How to Explain Systematic Error in a Projectile Range Experiment

Learn how to identify and explain systematic error in a projectile range experiment. Explore hypothetical datasets, measurement uncertainty, calibration errors, model limitations, and worked HSC Physics questions.

Five balls are launched horizontally from the same bench. Their measured ranges are 1.27 m, 1.28 m, 1.29 m, 1.28 m, and 1.28 m. The theoretical model predicts a range of 1.21 m.

The measurements are remarkably consistent, but every result is too large. Has repeating the experiment made it reliable, or has the experiment produced five equally biased measurements?

This is exactly the problem you need to recognise when explaining systematic error in a projectile range experiment.

To explain systematic error, identify a consistent bias in the measurements, show how it affects the experimental results, and connect it to a plausible cause in the apparatus or procedure. Then explain how you would test for or correct that cause. Simply writing “human error” or “inaccurate measurements” is not enough.

01What systematic error looks like in a projectile experiment

Start by separating two ideas: how closely your measurements agree with each other, and how closely they agree with the true value.

Imagine throwing a basketball at a hoop. If you consistently aim 10 cm to the left, your throws might land very close together. They are repeatable, but they are not accurate.

A projectile experiment can have the same problem. Repeated launches might produce almost identical landing positions, even though the measuring equipment or experimental arrangement shifts every result away from the correct value.

Systematic error versus random error

A systematic error introduces a consistent bias into measurements. It tends to push results in the same direction, making them consistently too high or too low.

A random error causes unpredictable variation between measurements. Individual results may be slightly higher or lower, even when the experimental conditions remain unchanged.

FeatureSystematic errorRandom error
Effect on measurementsConsistent biasUnpredictable variation
Repeated trialsDoes not remove the biasHelps reveal and reduce its influence
Effect on the meanShifts the mean away from the true valueProduces scatter around the mean
ExampleMeasuring every range from an incorrectly positioned zero markSmall differences in identifying the centre of each impact mark
ImprovementCalibrate or correct the apparatus and procedureRepeat measurements and improve measurement precision

The important distinction is that repeating an experiment does not automatically remove systematic error.

If a measuring tape starts 5 cm away from the correct origin, repeating the measurement 20 times will not fix its starting position. You’ll simply obtain 20 consistently biased measurements.

How to describe the direction of the error

A strong HSC explanation should identify whether the systematic error makes the measured range larger or smaller.

Choose the horizontal direction of projectile motion as positive. Define the difference between the experimental and predicted ranges as:

\[ \Delta R=R_{\text{exp}}-R_{\text{pred}} \]

where:

  • \(R_{\text{exp}}\) is the experimentally measured horizontal range, in metres.
  • \(R_{\text{pred}}\) is the theoretically predicted horizontal range, in metres.
  • \(\Delta R\) is the difference between the two values, in metres.

This gives a useful sign convention:

  • If \(\Delta R>0\), the measured range is greater than predicted.
  • If \(\Delta R<0\), the measured range is smaller than predicted.
  • If \(\Delta R\approx0\), the results agree within the relevant uncertainty.

A positive difference does not automatically identify the cause. It tells you the direction of the discrepancy, which is the starting point for investigating it.

02Collecting reliable projectile range data

Before diagnosing systematic error, make sure the experiment measures the quantity predicted by the physics model.

Consider a ball launched horizontally from the edge of a bench.

The ball leaves the launcher with horizontal velocity \(v_x\). It falls under gravity while travelling forwards, eventually making its first contact with the floor.

For this experiment, horizontal range is the horizontal displacement from the point directly beneath the launch position to the ball’s first landing position.

It is not the curved distance travelled through the air, and it is not the distance the ball travels after bouncing or rolling.

What should you measure?

A useful experimental setup needs three main measurements.

MeasurementSymbolWhat it represents
Launch height\(h\)Vertical distance from the ball’s centre at release to the floor
Initial horizontal velocity\(v_x\)Horizontal velocity as the ball leaves the launcher
Experimental range\(R_{\text{exp}}\)Horizontal distance from the origin to the first landing position

For a horizontal launch, assuming negligible air resistance and constant gravitational acceleration, the predicted range is:

\[ R_{\text{pred}}=v_x\sqrt{\frac{2h}{g}} \]

Here, \(g\) is gravitational acceleration, approximately \(9.8\text{ m s}^{-2}\).

