How to Test Whether Horizontal Projectile Speed Is Constant

A ball rolls off a laboratory bench. You film its flight and mark its horizontal position every 0.05 seconds. The marks get progressively farther apart. Does that mean the ball is speeding up horizontally?

Not necessarily. Before drawing that conclusion, you need to check whether the time intervals are equal, whether your position measurements are accurate, and whether the camera introduces distortion.

The most direct way to test whether horizontal projectile speed is constant is to measure horizontal displacement over equal time intervals. If the horizontal displacements are approximately equal, the horizontal velocity is consistent with being constant. A stronger test is to plot horizontal position against time: a straight line with a constant gradient supports the model of zero horizontal acceleration.

This is an experimental test, not a visual guess. You need to collect data, account for uncertainty, and decide whether any apparent change in speed is large enough to challenge the constant-velocity model.

## 1. What exactly are you testing?

For an ideal projectile, gravity acts vertically downwards. There is no horizontal force once the projectile has left the launcher (assuming air resistance is negligible).

Newton's second law gives:

\[
\sum F_x=ma_x
\]

If the net horizontal force is zero, then:

\[
a_x=0
\]

Therefore, horizontal velocity remains constant:

\[
v_x=u_x=\text{constant}
\]

Here:

- \(v_x\) is horizontal velocity at a particular time, in \(\text{m s}^{-1}\).
- \(u_x\) is the initial horizontal velocity, in \(\text{m s}^{-1}\).
- \(a_x\) is horizontal acceleration, in \(\text{m s}^{-2}\).

This is the model you are testing. It predicts that horizontal position changes uniformly with time:

\[
x=x_0+u_xt
\]

where \(x_0\) is the initial horizontal position and \(t\) is elapsed time.

Notice that the model does not predict that the projectile's total speed remains constant. Its vertical velocity changes because gravity accelerates it downwards. Only the horizontal component is predicted to remain constant.

[IMAGE: A labelled side-view illustration of a ball leaving a horizontal launcher at the edge of a laboratory bench. Show the curved projectile trajectory, equally spaced horizontal position markers at equal time intervals, a horizontal x-axis, a downward y-axis, and gravity acting vertically downwards. Include a small inset comparing constant horizontal velocity with increasing downward velocity. Keep the physical apparatus and projectile path recognisable.]

## 2. Collecting data to test horizontal velocity

There are several ways to measure horizontal motion. For an HSC investigation, video analysis is particularly useful because it lets you estimate position at multiple points throughout the flight.

### Using video analysis

Set up a camera perpendicular to the plane of motion. Position a calibration ruler or metre rule in the same plane as the projectile's path.

Record the projectile moving across the field of view. Then identify its horizontal position at regular time intervals using the video frames.

For example, if the recording runs at 50 frames per second, consecutive frames are separated by:

\[
\Delta t=\frac{1}{50}=0.020\text{ s}
\]

You could record every fifth frame to obtain measurements separated by \(0.100\text{ s}\).

The important conditions are:

- Keep the camera stationary.
- Place the calibration scale in the plane of the projectile's motion.
- Use equal time intervals between measurements.
- Keep the projectile visible throughout the selected section of its flight.
- Use the same reference point on the projectile for every position measurement.

A camera positioned at an angle can make a straight horizontal displacement appear shorter or longer than it really is. Parallax can also introduce systematic errors if the ruler and projectile are at different distances from the camera.

For more detail on selecting a suitable viewpoint, 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/).

[IMAGE: An annotated experimental setup viewed side-on, showing a fixed camera on a tripod perpendicular to the projectile's plane of motion, a metre rule positioned alongside the flight path, a ball launcher on a bench, and the projectile crossing the camera's field of view. Use sight lines to show why the camera, calibration ruler, and projectile must be aligned in the same plane.]

### What should you record?

Consider this hypothetical dataset for a ball launched horizontally. The initial horizontal position is defined as \(x=0\).

The following measurements are illustrative, not results from an actual experiment.

| Time, \(t\) (s) | Horizontal position, \(x\) (m) |
|---:|---:|
| 0.00 | 0.000 |
| 0.10 | 0.202 |
| 0.20 | 0.401 |
| 0.30 | 0.603 |
| 0.40 | 0.800 |
| 0.50 | 1.002 |

Look at the changes in position. They are approximately \(0.20\text{ m}\) for every \(0.10\text{ s}\).

