How to Find the Time Origin in Projectile Video Analysis

A projectile-video analysis can produce a convincing parabolic trajectory and still give you the wrong launch velocity or flight time. One common reason is that the stopwatch starts at the wrong moment.

Set \(t=0\) at the instant the projectile loses contact with the launcher and begins free flight. This is the time origin for analysing motion from launch.

There is one complication: the projectile might leave the launcher between two video frames. You therefore need to estimate the release time from the frames surrounding that event. If the release is not visible, you can choose another time origin, but you must distinguish it from the actual launch time.

01Choose the physical event that starts the clock

Imagine recording a ball launched from a projectile launcher. Your camera starts recording two seconds before the ball moves.

Should the ball’s launch time be two seconds?

No. The camera’s recording time and the projectile’s flight time are different measurements.

Think of a race. A camera operator might start recording well before the starting gun, but the runners’ race times begin at the starting signal. Similarly, projectile time begins when the ball becomes free of the launching mechanism, not when the camera starts recording.

The analogy has one limitation: a starting signal might be recorded at a precise instant, whereas a video usually captures the release somewhere between successive frames.

For a typical projectile launcher, define release as the instant the ball is no longer in contact with the launcher and is no longer being pushed by it.

This matters because the projectile-motion model assumes that the launching force has stopped acting. Once the ball is in free flight, we can model its motion using gravity, assuming air resistance is negligible.

02Set up the video so you can identify release

The easiest way to find the time origin is to make sure the release event is visible in the first place.

Consider a standard school projectile-motion experiment.

The apparatus consists of:

• A projectile launcher, with its exit point visible in the recording.
• A ball launched into a vertical plane.
• A stationary camera recording from the side.
• A calibration ruler positioned in the plane of motion.
• A recording that begins before the projectile is launched.

You want to see the ball while it is still in contact with the launcher and immediately after it becomes airborne.

A suitable camera position also helps you track the projectile’s centre accurately. For the full camera arrangement, see How to Choose a Camera Position for Projectile Motion.

Record the actual capture frame rate

Suppose your camera records at 120 frames per second (fps).

The time between consecutive frames is:

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

where:

• \(\Delta t\) is the time between frames, in seconds.
• \(f\) is the recording frame rate, in frames per second.

Therefore:

\[
\Delta t=\frac{1}{120}
=0.00833\text{ s}
\]

Each frame is approximately \(0.00833\text{ s}\) apart.

For slow-motion recordings, use the actual capture frame rate, not simply the playback frame rate. A video recorded at 120 fps might play back at 30 fps, but its original frames were captured at 120 fps.

If the recording uses a variable frame rate, use the individual frame timestamps rather than assuming every frame has the same time interval.

03Find the release time between two frames

Now suppose you inspect the video frame by frame.

You find:

• Frame 240: the ball is still touching the launcher.
• Frame 241: the ball is completely separated from the launcher.

When did release occur?

It happened somewhere between the two frame timestamps.

If we assume the release was equally likely to occur anywhere within that interval, a reasonable estimate is its midpoint.

Let \(n_0\) represent the estimated release frame.

Then:

\[
n_0=\frac{240+241}{2}
\]

\[
n_0=240.5
\]

Notice that the estimated release frame is not an integer. That’s perfectly acceptable. The physical event does not need to coincide with a recorded frame.

How uncertain is this estimate?

The release could have occurred anywhere within the interval.

At 120 fps, the interval is \(0.00833\text{ s}\). Using its midpoint gives a maximum timing difference of half that interval:

\[
\delta t=\frac{1}{2f}
\]

\[
\delta t=\frac{1}{240}
\]

\[
\delta t=0.00417\text{ s}
\]

Therefore, the estimated release time has an uncertainty of approximately:

\[
\boxed{\pm0.00417\text{ s}}
\]

This is the uncertainty associated with locating release between those two frames. It does not include other uncertainties, such as motion blur or an incorrectly specified frame rate.

If the release is obscured for several frames, the uncertainty will be larger.

04Convert frame numbers into time after release

Once you have estimated the release frame, calculating elapsed time is straightforward.

Use:

\[
\boxed{t=\frac{n-n_0}{f}}
\]

where:

• \(t\) is elapsed time after release, in seconds.
• \(n\) is the frame number being analysed.
• \(n_0\) is the estimated release frame.
• \(f\) is the actual capture frame rate, in fps.