This equation combines the vertical and horizontal parts of projectile motion. The height determines the flight time, while the horizontal velocity determines how far the ball travels during that time.

If you need to revise the derivation or understand how changing the launch conditions affects the range, see Horizontal Range of a Projectile: HSC Physics Guide.

A reliable data collection procedure

For a school investigation, follow a consistent sequence.

  1. Secure the launcher so it cannot move between trials.
  2. Check that the launcher is horizontal using a suitable level or alignment method.
  3. Measure the launch height from the centre of the ball at the instant of release.
  4. Use a plumb line to identify the point on the floor directly beneath the launch position. Mark this as \(x=0\).
  5. Place an impact-marking surface where the ball is expected to land.
  6. Measure the horizontal distance from the origin to the centre of the first impact mark.
  7. Repeat the launch at least five times, keeping the launch conditions unchanged.
  8. Record every result, including unusual measurements, rather than selecting only the readings that agree with the prediction.

A photogate can measure the launch speed, while calibrated video can help determine the launch conditions and landing position.

If video is used, the camera must be positioned correctly. An oblique viewing angle can distort apparent distances and introduce a consistent measurement bias. See How to Choose a Camera Position for Projectile Motion for a closer explanation.

Why measure the first impact?

Suppose a ball strikes the floor at 1.20 m but bounces and eventually comes to rest at 1.31 m.

If you measure 1.31 m, you have not measured the projectile’s landing position. You have measured its landing position plus the horizontal distance travelled after impact.

This creates a consistent overestimate if the same procedure is used for every trial.

The solution is to record the first point of contact. Carbon paper, an impact sensor, or frame-by-frame video can help identify it.

The key is to measure the same physical event that the theoretical model predicts.

03How to identify systematic error using experimental data

Repeated measurements help you identify the spread of your results. Comparing their mean with the predicted value helps you identify a possible bias.

Consider the following hypothetical dataset.

Illustrative data: horizontal projectile experiment

The ball is launched horizontally from a height of \(0.750\text{ m}\). Its measured horizontal launch speed is \(3.10\text{ m s}^{-1}\).

Five trials produce these landing positions:

TrialMeasured horizontal range (m)
11.27
21.28
31.29
41.28
51.28

Calculate the mean experimental range

The mean is calculated by adding all measurements and dividing by the number of trials.

\[ \bar R_{\text{exp}}= \frac{\sum R_i}{n} \]

where \(R_i\) represents an individual range measurement and \(n\) is the number of trials.

Substituting:

\[ \bar R_{\text{exp}}= \frac{1.27+1.28+1.29+1.28+1.28}{5} \]

\[ \boxed{\bar R_{\text{exp}}=1.28\text{ m}} \]

The average experimental range is 1.28 m.

Estimate the spread of the measurements

For a small set of repeated measurements, a practical estimate of the absolute uncertainty is half the range of the readings:

\[ \Delta R=\frac{R_{\max}-R_{\min}}{2} \]

The largest reading is 1.29 m, and the smallest is 1.27 m.

\[ \Delta R=\frac{1.29-1.27}{2} \]

\[ \boxed{\Delta R=0.01\text{ m}} \]

The experimental result can therefore be reported as:

\[ R_{\text{exp}}=1.28\pm0.01\text{ m} \]

This half-range method gives a simple estimate of the observed spread. It is not a formal 95% confidence interval or the standard error of the mean. If an investigation specifies a different statistical method, use that method.

Calculate the predicted range

For a horizontal launch, the vertical displacement is:

\[ -h=-\frac12gt^2 \]

Rearranging gives the flight time:

\[ t=\sqrt{\frac{2h}{g}} \]

Substitute the launch height:

\[ t=\sqrt{\frac{2(0.750)}{9.8}} \]

\[ t=0.391\text{ s} \]

The predicted horizontal range is:

\[ R_{\text{pred}}=v_xt \]

\[ R_{\text{pred}}=3.10(0.391) \]

\[ \boxed{R_{\text{pred}}=1.21\text{ m}} \]

For this illustration, suppose the uncertainty in the predicted range, based on the launch-speed and height measurements, is estimated as \(0.02\text{ m}\).

The predicted result is therefore:

\[ R_{\text{pred}}=1.21\pm0.02\text{ m} \]

Compare the experimental and predicted values

Using the positive-forward sign convention:

\[ \Delta R=R_{\text{exp}}-R_{\text{pred}} \]

\[ \Delta R=1.28-1.213 \]

\[ \boxed{\Delta R=+0.067\text{ m}} \]

The experimental range is approximately 6.7 cm greater than predicted.