That is the pattern expected for constant horizontal velocity.

Calculate the average horizontal velocity over each interval using:

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

For the first interval:

\[
v_x=\frac{0.202-0.000}{0.10-0.00}
=2.02\text{ m s}^{-1}
\]

For the second interval:

\[
v_x=\frac{0.401-0.202}{0.20-0.10}
=1.99\text{ m s}^{-1}
\]

The remaining intervals give:

| Time interval (s) | Horizontal displacement (m) | Average horizontal velocity (\(\text{m s}^{-1}\)) |
|---|---:|---:|
| 0.00-0.10 | 0.202 | 2.02 |
| 0.10-0.20 | 0.199 | 1.99 |
| 0.20-0.30 | 0.202 | 2.02 |
| 0.30-0.40 | 0.197 | 1.97 |
| 0.40-0.50 | 0.202 | 2.02 |

The calculated velocities vary slightly, but they cluster around \(2.00\text{ m s}^{-1}\).

Does this variation prove that the projectile is accelerating horizontally? No. Small differences can arise from the resolution of the video, selecting the projectile's centre, and reading the position scale. You need to compare the variation with the measurement uncertainty before making a conclusion.

[IMAGE: Two-panel scientific illustration. Left panel: the ball's horizontal positions at six equal time intervals, with nearly equal horizontal gaps. Right panel: a graph of the illustrative horizontal position data against time, showing six points close to a straight best-fit line. Label axes with units and show the constant gradient as horizontal velocity.]

## 3. Use a position-time graph for a stronger test

Calculating the velocity for each interval is useful, but a position-time graph often makes the overall pattern easier to assess.

Plot horizontal position \(x\) on the vertical axis and time \(t\) on the horizontal axis.

For constant horizontal velocity:

\[
x=x_0+u_xt
\]

This has the same mathematical form as a straight-line equation:

\[
y=c+mx
\]

The gradient of the position-time graph represents horizontal velocity.

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

If the graph is approximately a straight line, the data support constant horizontal velocity.

If the graph curves consistently upwards or downwards, that suggests horizontal acceleration.

There is an important distinction here. Random scatter around a straight line does not automatically indicate acceleration. A systematic curve is much stronger evidence that the constant-velocity model is inadequate.

### Interpreting the graph

[IMAGE: Three side-by-side horizontal position versus time graphs with clearly labelled axes and units. First: points scattered slightly around a straight line, representing constant horizontal velocity within uncertainty. Second: a curve that becomes progressively steeper, representing increasing horizontal velocity. Third: a curve that becomes progressively less steep, representing decreasing horizontal velocity. Include straight best-fit lines only on the first graph.]

| Graph pattern | Physical interpretation |
|---|---|
| Straight line with positive gradient | Constant positive horizontal velocity |
| Straight line with zero gradient | Zero horizontal velocity |
| Curve with increasing gradient | Increasing horizontal velocity |
| Curve with decreasing gradient | Decreasing horizontal velocity |

The gradient matters more than the height of the graph. A projectile can move from \(x=0.2\text{ m}\) to \(x=1.0\text{ m}\) while maintaining exactly the same horizontal velocity. Its position changes, but its speed does not.

### Worked example: finding horizontal velocity from a graph

A position-time graph for a projectile has a best-fit straight line passing through the points \((0.10\text{ s}, 0.25\text{ m})\) and \((0.50\text{ s}, 1.05\text{ m})\).

Calculate the horizontal velocity.

**Step 1**: Identify the physical meaning of the gradient.

The gradient of a horizontal position-time graph represents horizontal velocity.

**Step 2**: Calculate the gradient using two points on the best-fit line.

\[
\begin{aligned}
v_x&=\frac{\Delta x}{\Delta t}\\
&=\frac{1.05-0.25}{0.50-0.10}\\
&=\frac{0.80}{0.40}\\
&=2.0\text{ m s}^{-1}
\end{aligned}
\]

**Step 3**: Interpret the result.

The projectile's horizontal velocity is approximately \(2.0\text{ m s}^{-1}\) throughout the measured interval, provided the straight-line model is consistent with the scatter and uncertainty of the data.