The important part is subtracting the release frame before dividing by the frame rate.

05What should your projectile data look like?

Once the time origin is correct, you can investigate whether the measured positions agree with the projectile-motion model.

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

For an ideal projectile:

\[
x=x_0+u_xt
\]

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

where:

• \(x_0\) and \(y_0\) are the initial coordinates at release, in metres.
• \(u_x\) and \(u_y\) are the initial horizontal and vertical velocity components, in \(\mathrm{m\,s^{-1}}\).
• \(t\) is time after release, in seconds.
• \(g=9.80\text{ m\,s}^{-2}\) is the magnitude of gravitational acceleration.

The horizontal position changes linearly with time, while the vertical position follows a quadratic relationship.

An illustrative projectile dataset

Suppose a hypothetical experiment uses the following conditions:

• Recording frame rate: \(120\text{ fps}\).
• Estimated release frame: \(240.5\).
• Initial horizontal position: \(0.00\text{ m}\).
• Initial vertical position: \(1.20\text{ m}\).
• Initial horizontal velocity: \(3.00\text{ m\,s}^{-1}\).
• Initial vertical velocity: \(2.40\text{ m\,s}^{-1}\).
• Gravitational acceleration: \(9.80\text{ m\,s}^{-2}\) downwards.

The following is a constructed illustrative dataset, not experimental results.

Notice something important: the first recorded free-flight position has a time of \(0.0042\text{ s}\), not zero.

That’s correct.

The estimated release happened at frame 240.5, but the first fully separated frame was frame 241.

You should not force the first measured position to have a time of zero just because it is the first row in your table.

Doing so would change the reference point for your analysis.

06What if the release is not visible?

Sometimes the camera only captures the projectile after it has left the launcher.

Perhaps the launcher is outside the camera’s field of view. Perhaps another object blocks the release point.

You can still analyse the projectile, but you need to be honest about what your data can establish.

There are two useful approaches.

Option 1: Use the first reliable free-flight frame

You can define the first reliably tracked frame as your new time origin.

For example, if the first frame you can confidently track is frame 300, define:

\[
t’=0\qquad\text{at frame 300}
\]

Then:

\[
t’=\frac{n-300}{f}
\]

This gives you a perfectly valid time coordinate for analysing the recorded free-flight motion.

However, it is not necessarily the time since launch.

Your initial position and velocity now refer to the projectile at frame 300, rather than at the instant it left the launcher.

If you want the actual launch velocity or total flight time, you need to determine the missing time interval.

Option 2: Extrapolate backwards using the horizontal motion

There is another possibility.

Suppose the launcher itself is outside the recording, but you know the exact horizontal coordinate of its exit point.

If horizontal acceleration is negligible, the projectile travels at approximately constant horizontal velocity:

\[
x=x_0+u_xt
\]

You can use measured horizontal positions to estimate when the projectile crossed the known release position.

This is called extrapolation.

However, it depends on the assumption that horizontal velocity remained constant throughout free flight. It is not as direct as observing the release itself.

We will use this method in a practice question below.

How would you improve an experiment that misses the release?

Imagine an initial recording that shows only the airborne ball.

You can track its position accurately, but you have no way to distinguish between a projectile released three frames before the first measurement and one released ten frames before it.

A better experimental design is to:

1. Widen the camera’s field of view to include the launcher exit.
2. Begin recording before the projectile is launched.
3. Identify the last frame showing contact and the first frame showing separation.
4. Estimate the release time from that interval.

If the release point genuinely cannot be filmed, a synchronised timing marker or sensor at the exit plane can provide an independent reference. Any delay between the sensor event and actual release must be considered.

The key lesson is that a better curve fit cannot replace a missing timing reference.

07Changing the time origin changes the initial conditions

Here is a subtle point that can catch out otherwise strong students.

Choosing a different time origin does not change the projectile’s actual motion.

It changes the values you call its initial position and initial velocity.

Suppose you originally define \(t=0\) at release.

Now imagine choosing a new time origin \(0.050\text{ s}\) later.

The projectile has already moved during that interval. Its position and velocity are therefore different.

Its acceleration, however, is still the same.

08The contrasting mistake: Using the start of the video

Consider another hypothetical recording.

The camera begins recording at frame 210, but the projectile is not released until approximately frame 240.5.