The predicted uncertainty interval extends from 1.19 m to 1.23 m. The experimental interval extends from 1.27 m to 1.29 m.

These intervals do not overlap.

This is evidence of a systematic discrepancy between the experiment and the theoretical prediction. The small spread suggests that random variation alone is unlikely to explain the difference.

However, there is an important qualification: the data do not prove that the measuring equipment is responsible. A consistent disagreement could also result from incorrect launch conditions or a limitation of the theoretical model.

You need further evidence before identifying the cause.

04Common sources of systematic error in a projectile range experiment

Different systematic errors produce different patterns. Recognising those patterns helps you decide what to investigate.

1. Incorrect measurement origin

Suppose the true origin is directly beneath the launch point, but the measuring tape begins 4 cm behind it.

Every measured range will be 4 cm too large.

For example:

True range (m)Measured range (m)
0.800.84
1.001.04
1.201.24
1.401.44

The difference is constant:

\[ R_{\text{measured}}-R_{\text{true}}=+0.04\text{ m} \]

This is an additive systematic error. It shifts every measurement by approximately the same absolute amount.

A plumb line provides an independent way to locate the correct origin. Check that the zero mark of the measuring tape aligns with this point.

2. Incorrect measuring instrument calibration

A measuring tape or ruler might have an incorrectly calibrated scale.

Unlike a misplaced zero mark, a scale calibration error often produces a discrepancy that increases with the distance measured.

For example, imagine that a tape consistently reads 2% too high.

True distance (m)Measured distance (m)
0.500.51
1.001.02
1.501.53

The absolute error increases with distance, but the percentage error remains approximately constant.

This is a proportional systematic error.

Check the instrument against a known reference length. If the measured distance is consistently incorrect, recalibrate or replace the instrument.

3. Launcher not horizontal

A horizontal-launch experiment assumes the projectile leaves the launcher with zero initial vertical velocity.

If the launcher is tilted slightly upwards or downwards, this assumption is no longer correct.

An upward tilt gives the projectile an initial upward velocity, increasing its flight time. A downward tilt gives it an initial downward velocity, decreasing its flight time.

For the same horizontal speed and release height:

  • An upward tilt generally increases the horizontal range.
  • A downward tilt generally decreases the horizontal range.

This is a systematic experimental problem because the same incorrect launch angle can affect every trial.

It also creates a model mismatch: the horizontal-launch equation is being applied to a projectile that does not actually have zero initial vertical velocity.

Check the launcher alignment before collecting data, and measure the actual launch angle if necessary.

4. Measuring the wrong landing position

A ball may bounce, roll, or slide after its first contact with the floor.

If the experiment records the final resting position instead of the first impact position, the measured range will generally be too large.

This is not a random measurement error. It is a consistent procedural error caused by measuring the wrong event.

Use an impact-marking surface or video analysis to identify the first contact point.

5. Camera perspective and parallax

A camera placed at an angle to the projectile’s plane of motion can distort apparent horizontal distances.

This can happen even if the camera records every trial consistently.

For example, if the calibration ruler is placed in a different plane from the projectile, the apparent scale in the video may not represent the actual distance travelled.

Position the camera perpendicular to the plane of motion and place the calibration scale in that same plane.

05Systematic error or limitation of the physics model?

This is one of the most important distinctions in an HSC experimental analysis.

A systematic measurement error means the experimental procedure or equipment introduces a consistent bias.

A limitation of the physics model means the mathematical model does not include every physical effect influencing the projectile.

These can produce similar-looking results, but they are not the same problem.

Air resistance: when the ideal model is incomplete

The standard projectile model assumes that air resistance is negligible.

In reality, a projectile moving through air experiences drag. This force acts opposite to its motion relative to the air.

For many ordinary school experiments, the effect is small enough to ignore. However, it can become significant for light objects, large surface areas, or sufficiently long flight times.

A lighter ball generally experiences a greater acceleration due to the same drag force than a heavier ball of identical shape and size.

Consequently, the lighter ball may travel a shorter horizontal distance than predicted by the ideal model.

This does not necessarily mean the experiment was performed incorrectly. It may mean the model is incomplete for the conditions being tested.