The gradient gives the velocity. The fact that the graph rises does not mean the projectile is speeding up.

04How to use uncertainty before claiming acceleration

Your measurements will never be perfectly exact. The important question is whether the observed variation is larger than the uncertainty of the measurements.

Suppose your video analysis gives horizontal positions to the nearest \(0.01\text{ m}\). The uncertainty in each position reading is approximately:

\[ \Delta x_{\text{uncertainty}}=\pm0.005\text{ m} \]

When calculating displacement between two positions, both readings contribute uncertainty.

Using a conservative maximum-uncertainty approach:

\[ \text{Uncertainty in displacement} =\pm(0.005+0.005) =\pm0.010\text{ m} \]

If the time interval is \(0.100\text{ s}\), the uncertainty in the calculated average velocity is:

\[ \Delta v_x=\frac{0.010}{0.100} =\pm0.10\text{ m s}^{-1} \]

This gives us a way to decide whether the differences between measured velocities are meaningful.

05What would data showing horizontal acceleration look like?

Now consider a contrasting hypothetical dataset.

The ball is filmed using the same time intervals, but the measured positions are:

Calculate the average velocities:

The horizontal displacements become progressively smaller despite the equal time intervals.

If the position uncertainty is again \(\pm0.005\text{ m}\), each calculated velocity has an uncertainty of approximately \(\pm0.10\text{ m s}^{-1}\).

Compare the first and last intervals:

• First interval: \(2.30\pm0.10\text{ m s}^{-1}\).
• Last interval: \(1.70\pm0.10\text{ m s}^{-1}\).

Their uncertainty ranges do not overlap.

This is evidence of decreasing horizontal velocity, not merely random measurement variation.

If we approximate the interval velocities as instantaneous velocities at their respective midpoints, the average horizontal acceleration between the first and last intervals is:

\[ a_{x,\text{avg}} =\frac{1.70-2.30}{0.45-0.05} \]

\[ a_{x,\text{avg}}=-1.5\text{ m s}^{-2} \]

The negative sign indicates acceleration opposite to the direction of horizontal travel.

What might cause this?

One possibility is air resistance. Drag acts opposite to the projectile’s motion, reducing its horizontal velocity as it travels.

However, do not jump straight to that explanation. A camera calibration problem, a changing time interval, or another horizontal force could also produce an apparent change.

The measurements tell you that the constant-velocity model is not adequately describing the observed data. Further testing is needed to identify why.

06A common misconception: constant horizontal speed does not mean constant total speed

Imagine a ball launched horizontally at \(2.00\text{ m s}^{-1}\).

Under the ideal projectile model, its horizontal velocity remains:

\[ v_x=2.00\text{ m s}^{-1} \]

But gravity continuously increases its downward velocity.

Take upwards as positive. After \(0.40\text{ s}\):

\[ v_y=u_y-gt \]

Since the ball was launched horizontally, \(u_y=0\):

\[ v_y=0-(9.8)(0.40) \]

\[ v_y=-3.92\text{ m s}^{-1} \]

The total speed is the magnitude of the velocity vector:

\[ v=\sqrt{v_x^2+v_y^2} \]

\[ v=\sqrt{(2.00)^2+(-3.92)^2} \]

\[ v=4.40\text{ m s}^{-1} \]

So the projectile’s total speed has increased from \(2.00\text{ m s}^{-1}\) to \(4.40\text{ m s}^{-1}\), even though its horizontal speed has remained constant.

This is not a contradiction. The projectile is accelerating because its velocity vector is changing vertically.

The ideal model predicts:

\[ a_x=0,\qquad a_y=-g \]

It does not predict that the total acceleration or total speed is zero or constant.

If your video shows equal horizontal spacing but increasingly large vertical displacements, that is exactly the pattern you should expect.

07What can make your experimental result misleading?

Even a carefully collected dataset can produce an incorrect conclusion if the measurement setup introduces systematic errors.

The most important issues for this investigation are:

There is an important distinction between a constant measurement error and one that changes during the experiment.

A constant error in the horizontal origin shifts every position by the same amount. It changes the graph’s intercept, but not its gradient.