A student decides to use frame 210 as \(t=0\).

They then analyse the projectile at frame 277.

Their calculation is:

\[
t=\frac{277-210}{120}
\]

\[
t=0.5583\text{ s}
\]

But the actual time after release is:

\[
t=\frac{277-240.5}{120}
\]

\[
t=0.3042\text{ s}
\]

The student’s time is approximately \(0.2542\text{ s}\) too large.

Why might the student’s results still look reasonable?

This is where the misconception becomes interesting.

Suppose the student calculates the time taken to travel between frames 253 and 277.

Using the correct origin:

\[
\Delta t=\frac{277-253}{120}
=0.200\text{ s}
\]

Using frame 210 as the time origin:

\[
\Delta t=0.5583-0.3583
\]

\[
\Delta t=0.200\text{ s}
\]

Both methods give the same time interval.

Why?

Because subtracting the same constant time offset from every measurement does not change the difference between two times.

This means the student might calculate a perfectly reasonable average velocity between two measured positions, even though their claimed time since launch is wrong.

The consequences depend on what they calculate:

There is an additional problem if the student includes frames from before release in the projectile-motion fit.

While the ball is still in contact with the launcher, the launching force is acting on it. The projectile-motion model does not describe that part of its motion.

A single quadratic fit across both contact and free-flight motion is therefore not a valid way to test the projectile model.

09Account for timing uncertainty and model limitations

Even if you identify the correct event, your time origin is still an estimate.

It is important to distinguish uncertainty in the timing from uncertainty in the position measurements.

1. Frame-rate uncertainty

If your camera records at 60 fps, the time between frames is:

\[
\Delta t=\frac{1}{60}
=0.0167\text{ s}
\]

If release occurs somewhere between two frames, the midpoint estimate has a timing uncertainty of approximately:

\[
\delta t=\frac{0.0167}{2}
\]

\[
\delta t=\pm0.0083\text{ s}
\]

Compare this with a 120 fps recording, where the corresponding uncertainty is approximately \(\pm0.0042\text{ s}\).

A higher frame rate gives you a narrower timing interval, provided the actual frame rate is known and the release can be identified clearly.

2. Motion blur and unclear release

A projectile might be blurred across several frames.

You might also struggle to identify the exact instant it separates from a launcher.

If the event is only bracketed across four frame intervals, your midpoint estimate has an uncertainty of approximately two frame intervals in either direction.

At 120 fps:

\[
\delta t=\frac{2}{120}
\]

\[
\delta t=\pm0.0167\text{ s}
\]

Do not report an artificially precise release time just because your calculations produce several decimal places.

3. Position uncertainty is a separate issue

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

That introduces uncertainty into its measured position.

However, even perfect pixel measurements cannot tell you exactly when release occurred if the event falls between frames.

These are different sources of uncertainty:

• Position uncertainty affects the measured coordinates.
• Timing uncertainty affects the assigned time values.
• Calibration uncertainty affects the conversion between pixels and metres.
• Frame-rate uncertainty affects the conversion between frames and seconds.

Improving one does not automatically remove the others.

4. The projectile model has a limited starting point

The equations for projectile motion apply after the projectile becomes free of the launching mechanism.

They assume:

• Air resistance is negligible.
• Gravitational acceleration is approximately constant.
• The camera provides a consistent reference frame.
• The measured object is following the assumed two-dimensional trajectory.

If the ball is still being pushed by the launcher, the projectile model does not yet apply.

If air resistance is significant, the measured trajectory might also depart from the ideal equations.

A high-quality curve fit is not proof that the physical assumptions are correct. You must consider which part of the motion you have actually measured.

10A practical decision rule

When analysing a projectile video, use the following approach.

The distinction between an estimated launch time and a tracking time is particularly important when calculating total flight time.

If your first measured point occurs after release, the time from that point to landing is not the projectile’s complete flight time.

You need to account for the missing interval.

For the next stage of the analysis, see Projectile Time of Flight for Level and Uneven Landings.

11Questions and solutions

Question 1

A projectile is recorded using a camera operating at 50 fps.

Frame 89 shows the projectile still touching the launcher. Frame 90 shows it completely separated.

Assume the release occurred halfway between the two frame timestamps.

Calculate:

(a) The estimated release frame.

(b) The time between consecutive frames.