A useful diagnostic comparison

Experimental observationPossible explanationUseful test
All measured ranges are too large by approximately the same distanceIncorrect zero positionCheck the origin using a plumb line
Percentage error increases with measured distanceScale calibration errorCheck against known reference lengths
All ranges are consistently too short or too long, and the launcher is misalignedIncorrect launch angleMeasure and correct the launch angle
The measured final resting position exceeds the predicted landing positionBounce or rolling after impactRecord the first contact point
Video ranges disagree with direct measurementsCamera perspective or calibration errorReposition the camera and calibrate in the projectile’s plane
A light projectile falls short of the ideal prediction more than a heavier projectile of the same sizeAir resistance may be significantCompare projectiles with different masses under matched launch conditions

The last observation is particularly useful. It suggests a limitation of the ideal projectile model rather than a simple measuring-tape error.

A consistent difference between experimental and theoretical results should therefore be investigated, not automatically labelled as systematic measurement error.

06Worked examples: diagnosing calibration errors

The previous dataset showed how to recognise a discrepancy. Now you need to work out what could have caused it.

Two calibration patterns are particularly useful: a constant absolute error and a constant percentage error.

Worked example: Is the measuring tape’s zero position incorrect?

A student checks a measuring tape against a calibrated reference ruler. The following hypothetical results are recorded.

Reference distance (m)Tape reading (m)
0.500.54
1.001.04
1.501.54

During a projectile experiment, the tape records a horizontal range of 1.36 m. Identify the systematic error and calculate the corrected range.

Step 1

\[ \text{Error}= R_{\text{measured}}-R_{\text{true}} \]

For the three calibration measurements:

\[ \begin{aligned} 0.54-0.50&=+0.04\text{ m}\\ 1.04-1.00&=+0.04\text{ m}\\ 1.54-1.50&=+0.04\text{ m} \end{aligned} \]

Step 2

The tape reads 0.04 m too high at every reference distance.

Because the absolute error remains constant, the results suggest an additive zero-position error rather than a scale calibration error.

Step 3

The correction must be subtracted from the measured range:

\[ R_{\text{corrected}}=1.36-0.04 \]

\[ \boxed{R_{\text{corrected}}=1.32\text{ m}} \]

The corrected experimental range is 1.32 m.

Step 4

The tape’s zero mark is positioned 4 cm behind the true origin. Consequently, every measured horizontal distance includes an extra 4 cm.

This error affects every trial in the same direction. Repeating the experiment cannot remove it.

Worked example: Is the measuring scale incorrectly calibrated?

A second student checks a measuring instrument against known distances.

True distance (m)Measured distance (m)
0.500.51
1.001.02
1.501.53

The instrument records a projectile’s horizontal range as 1.53 m. Calculate the corrected range.

Step 1

For the first measurement:

\[ \frac{0.51}{0.50}=1.02 \]

Check the other measurements:

\[ \frac{1.02}{1.00}=1.02 \]

\[ \frac{1.53}{1.50}=1.02 \]

The instrument consistently reads 1.02 times the true distance.

Step 2

The instrument has a scale error of 2%.

Unlike the previous example, the absolute error increases as the distance increases. The percentage error remains constant.

Step 3

Divide the measured value by the calibration factor:

\[ R_{\text{corrected}}= \frac{R_{\text{measured}}}{1.02} \]

\[ R_{\text{corrected}}= \frac{1.53}{1.02} \]

\[ \boxed{R_{\text{corrected}}=1.50\text{ m}} \]

The corrected experimental range is 1.50 m.

Step 4

The instrument exaggerates every measured distance by 2%. Dividing by 1.02 reverses that proportional error.

Notice the difference between the two examples:

  • A zero-position error requires subtracting a constant distance.
  • A scale calibration error requires correcting by a constant scale factor.

Applying the wrong correction would introduce another systematic error rather than remove the original one.

07A practical method for explaining systematic error in an HSC response

When an exam question asks you to identify or explain systematic error, build your response around five points.

StageWhat to include
1. Identify the discrepancyState whether experimental ranges are consistently greater or smaller than predicted.
2. Use the dataCompare the mean experimental range, its uncertainty, and the predicted range.
3. Identify a possible causeName a specific issue, such as an incorrect origin, misaligned launcher, or calibration error.
4. Explain the mechanismDescribe exactly how that issue shifts the measured or predicted range.
5. Suggest an improvementExplain how calibration, alignment, or an independent measurement would reduce the error.