A constant scale-calibration error changes the numerical value of the measured velocity. However, it does not turn a genuinely straight position-time graph into a curved one.

A scale that changes across the image is more dangerous. It can make a constant-speed projectile appear to accelerate.

Similarly, a fixed error in the camera’s frame-rate calibration changes all calculated velocities by the same proportion. It does not necessarily create an apparent change in velocity.

A variable-frame-rate recording is different. Equal numbers of frames may represent unequal elapsed times. In that situation, use the actual timestamps rather than assuming every frame interval is identical.

For more guidance on setting up the recording, see How to Choose a Camera Position for Projectile Motion.

The limitations of the ideal projectile model

The prediction \(a_x=0\) assumes negligible horizontal forces.

In a real experiment, air resistance can act against the projectile’s horizontal motion. The projectile may therefore slow down horizontally.

Wind can also change horizontal velocity. A spinning projectile can experience additional aerodynamic forces.

A small upward or downward launch angle is not, by itself, evidence of horizontal acceleration. It changes the initial vertical velocity, but the horizontal component can still remain constant if there is no horizontal force.

This distinction matters when interpreting experimental data. A mismatch with the ideal model does not automatically mean your measurements are wrong. It may mean the assumptions of the model are no longer sufficiently accurate.

08A reliable decision process

When asked to test whether horizontal projectile speed is constant, follow this sequence.

Step 1: Establish the horizontal coordinate.

Choose the launch point as \(x=0\), and take the positive horizontal direction to be the direction of travel.

Step 2: Collect position measurements.

Record horizontal position at several equally spaced time intervals. Check the camera’s actual frame timing.

Step 3: Calculate average horizontal velocities.

Use:

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

Repeat for each interval.

Step 4: Account for uncertainty.

Estimate the uncertainty in each displacement and time interval. Calculate the corresponding uncertainty in velocity.

Step 5: Inspect the position-time graph.

Look for a straight-line relationship and a constant gradient. Check whether deviations from the line are random or show a systematic pattern.

Step 6: Make a qualified conclusion.

Decide whether the results are:

• consistent with constant horizontal velocity;
• evidence of changing horizontal velocity; or
• inconclusive because the measurement uncertainty is too large.

A strong experimental conclusion might read:

The horizontal position-time graph was approximately linear, with a constant gradient over the measured interval. The calculated horizontal velocities were consistent within experimental uncertainty. Therefore, the results support the prediction of constant horizontal velocity under the assumption that air resistance is negligible.

This is much better than writing that the horizontal speed was constant simply because the projectile appeared to travel smoothly.

06Questions and solutions

Question 1

A ball is launched horizontally from a bench. Its horizontal position is recorded at intervals of \(0.10\text{ s}\).

The following illustrative measurements are obtained:

Each position measurement has an uncertainty of \(\pm0.002\text{ m}\).

1. Calculate the average horizontal velocity for each interval.
2. Estimate the uncertainty in each velocity.
3. State whether the results support constant horizontal velocity.

Solution 1

The measured horizontal velocity is \(1.50\pm0.04\text{ m s}^{-1}\) for every interval, so the results are consistent with constant horizontal velocity.

For each interval:

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

For the first interval:

\[ v_x=\frac{0.150-0.000}{0.100} =1.50\text{ m s}^{-1} \]

The other intervals have the same displacement and time, so they also give \(1.50\text{ m s}^{-1}\).

The uncertainty in each displacement is:

\[ \Delta x=\pm(0.002+0.002) =\pm0.004\text{ m} \]

Therefore:

\[ \Delta v_x=\frac{0.004}{0.100} =\pm0.04\text{ m s}^{-1} \]

All three intervals give:

\[ v_x=1.50\pm0.04\text{ m s}^{-1} \]

The measurements show no resolved change in horizontal velocity. They support the constant-horizontal-velocity model within the experimental uncertainty.

Question 2

A student records the horizontal position of a projectile at equal time intervals.

The illustrative data are:

Each position reading has an uncertainty of \(\pm0.001\text{ m}\). Assume the time intervals are accurately known.

1. Calculate the average horizontal velocity in each interval.
2. Estimate the uncertainty in each velocity.
3. Estimate the average horizontal acceleration between the midpoints of the first and last intervals.
4. Does the data support constant horizontal velocity?

