortion.
• The projectile is a rigid sphere, and its apparent diameter can be measured accurately.
• The camera’s original calibration is correct at \(t=0\).
• The projectile is subject to negligible air resistance.

(a) Determine the relative image-scale factor at \(t=0.20\text{ s}\) and \(t=0.40\text{ s}\), using the apparent diameter of the ball.

(b) Correct the horizontal and vertical position measurements.

(c) Calculate the horizontal velocity during each time interval.

(d) Estimate the gravitational acceleration using the corrected vertical positions.

(e) Explain why the uncorrected measurements might lead a student to conclude that air resistance is significant. Explain why the corrected measurements support a different conclusion.

Solution 5

After correcting for the changing image scale, the projectile has a constant horizontal velocity of \(3.00\text{ m s}^{-1}\) and a vertical acceleration of approximately \(9.82\text{ m s}^{-2}\) downwards.

The apparent horizontal deceleration and reduced gravitational acceleration are therefore explained by perspective distortion rather than a failure of the ideal projectile model.

(a) Calculate the relative image-scale factors.

For a sphere of constant physical diameter, its apparent image diameter is inversely proportional to camera depth.

Therefore, the relative scale factor is:

\[ k=\frac{d_{\text{image}}}{d_{\text{initial}}} \]

where \(d_{\text{image}}\) is the apparent diameter at a particular time and \(d_{\text{initial}}\) is the initial apparent diameter.

At \(t=0.20\text{ s}\):

\[ k_1=\frac{19.0}{20.0} \]

\[ \boxed{k_1=0.950} \]

At \(t=0.40\text{ s}\):

\[ k_2=\frac{18.0}{20.0} \]

\[ \boxed{k_2=0.900} \]

The projectile’s apparent image scale has decreased as it moves further away.

(b) Correct the position measurements.

The apparent coordinates are related to the actual coordinates by:

\[ x_{\text{app}}=kx_{\text{true}} \]

\[ y_{\text{app}}=ky_{\text{true}} \]

Therefore:

\[ x_{\text{true}}=\frac{x_{\text{app}}}{k} \]

\[ y_{\text{true}}=\frac{y_{\text{app}}}{k} \]

At \(t=0.20\text{ s}\):

\[ x_{\text{true}}=\frac{0.570}{0.950} \]

\[ x_{\text{true}}=0.600\text{ m} \]

And:

\[ y_{\text{true}}=\frac{0.954}{0.950} \]

\[ y_{\text{true}}=1.004\text{ m} \]

At \(t=0.40\text{ s}\):

\[ x_{\text{true}}=\frac{1.080}{0.900} \]

\[ x_{\text{true}}=1.200\text{ m} \]

And:

\[ y_{\text{true}}=\frac{0.374}{0.900} \]

\[ y_{\text{true}}=0.416\text{ m} \]

The corrected dataset is:

(c) Calculate the horizontal velocities.

For the first interval:

\[ v_{x1}=\frac{0.600-0}{0.20} \]

\[ v_{x1}=3.00\text{ m s}^{-1} \]

For the second interval:

\[ v_{x2}=\frac{1.200-0.600}{0.20} \]

\[ v_{x2}=3.00\text{ m s}^{-1} \]

Therefore:

\[ \boxed{v_x=3.00\text{ m s}^{-1}} \]

The corrected horizontal velocity is constant, as expected when air resistance is negligible.

(d) Calculate the vertical acceleration.

Use the three-position equation:

\[ g=-\frac{y_3-2y_2+y_1}{(\Delta t)^2} \]

Substitute the corrected values:

\[ g= -\frac{0.41556-2(1.00421)+1.200} {(0.20)^2} \]

\[ g=\frac{0.39286}{0.0400} \]

\[ \boxed{g\approx9.82\text{ m s}^{-2}} \]

The result is very close to the expected gravitational acceleration of approximately \(9.80\text{ m s}^{-2}\).

The small difference is consistent with rounding in the illustrative measurements.

(e) Explain the misleading original result.

If the student had used the uncorrected coordinates, the horizontal velocities would have appeared to be:

First interval:

\[ v_{x1}=\frac{0.570}{0.20} =2.85\text{ m s}^{-1} \]

Second interval:

\[ v_{x2}=\frac{1.080-0.570}{0.20} =2.55\text{ m s}^{-1} \]

The apparent horizontal velocity decreases.

Similarly, using the uncorrected vertical positions gives:

\[ g_{\text{app}}= -\frac{0.374-2(0.954)+1.200}{0.0400} \]

\[ g_{\text{app}}=8.35\text{ m s}^{-2} \]

These results might tempt the student to conclude that air resistance is reducing the horizontal velocity and changing the measured gravitational acceleration.

However, the apparent diameter of the spherical projectile provides evidence that its camera depth is changing.

Correcting for this changing image scale restores constant horizontal velocity and a vertical acceleration close to the expected value.

There is an important limitation: this correction relies on the sphere having a constant physical diameter and the camera maintaining a stable optical configuration. If the projectile were an irregular object that rotated during flight, its apparent dimensions might change for reasons unrelated to depth.

The useful lesson is that a camera measurement must be checked against the geometry of the recording before it is used to make claims about the forces acting on a projectile.

09What should you investigate next?

Once parallax has been controlled or corrected, your recorded positions become much more useful for testing projectile motion.

You can then:

• Plot horizontal position against time to investigate whether horizontal velocity is constant.
• Plot vertical position against time to investigate whether vertical acceleration is constant.
• Use corrected vertical measurements to estimate gravitational acceleration.
• Compare experimental results with theoretical projectile predictions.

For the next stage of video analysis, see How to Find g from a Projectile Video: HSC Physics Guide.

The key experimental decision is to establish whether an unexpected trajectory represents actual projectile behaviour or an error introduced when the camera converts three-dimensional motion into two-dimensional measurements.