For example, a strong response might read:

The experimental horizontal range is consistently greater than the theoretical prediction. This suggests a positive systematic discrepancy rather than random variation alone. One possible cause is an incorrectly positioned measuring origin, which would add a consistent distance to every measured range. The origin should be checked using a plumb line, and the measuring tape should be calibrated against a known reference distance. Repeated trials alone would not remove this error.

Notice how the explanation connects the cause to its physical effect.

Compare that with:

The results are inaccurate due to human error.

The second statement identifies neither the source nor the direction of the error. It also gives no meaningful improvement.

Be careful not to claim that a particular systematic error definitely occurred unless the experimental evidence supports it. If you cannot identify the exact cause, describe it as a possible source and explain how you would investigate it.


08Questions and solutions

The following questions move from basic data interpretation to more demanding experimental analysis. They use hypothetical data and assume negligible air resistance unless stated otherwise.

Question 1

A student launches a ball horizontally from a bench. The theoretical model predicts a horizontal range of \(0.95\pm0.02\text{ m}\).

Five experimental trials produce the following measurements:

TrialHorizontal range (m)
11.01
21.00
31.02
41.01
51.00

(a) Calculate the mean experimental range.

(b) Estimate the absolute uncertainty using the half-range method.

(c) Calculate the difference between the experimental and predicted ranges.

(d) Explain what the results suggest about systematic error.

Solution 1

The mean experimental range is \(1.008\text{ m}\), with an estimated half-range uncertainty of \(0.010\text{ m}\). The experimental result is consistently greater than the theoretical prediction, suggesting a positive systematic discrepancy.

(a) Calculate the mean.

Add the five measurements and divide by five:

\[ \bar R= \frac{1.01+1.00+1.02+1.01+1.00}{5} \]

\[ \boxed{\bar R=1.008\text{ m}} \]

(b) Calculate the half-range uncertainty.

The maximum measurement is 1.02 m, and the minimum is 1.00 m.

\[ \Delta R=\frac{1.02-1.00}{2} \]

\[ \boxed{\Delta R=0.010\text{ m}} \]

Therefore:

\[ R_{\text{exp}}=1.008\pm0.010\text{ m} \]

(c) Calculate the difference.

Using the positive-forward sign convention:

\[ \Delta R=R_{\text{exp}}-R_{\text{pred}} \]

\[ \Delta R=1.008-0.95 \]

\[ \boxed{\Delta R=+0.058\text{ m}} \]

The measured range is 5.8 cm greater than predicted.

(d) Explain the result.

The five measurements are closely grouped, but their mean is noticeably greater than the predicted value.

The predicted uncertainty interval is 0.93-0.97 m, while the experimental interval is 0.998-1.018 m. These intervals do not overlap.

This suggests a systematic discrepancy that cannot be explained by the estimated measurement spread alone.

Possible causes include an incorrectly positioned origin or a consistent error in the launch conditions. Further calibration and alignment checks are needed to identify the actual cause.

Question 2

A ball is launched from a height of \(1.20\text{ m}\) above the floor. Its initial speed is \(4.0\text{ m s}^{-1}\), but the launcher is tilted \(4.0^\circ\) below the horizontal.

A student incorrectly assumes that the launcher is perfectly horizontal when calculating the predicted range.

Assume:

  • Gravitational acceleration is \(9.8\text{ m s}^{-2}\).
  • Air resistance is negligible.
  • The ball lands on a horizontal floor.

(a) Calculate the horizontal range predicted using the student’s horizontal-launch assumption.

(b) Calculate the actual horizontal range, accounting for the downward launch angle.

(c) Explain why the student’s prediction differs from the actual range. Is this random measurement error, systematic experimental error, or a limitation of the model being used?

Solution 2

The student’s horizontal-launch model predicts approximately \(1.98\text{ m}\), while the actual range is approximately \(1.86\text{ m}\). The incorrect horizontal-launch assumption overestimates the range by approximately \(0.12\text{ m}\).

(a) Calculate the student’s predicted range.

For a horizontal launch, the initial vertical velocity is assumed to be zero.

The flight time is:

\[ t=\sqrt{\frac{2h}{g}} \]

Substitute the values:

\[ t=\sqrt{\frac{2(1.20)}{9.8}} \]

\[ t=0.495\text{ s} \]

The student assumes the full initial speed is horizontal, so:

\[ R_{\text{pred}}=ut \]

\[ R_{\text{pred}}=4.0(0.495) \]

\[ \boxed{R_{\text{pred}}=1.98\text{ m}} \]

(b) Calculate the actual range.