Solution 2

The interval velocities increase from \(2.00\) to \(2.20\text{ m s}^{-1}\), giving an estimated average horizontal acceleration of \(+0.50\text{ m s}^{-2}\). The data do not support constant horizontal velocity within the stated uncertainty.

Calculate the interval displacements:

\[ 0.200,\ 0.205,\ 0.210,\ 0.215,\ 0.220\text{ m} \]

Each time interval is \(0.100\text{ s}\).

Therefore:

\[ \begin{aligned} v_{x,1}&=\frac{0.200}{0.100}=2.00\\\ v_{x,2}&=\frac{0.205}{0.100}=2.05\\\ v_{x,3}&=\frac{0.210}{0.100}=2.10\\\ v_{x,4}&=\frac{0.215}{0.100}=2.15\\\ v_{x,5}&=\frac{0.220}{0.100}=2.20 \end{aligned} \]

All values are in \(\text{m s}^{-1}\).

The uncertainty in each displacement is:

\[ \Delta x=\pm(0.001+0.001) =\pm0.002\text{ m} \]

Hence:

\[ \Delta v_x=\frac{0.002}{0.100} =\pm0.02\text{ m s}^{-1} \]

The first and last velocities are therefore:

\[ v_{x,1}=2.00\pm0.02\text{ m s}^{-1} \]

\[ v_{x,5}=2.20\pm0.02\text{ m s}^{-1} \]

Their uncertainty ranges do not overlap.

To estimate average horizontal acceleration, use the midpoints of the first and last intervals:

\[ t_1=0.05\text{ s},\qquad t_2=0.45\text{ s} \]

Then:

\[ a_{x,\text{avg}}=\frac{\Delta v_x}{\Delta t} \]

\[ a_{x,\text{avg}} =\frac{2.20-2.00}{0.45-0.05} \]

\[ a_{x,\text{avg}}=+0.50\text{ m s}^{-2} \]

Using the conservative velocity uncertainties, the approximate uncertainty in this acceleration is:

\[ \Delta a_x=\frac{0.02+0.02}{0.40} =0.10\text{ m s}^{-2} \]

So:

\[ a_{x,\text{avg}}=+0.50\pm0.10\text{ m s}^{-2} \]

The measurements suggest a genuine increase in horizontal velocity, assuming the calibration and timing are reliable.

The tempting conclusion is that gravity must be responsible for the change. However, gravity acts vertically, not horizontally, in the ideal projectile model. A horizontal acceleration would require an additional horizontal force or a problem with the measurement method.

Question 3

A projectile is launched horizontally with an initial horizontal velocity of \(2.40\text{ m s}^{-1}\).

Video analysis shows that it travels \(0.240\text{ m}\) horizontally during every \(0.100\text{ s}\) interval.

Assume negligible air resistance and use \(g=9.8\text{ m s}^{-2}\).

After \(0.60\text{ s}\):

1. Calculate the projectile’s horizontal velocity.
2. Calculate its vertical velocity.
3. Calculate its total speed.
4. Explain why the increase in total speed does not contradict constant horizontal velocity.

Solution 3

The horizontal velocity remains \(2.40\text{ m s}^{-1}\), while the vertical velocity becomes \(-5.88\text{ m s}^{-1}\). The total speed increases to approximately \(6.35\text{ m s}^{-1}\).

The horizontal velocity is:

\[ v_x=\frac{0.240}{0.100} =2.40\text{ m s}^{-1} \]

The repeated equal horizontal displacements show that this component remains constant.

For vertical motion:

\[ v_y=u_y+ a_yt \]

Since the projectile is launched horizontally:

\[ u_y=0,\qquad a_y=-9.8\text{ m s}^{-2} \]

Therefore:

\[ v_y=0-(9.8)(0.60) \]

\[ v_y=-5.88\text{ m s}^{-1} \]

The negative sign indicates downward motion.

The total speed is calculated using the perpendicular velocity components:

\[ v=\sqrt{v_x^2+v_y^2} \]

\[ v=\sqrt{(2.40)^2+(-5.88)^2} \]

\[ v=\sqrt{40.3344} \]

\[ v=6.35\text{ m s}^{-1} \]

The tempting assumption is that if an object speeds up, its horizontal velocity must also increase.