(c) The elapsed time after release at frame 104.

(d) The uncertainty in the estimated elapsed time due to locating the release between frames.

Solution 1

The estimated release frame is 89.5, and the elapsed time at frame 104 is \(0.290\pm0.010\text{ s}\).

(a) Estimated release frame

\[
n_0=\frac{89+90}{2}
\]

\[
\boxed{n_0=89.5}
\]

(b) Time between frames

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

\[
\Delta t=\frac{1}{50}
\]

\[
\boxed{\Delta t=0.020\text{ s}}
\]

(c) Elapsed time at frame 104

Use:

\[
t=\frac{n-n_0}{f}
\]

Substitute:

\[
t=\frac{104-89.5}{50}
\]

\[
t=\frac{14.5}{50}
\]

\[
\boxed{t=0.290\text{ s}}
\]

(d) Timing uncertainty

The release is estimated to within half a frame:

\[
\delta t=\frac{0.5}{50}
\]

\[
\boxed{\delta t=0.010\text{ s}}
\]

Therefore, the result is:

\[
\boxed{t=0.290\pm0.010\text{ s}}
\]

This uncertainty only accounts for the release event being located between two frames. It does not include uncertainty in the camera’s actual frame rate.

Question 2

A camera records a projectile experiment at 120 fps. Recording begins at frame 0.

The ball is still in contact with the launcher at frame 300 and is completely separated at frame 301.

A student records the projectile’s positions at frames 325 and 349.

The student incorrectly uses:

\[
t=\frac{n}{120}
\]

to calculate time after launch.

(a) Calculate the correct elapsed time after release at frames 325 and 349, using the midpoint release estimate.

(b) Calculate the time interval between the two measurements.

(c) Explain why the student’s average velocity calculated from the displacement between these two frames could still be correct, despite using the wrong time origin.

Solution 2

The correct elapsed times are approximately \(0.204\text{ s}\) and \(0.404\text{ s}\), and the interval between the measurements is \(0.200\text{ s}\).

(a) Correct elapsed times

First, estimate the release frame:

\[
n_0=\frac{300+301}{2}=300.5
\]

At frame 325:

\[
t_1=\frac{325-300.5}{120}
\]

\[
\boxed{t_1=0.2042\text{ s}}
\]

At frame 349:

\[
t_2=\frac{349-300.5}{120}
\]

\[
\boxed{t_2=0.4042\text{ s}}
\]

(b) Time interval

Subtract the two elapsed times:

\[
\Delta t=t_2-t_1
\]

\[
\Delta t=0.4042-0.2042
\]

\[
\boxed{\Delta t=0.200\text{ s}}
\]

(c) Why could the average velocity still be correct?

The student uses the camera’s recording start as the time origin instead of the projectile’s release.

Their calculated times are:

\[
t_1=\frac{325}{120}=2.7083\text{ s}
\]

\[
t_2=\frac{349}{120}=2.9083\text{ s}
\]

However, the difference is still:

\[
2.9083-2.7083=0.200\text{ s}
\]

The incorrect time origin adds the same constant offset to every measurement.

When calculating the difference between two times, that offset cancels.

Therefore, an average velocity calculated using the displacement between these frames can still be correct, provided the position measurements are accurate.

However, the student cannot correctly interpret either timestamp as time after launch.

Question 3

A projectile is recorded at 100 fps, but the camera does not capture the launcher.

The known horizontal coordinate of the launcher exit plane is \(x=0\).

The following hypothetical measurements are obtained:

Assume horizontal velocity is constant after release.

(a) Calculate the projectile’s horizontal velocity.

(b) Use the measurements to estimate the frame at which the projectile crossed the launcher exit plane.

(c) Explain why this estimate is less reliable than directly identifying release in the video.

Solution 3

The estimated horizontal velocity is \(1.00\text{ m\,s}^{-1}\), and the projectile crossed the known exit plane at approximately frame 182.

(a) Calculate horizontal velocity

Use the first and last measurements:

\[
\Delta x=0.70-0.30
\]

\[
\Delta x=0.40\text{ m}
\]

The time interval is:

\[
\Delta t=\frac{252-212}{100}
\]

\[
\Delta t=0.40\text{ s}
\]

Therefore:

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

\[
v_x=\frac{0.40}{0.40}
\]

\[
\boxed{v_x=1.00\text{ m\,s}^{-1}}
\]

(b) Estimate the release frame

At frame 212, the projectile is \(0.30\text{ m}\) beyond the exit plane.