The initial velocity must be resolved into horizontal and vertical components.

Take upwards as positive.

The horizontal component is:

\[ v_x=4.0\cos4.0^\circ \]

\[ v_x=3.99\text{ m s}^{-1} \]

The vertical component is:

\[ v_y=-4.0\sin4.0^\circ \]

\[ v_y=-0.279\text{ m s}^{-1} \]

The negative sign indicates that the initial vertical velocity is downwards.

Now use the vertical displacement equation:

\[ \Delta y=v_yt-\frac12gt^2 \]

The ball starts 1.20 m above the floor, so:

\[ -1.20=-0.279t-4.9t^2 \]

Rearranging:

\[ 4.9t^2+0.279t-1.20=0 \]

Using the positive quadratic solution:

\[ t=\frac{-0.279+ \sqrt{(0.279)^2+4(4.9)(1.20)}}{9.8} \]

\[ t=0.467\text{ s} \]

Now calculate the horizontal range:

\[ R=v_xt \]

\[ R=3.99(0.467) \]

\[ \boxed{R=1.86\text{ m}} \]

(c) Explain the discrepancy.

The downward launch angle gives the projectile an initial downward velocity.

This means the ball reaches the floor sooner than it would if launched horizontally. Although its horizontal speed is almost unchanged, it has less time to travel forwards.

The actual range is therefore approximately:

\[ 1.98-1.86=0.12\text{ m} \]

shorter than the student’s prediction.

This is a consistent experimental setup error because the launcher is not aligned as intended. It also means the student’s theoretical model uses an incorrect initial condition.

It is not random measurement error. Repeating the experiment with the same launcher angle would reproduce the discrepancy.

Question 3

A student predicts that a ball launched horizontally will first contact the floor at a horizontal distance of \(1.25\pm0.02\text{ m}\).

The student performs five trials but measures where the ball eventually comes to rest.

The recorded positions are:

TrialFinal resting position (m)
11.35
21.36
31.37
41.36
51.36

A video analysis later reveals that the ball first contacts the floor at approximately 1.25 m, then rolls a further 0.11 m before stopping.

(a) Calculate the mean final resting position.

(b) Estimate its half-range uncertainty.

(c) Explain why the original measurements appear to disagree with the theoretical prediction.

(d) Identify the appropriate experimental improvement.

Solution 3

The mean final resting position is \(1.36\text{ m}\), with an estimated half-range uncertainty of \(0.01\text{ m}\). The discrepancy is caused by measuring the ball’s final resting position rather than its first landing position.

(a) Calculate the mean.

\[ \bar R= \frac{1.35+1.36+1.37+1.36+1.36}{5} \]

\[ \boxed{\bar R=1.36\text{ m}} \]

(b) Calculate the half-range uncertainty.

\[ \Delta R=\frac{1.37-1.35}{2} \]

\[ \boxed{\Delta R=0.01\text{ m}} \]

The original result is:

\[ R_{\text{exp}}=1.36\pm0.01\text{ m} \]

(c) Explain the discrepancy.

The theoretical model predicts the position where the projectile first contacts the floor.

The student instead measures the position where the ball eventually stops.

The ball travels an additional 0.11 m after its initial impact, so every measurement is consistently too large.

The experimental method is measuring a different physical quantity from the one predicted by the model.

This is a systematic procedural error, not evidence that the projectile-motion equations have failed.

The corrected first-contact measurement is approximately:

\[ \boxed{R_{\text{exp}}=1.25\text{ m}} \]

This agrees with the theoretical prediction within the stated uncertainty.

(d) Recommend an improvement.

Use carbon paper, an impact sensor, or calibrated high-speed video to record the ball’s first contact with the floor.

Do not use the final resting position as a substitute for the landing position.

The distinction is important: highly repeatable measurements can still be systematically wrong if the wrong event is being measured.

Question 4

A student performs six consecutive launches using the same horizontal projectile launcher.

The launch height is kept constant. A photogate measures the horizontal launch speed immediately before each ball leaves the launcher.

The following hypothetical data are collected:

TrialLaunch speed (m s\(^{-1}\))Measured range (m)
12.901.10
22.961.12
33.021.14
43.081.16
53.141.18
63.201.20

(a) Calculate the percentage increase in launch speed between the first and sixth trials.

(b) Calculate the percentage increase in measured range over the same trials.