That is incorrect because speed is the magnitude of the complete velocity vector. Gravity changes the vertical component, increasing the total speed while leaving the horizontal component unchanged.

Question 4

A student uses video analysis to test whether a projectile has constant horizontal velocity.

The camera is positioned at an angle to the plane of motion, causing the apparent image scale to change across the recording.

During four equal time intervals of \(0.100\text{ s}\), the following pixel displacements are measured:

Each calibration interval represents an actual horizontal distance of \(0.200\text{ m}\).

The student uses the first calibration measurement, \(200\text{ pixels per metre}\), for the entire video.

1. Calculate the four apparent horizontal velocities using this single calibration.
2. Recalculate the velocities using the local calibration measurements.
3. Explain whether the projectile actually demonstrates horizontal acceleration.
4. Suggest an improvement to the experimental setup.

Solution 4

The single-scale method produces apparent velocities of \(2.0\), \(2.1\), \(2.2\), and \(2.3\text{ m s}^{-1}\). However, local calibration gives \(2.0\text{ m s}^{-1}\) for every interval, supporting constant horizontal velocity.

Using the first calibration measurement:

\[ \text{Scale}=\frac{40\text{ pixels}}{0.200\text{ m}} =200\text{ pixels m}^{-1} \]

Applying this scale to every interval:

\[ v_{x,1}=\frac{40/200}{0.100} =2.0\text{ m s}^{-1} \]

\[ v_{x,2}=\frac{42/200}{0.100} =2.1\text{ m s}^{-1} \]

\[ v_{x,3}=\frac{44/200}{0.100} =2.2\text{ m s}^{-1} \]

\[ v_{x,4}=\frac{46/200}{0.100} =2.3\text{ m s}^{-1} \]

This appears to show increasing horizontal velocity.

However, the local calibration measurements reveal that the image scale changes across the recording.

For each interval, the projectile displacement matches the corresponding calibration interval. Each therefore represents the same actual horizontal displacement:

\[ \Delta x=0.200\text{ m} \]

The corrected velocities are:

\[ v_x=\frac{0.200}{0.100} =2.0\text{ m s}^{-1} \]

for all four intervals.

The apparent acceleration was caused by using an incorrect global scale factor.

The tempting route is to assume that the same number of pixels always represents the same physical distance. That assumption only works when the image scale is uniform or has been appropriately corrected.

An improvement would be to position the camera perpendicular to the plane of motion and place the calibration ruler in the same plane as the projectile. If some perspective distortion remains, use calibration markers across the full measurement region rather than relying on one measurement near the launch point.

Question 5

A student analyses a projectile using a high-speed camera.

The camera records frame numbers at intervals of five frames. Its nominal frame rate is \(50\text{ frames s}^{-1}\), so the student assumes that every selected interval represents \(0.100\text{ s}\).

However, the recording uses variable frame timing. The actual timestamps in the video file are:

After correcting the camera’s spatial calibration, the projectile’s measured horizontal positions are:

Each position measurement has an uncertainty of \(\pm0.002\text{ m}\), and each timestamp has an uncertainty of \(\pm0.001\text{ s}\). Assume independent endpoint uncertainties and use a conservative maximum-uncertainty approach.

The student argues:

Every five frames, the projectile travels exactly \(0.200\text{ m}\). Therefore, its horizontal speed must be constant.

Evaluate this conclusion.

1. Calculate the actual time interval for each pair of consecutive positions.
2. Calculate the average horizontal velocity in each interval.
3. Estimate the uncertainty in the first and last velocity measurements using:

\[ \frac{\Delta v_x}{v_x} \approx \frac{\Delta(\Delta x)}{\Delta x} + \frac{\Delta(\Delta t)}{\Delta t} \]

1. Determine whether the first and last velocity measurements are consistent within uncertainty.
2. Explain what the results suggest about the ideal projectile model, and identify an additional experiment that could help establish the cause of the observed change.

Solution 5

The student’s conclusion is incorrect. Although the projectile travels equal horizontal distances between the selected frames, the actual time intervals increase. The measured horizontal velocity therefore decreases from \(2.00\text{ m s}^{-1}\) to approximately \(1.74\text{ m s}^{-1}\).