Using constant horizontal velocity, the time required to travel this distance is:

\[
\Delta t=\frac{0.30}{1.00}
\]

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

Convert this interval into frames:

\[
\Delta n=0.30(100)
\]

\[
\Delta n=30
\]

Therefore, the estimated exit-plane crossing occurred at:

\[
n_0=212-30
\]

\[
\boxed{n_0=182}
\]

This gives an estimated launch time of frame 182.

(c) Explain the limitation

This method relies on extrapolating the measured horizontal motion backwards.

It assumes that:

• The horizontal velocity remained constant after release.
• The known \(x=0\) coordinate corresponds to the actual release plane.
• The position calibration is accurate.
• The camera’s frame timing is reliable.

If horizontal acceleration is present, the extrapolated crossing time might be wrong.

Directly observing release is preferable because it provides an actual visual reference rather than an estimate based on extending a mathematical model beyond the measured data.

Question 4

A projectile is filmed at 240 fps.

The release event is bracketed between frames 511 and 512. The landing event is bracketed between frames 703 and 704.

Assume both event times are estimated using the midpoint of their respective frame intervals.

(a) Calculate the estimated total flight time.

(b) Calculate the maximum uncertainty in the flight time using a conservative worst-case approach.

(c) Explain why the uncertainty in total flight time is not simply half a frame interval.

Solution 4

The estimated total flight time is \(0.800\text{ s}\), with a conservative timing uncertainty of approximately \(\pm0.0042\text{ s}\).

(a) Calculate the total flight time

The estimated release frame is:

\[
n_r=\frac{511+512}{2}=511.5
\]

The estimated landing frame is:

\[
n_l=\frac{703+704}{2}=703.5
\]

Therefore:

\[
T=\frac{n_l-n_r}{f}
\]

\[
T=\frac{703.5-511.5}{240}
\]

\[
T=\frac{192}{240}
\]

\[
\boxed{T=0.800\text{ s}}
\]

(b) Calculate the maximum uncertainty

Each event has an uncertainty of half a frame.

However, total flight time depends on two uncertain events:

• The release time.
• The landing time.

For a conservative worst-case calculation, these uncertainties can add.

Therefore:

\[
\delta T=\frac{0.5+0.5}{240}
\]

\[
\delta T=\frac{1}{240}
\]

\[
\boxed{\delta T=0.00417\text{ s}}
\]

The estimated flight time is:

\[
\boxed{T=0.800\pm0.004\text{ s}}
\]

(c) Explain why the uncertainties add

The shortest possible flight occurs if release happens as late as possible and landing happens as early as possible.

The longest possible flight occurs if release happens as early as possible and landing happens as late as possible.

Both event uncertainties therefore contribute to the uncertainty in the time difference.

Using only half a frame would account for just one of the two uncertain events.

Question 5

A projectile experiment is recorded at 120 fps.

The video shows:

• Frame 240: the projectile is still touching the launcher.
• Frame 241: the projectile is completely separated.
• Frame 335: the projectile is still airborne.
• Frame 336: the projectile has contacted the ground.

Three students analyse the same recording.

Student A uses the midpoint of the release interval as \(t=0\), and the midpoint of the landing interval to estimate total flight time. Their vertical-motion fit is:

\[
y=1.20+2.40t-4.90t^2
\]

They report an initial vertical velocity of \(2.40\text{ m\,s}^{-1}\).

Student B uses frame 241 as \(t’=0\). Their vertical-motion fit is:

\[
y=1.210+2.359t’-4.90t’^2
\]

They report an initial vertical velocity of \(2.359\text{ m\,s}^{-1}\), and a total flight time of \(0.7875\text{ s}\).

Student C includes frames 230-240, during which the projectile is still in contact with the launcher, in a single quadratic fit with the free-flight data. Their fit has a high coefficient of determination, \(R^2\), and produces an acceleration of approximately \(-9.8\text{ m\,s}^{-2}\).

Evaluate the three analyses.

In your answer:

(a) Calculate the estimated total flight time using Student A’s time origin.

(b) Explain the difference between Student A’s and Student B’s initial vertical velocities.

(c) Determine whether Student B’s reported flight time represents the complete flight.