(c) Explain why the data suggest a systematic change in the experimental conditions rather than ordinary random variation.

(d) Explain why calculating a single mean range and half-range uncertainty would not provide a complete description of this dataset.

(e) Suggest an appropriate improvement to the investigation.

Solution 4

The launch speed increases by approximately 10.3%, while the measured range increases by approximately 9.1%. The matching upward trends indicate that the launch conditions are changing systematically during the experiment.

(a) Calculate the percentage increase in launch speed.

The percentage increase is:

\[ \%\text{ increase}= \frac{\text{final}-\text{initial}} {\text{initial}}\times100 \]

Substituting:

\[ \frac{3.20-2.90}{2.90}\times100 \]

\[ \boxed{10.3\%} \]

(b) Calculate the percentage increase in range.

\[ \%\text{ increase}= \frac{1.20-1.10}{1.10}\times100 \]

\[ \boxed{9.1\%} \]

(c) Explain the pattern.

For a horizontal launch from a constant height, the theoretical range is:

\[ R=v_x\sqrt{\frac{2h}{g}} \]

If \(h\) and \(g\) remain constant, the range is directly proportional to horizontal launch speed:

\[ R\propto v_x \]

The launch speed increases throughout the trial sequence, and the measured range increases in a similar pattern.

This suggests that the launcher is not producing a consistent launch speed, despite its setting remaining unchanged.

The experiment therefore contains a systematic drift in its launch conditions.

The data do not establish why the launcher speed is increasing. The launching mechanism, release procedure, or another changing condition would need to be investigated.

(d) Explain the limitation of using only the mean and half-range.

The mean range is:

\[ \bar R=\frac{1.10+1.12+1.14+1.16+1.18+1.20}{6} \]

\[ \boxed{\bar R=1.15\text{ m}} \]

The half-range estimate is:

\[ \Delta R=\frac{1.20-1.10}{2} \]

\[ \boxed{\Delta R=0.05\text{ m}} \]

Reporting \(1.15\pm0.05\text{ m}\) describes the spread, but hides the ordered increase in the measurements.

The results are not simply scattered randomly around a fixed value. They follow a clear upward trend.

A graph of launch speed and range against trial number would reveal this pattern.

(e) Recommend an improvement.

Investigate why the launch speed changes between trials and modify the procedure so that the speed remains consistent.

Use the photogate to monitor the launch speed for every trial. If the speed still varies, record the actual speed and calculate an individual predicted range for each launch rather than treating all launches as identical.

The investigation should also check that the launcher setting, release mechanism, and projectile position remain consistent.

Simply increasing the number of trials would not remove the underlying drift.

Question 5

An advanced projectile investigation compares two spheres with identical diameters and surface finishes:

  • A heavier sphere with a mass of 80 g.
  • A lighter sphere with a mass of 20 g.

Both spheres are launched horizontally. For each experimental configuration, the two spheres have the same initial horizontal speed and release height.

Three configurations are tested. They have progressively longer flight times, producing the following theoretical ranges when air resistance is neglected:

ConfigurationPredicted range (m)
A0.80
B1.20
C1.60

The experiment uses a floor measuring tape to record the mean landing positions. Each mean has an estimated half-range spread of \(0.01\text{ m}\).

The following hypothetical results are obtained:

SphereConfiguration A (m)Configuration B (m)Configuration C (m)
80 g0.841.241.64
20 g0.821.161.52

An independent calibration test establishes that the measuring tape’s zero mark is positioned 0.040 m behind the true projectile origin. Every recorded range is therefore 0.040 m greater than the actual horizontal displacement.

Assume:

  • The calibration offset applies equally to both spheres and all configurations.
  • The predicted ranges have uncertainties of \(0.01\text{ m}\).
  • The reported experimental spread estimates are \(0.01\text{ m}\).
  • The spheres have identical aerodynamic shapes and surface finishes.
  • The measured launch speeds and release heights are matched within each configuration.
  • There is no significant spin or wind.

(a) Correct all six experimental range measurements for the tape’s zero-position error.

(b) Calculate the difference between each corrected experimental range and its corresponding theoretical prediction.

(c) Explain why the heavier sphere agrees more closely with the ideal projectile model than the lighter sphere.

(d) A student argues that the entire disagreement is caused by the measuring tape, because the tape reads 0.040 m too high. Explain why this conclusion is incomplete.