Step 1: Calculate the actual time intervals.

Subtract consecutive timestamps:

\[ \begin{aligned} \Delta t_1&=0.100-0.000=0.100\text{ s}\\\ \Delta t_2&=0.205-0.100=0.105\text{ s}\\\ \Delta t_3&=0.315-0.205=0.110\text{ s}\\\ \Delta t_4&=0.430-0.315=0.115\text{ s} \end{aligned} \]

The intervals are not equal.

The student has confused equal frame-number differences with equal elapsed times.

Step 2: Calculate the average horizontal velocities.

Every physical displacement is \(0.200\text{ m}\).

Therefore:

\[ v_{x,1}=\frac{0.200}{0.100} =2.00\text{ m s}^{-1} \]

\[ v_{x,2}=\frac{0.200}{0.105} =1.90\text{ m s}^{-1} \]

\[ v_{x,3}=\frac{0.200}{0.110} =1.82\text{ m s}^{-1} \]

\[ v_{x,4}=\frac{0.200}{0.115} =1.74\text{ m s}^{-1} \]

The velocity decreases over the measured flight.

Step 3: Calculate the uncertainty in the first velocity.

The displacement uses two position measurements, each with uncertainty \(\pm0.002\text{ m}\).

Therefore:

\[ \Delta(\Delta x)=\pm0.004\text{ m} \]

The time interval uses two timestamps, each with uncertainty \(\pm0.001\text{ s}\).

Therefore:

\[ \Delta(\Delta t)=\pm0.002\text{ s} \]

For the first interval:

\[ \frac{\Delta v_x}{v_x} \approx \frac{0.004}{0.200} + \frac{0.002}{0.100} \]

\[ =0.020+0.020=0.040 \]

So:

\[ \Delta v_x=(2.00)(0.040) =0.08\text{ m s}^{-1} \]

The first measurement is:

\[ v_{x,1}=2.00\pm0.08\text{ m s}^{-1} \]

Step 4: Calculate the uncertainty in the last velocity.

For the last interval:

\[ \frac{\Delta v_x}{v_x} \approx \frac{0.004}{0.200} + \frac{0.002}{0.115} \]

\[ =0.020+0.0174=0.0374 \]

Therefore:

\[ \Delta v_x=(1.739)(0.0374) \]

\[ \Delta v_x\approx0.07\text{ m s}^{-1} \]

The last measurement is:

\[ v_{x,4}=1.74\pm0.07\text{ m s}^{-1} \]

Compare the uncertainty ranges:

• First interval: \(1.92\) to \(2.08\text{ m s}^{-1}\).
• Last interval: \(1.67\) to \(1.81\text{ m s}^{-1}\).

The ranges do not overlap.

The decrease in horizontal velocity is therefore larger than the estimated measurement uncertainty.

Step 5: Interpret the physical result.

The measurements are inconsistent with constant horizontal velocity, assuming the spatial calibration and timestamps are reliable.

A possible explanation is air resistance. Drag acts against the projectile’s horizontal motion, reducing its horizontal velocity.

However, the data do not prove that air resistance is responsible. Another horizontal force or an error in the timing system could produce a similar result.

An appropriate follow-up would be to repeat the experiment using an independently verified timing system, such as a constant-frame-rate recording checked against an external timing signal.

If the horizontal deceleration persists, repeat the experiment with projectiles of the same shape and size but different masses, keeping the launch conditions similar. A smaller drag-induced acceleration for the heavier projectile would support the air-resistance explanation.

The crucial lesson is that equal distances per frame do not establish constant speed. Physics depends on displacement per unit of actual elapsed time, not displacement per selected image.

09What to investigate next

Once you can establish whether horizontal velocity remains constant, you can test the other half of the projectile model: vertical motion.

For an ideal projectile, taking upwards as positive:

\[ a_y=-g \]

This predicts a quadratic relationship between vertical position and time, even while horizontal position changes linearly.

Combining the two measurements lets you test whether the complete trajectory agrees with the ideal projectile model, rather than judging its shape by eye.

For a broader explanation of how the two components work together, see Projectile Motion: Horizontal and Vertical Motion Explained.

The key experimental decision remains the same: distinguish a genuine change in motion from a change introduced by the way you measured it.