(d) Explain why Student C’s result cannot automatically be accepted, despite the high \(R^2\).

(e) Calculate the conservative worst-case uncertainty in the total flight time.

Solution 5

Student A’s analysis gives the appropriate estimated release-to-landing time. Student B has used a valid alternative time origin, but has misidentified the velocity and duration measured from that origin as launch quantities. Student C has applied the projectile model to data from a period when the launching force is still acting, so their fit is not a valid test of the free-flight model.

(a) Calculate the estimated total flight time

The estimated release frame is:

\[
n_r=\frac{240+241}{2}=240.5
\]

The estimated landing frame is:

\[
n_l=\frac{335+336}{2}=335.5
\]

Therefore:

\[
T=\frac{335.5-240.5}{120}
\]

\[
T=\frac{95}{120}
\]

\[
\boxed{T=0.7917\text{ s}}
\]

This is the estimated time from release to landing.

(b) Explain the difference between the initial vertical velocities

Student A measures time from release.

Student B starts their clock at frame 241, which is approximately half a frame after the estimated release time.

The time difference between their origins is:

\[
\delta t=\frac{0.5}{120}
\]

\[
\delta t=0.004167\text{ s}
\]

During this interval, gravity reduces the upward vertical velocity.

Using:

\[
v_y=u_y-gt
\]

we obtain:

\[
v_y=2.40-9.80(0.004167)
\]

\[
v_y=2.3592\text{ m\,s}^{-1}
\]

Therefore:

\[
\boxed{v_y\approx2.359\text{ m\,s}^{-1}}
\]

This agrees with Student B’s fitted coefficient.

The important distinction is that Student B has measured the vertical velocity at frame 241, not at the instant of release.

Their result is physically reasonable, but calling it the launch velocity is incorrect.

Both students can obtain the same gravitational acceleration because changing the time origin does not change the projectile’s acceleration.

(c) Determine whether Student B’s flight time is complete

Student B measures time from frame 241 to the estimated landing frame 335.5.

Their elapsed time is:

\[
T’=\frac{335.5-241}{120}
\]

\[
T’=\frac{94.5}{120}
\]

\[
\boxed{T’=0.7875\text{ s}}
\]

This is the time from the first fully separated frame to landing.

It is not the complete time from release.

To recover the estimated total flight time, add the missing interval:

\[
T=T’+0.004167
\]

\[
T=0.7875+0.004167
\]

\[
\boxed{T=0.7917\text{ s}}
\]

Student B’s calculation is correct for their chosen time origin. Their interpretation is incorrect because they have labelled the result as total flight time.

(d) Explain why Student C’s result is not automatically valid

Student C has included frames during which the projectile is still in contact with the launcher.

During this period, the launcher can exert a force on the projectile. Its acceleration is therefore not necessarily equal to gravitational acceleration alone.

The projectile-motion equations only apply once the projectile has become free of the launching mechanism.

A high \(R^2\) does not prove that the physical model is appropriate.

It only indicates how closely the fitted curve matches the included data. A mathematical fit can look convincing even when the data combine two different stages of motion.

Student C should identify the release event using the contact frames, but exclude those frames from the free-flight trajectory fit.

(e) Calculate the worst-case timing uncertainty

There are two uncertain events:

• Release is bracketed within one frame.
• Landing is also bracketed within one frame.

Each midpoint estimate has an uncertainty of half a frame.

Using the conservative worst-case approach:

\[
\delta T=\frac{0.5+0.5}{120}
\]

\[
\delta T=\frac{1}{120}
\]

\[
\boxed{\delta T=0.00833\text{ s}}
\]

Therefore, the estimated complete flight time is:

\[
\boxed{T=0.792\pm0.008\text{ s}}
\]

This uncertainty comes from locating the two events in the video. It does not include uncertainty in the actual frame rate, image blur, or the physical identification of release and landing.

The final distinction is important: two students can analyse exactly the same projectile and obtain different initial velocities without either measuring the motion incorrectly. They may simply have chosen different time origins.

What matters is that each student identifies what their zero represents and uses it consistently.

Once the time origin is established, the next step is to fit the horizontal and vertical position data to determine the projectile’s velocity components and gravitational acceleration. Those measurements can then be used to investigate the complete flight, including the time required to reach a landing point at the same or a different height.