(e) Identify the most defensible explanation for the remaining differences, and explain how the pattern across configurations supports your conclusion.

Solution 5

The tape calibration error must first be removed. After correction, the heavier sphere agrees with the ideal model within the stated uncertainty, while the lighter sphere shows increasingly short ranges. The remaining discrepancy is most consistent with air resistance having a greater effect on the lighter sphere.

(a) Correct the experimental measurements.

The measuring tape reads 0.040 m too high.

Therefore:

\[ R_{\text{corrected}}= R_{\text{measured}}-0.040 \]

For the 80 g sphere:

\[ \begin{aligned} A &: 0.84-0.04=0.80\text{ m}\\ B &: 1.24-0.04=1.20\text{ m}\\ C &: 1.64-0.04=1.60\text{ m} \end{aligned} \]

For the 20 g sphere:

\[ \begin{aligned} A &: 0.82-0.04=0.78\text{ m}\\ B &: 1.16-0.04=1.12\text{ m}\\ C &: 1.52-0.04=1.48\text{ m} \end{aligned} \]

The corrected results are:

SphereConfiguration A (m)Configuration B (m)Configuration C (m)
80 g0.801.201.60
20 g0.781.121.48

(b) Calculate the differences from the theoretical predictions.

Use:

\[ \Delta R=R_{\text{corrected}}-R_{\text{predicted}} \]

For the 80 g sphere:

ConfigurationCalculationDifference (m)
A\(0.80-0.80\)0.00
B\(1.20-1.20\)0.00
C\(1.60-1.60\)0.00

For the 20 g sphere:

ConfigurationCalculationDifference (m)
A\(0.78-0.80\)-0.02
B\(1.12-1.20\)-0.08
C\(1.48-1.60\)-0.12

The heavier sphere agrees with the ideal predictions.

The lighter sphere has increasingly negative differences, meaning its measured range becomes progressively shorter than predicted.

The difference in configuration A is at the boundary of the combined stated uncertainty limits. The discrepancies in configurations B and C are clearly larger than those limits.

(c) Explain why the heavier sphere follows the ideal model more closely.

Both spheres have the same diameter and shape. At the same speed, they experience approximately comparable aerodynamic drag forces.

However, Newton’s second law gives:

\[ F_{\text{net}}=ma \]

For a given drag force, a smaller mass experiences a greater acceleration due to that force.

The lighter sphere is therefore affected more strongly by air resistance than the heavier sphere.

As the flight time increases, the drag has more opportunity to change the sphere’s horizontal velocity. This helps explain why the lighter sphere’s range falls increasingly below the ideal prediction.

(d) Explain why the measuring tape is not the complete explanation.

The tape introduces a genuine systematic error. It adds 0.040 m to every recorded range.

However, the calibration correction removes this error from both sets of measurements.

After correction, the heavier sphere agrees with the theoretical predictions, while the lighter sphere still has a significant shortfall in configurations B and C.

A fixed zero-position error cannot explain this remaining difference because it would affect both spheres equally.

The student’s explanation accounts for the original measurement bias but not the remaining physical discrepancy.

(e) Identify the most defensible explanation.

The remaining differences are most consistent with the ideal projectile model neglecting significant air resistance for the lighter sphere.

Three observations support this conclusion:

  1. The independent calibration test identifies and allows correction of the same additive measurement error for both spheres.
  2. The heavier sphere agrees with the ideal model after this correction.
  3. The lighter sphere falls progressively shorter as flight time increases, despite matched launch conditions within each configuration.

The results suggest that air resistance has a greater influence on the lighter sphere because its smaller mass produces a greater acceleration under the same drag force.

This is a limitation of the ideal theoretical model, not evidence of another fixed measuring-tape error.

There is an important scientific distinction here: correcting a systematic measurement error does not guarantee that experimental results will agree perfectly with a simplified theoretical model. The remaining difference may reveal a physical effect that the model ignores.

In a real investigation, further checks would still be useful, including repeated calibration, independent video measurements, and tests for small differences in launch conditions.

09Where to take this next

Once you can recognise and explain systematic error, the next step is deciding whether an experimental result genuinely agrees with a theoretical prediction.

That requires comparing both values alongside their uncertainties, rather than simply calculating a percentage difference.

For the next stage, work through How to Compare Predicted and Experimental Landing Positions. It develops the uncertainty comparison needed to decide whether an apparent disagreement is significant, and how to interpret the result without confusing measurement error with limitations of the physics model.