How to Compare Predicted and Experimental Landing Positions
Learn how to compare theoretical and experimental projectile landing positions using repeated measurements and uncertainty intervals. Includes worked calculations, hypothetical datasets, and methods for identifying measurement errors and limitations of physics models.
A projectile-motion calculation predicts that a ball will first hit the floor 1.19 m from the point directly beneath its launch position. Five experimental trials give an average landing position of 1.21 m.
Has the projectile-motion model failed?
Not necessarily. The difference is 0.02 m, but you cannot decide whether that matters until you consider the uncertainty in both values.
To compare a predicted landing position with an experimental result, you need to:
- Calculate the predicted landing position using an appropriate physics model.
- Collect repeated experimental measurements of the actual landing position.
- Estimate the uncertainty in both results.
- Compare their uncertainty intervals.
- Investigate any disagreement to determine whether it comes from measurement error, experimental limitations, or an assumption in the physics model.
The important distinction is between a result that is different and a result that is significantly different.
01Establish exactly what you are comparing
Before calculating anything, establish a consistent coordinate system.
Consider a ball launched horizontally from the edge of a laboratory bench. It travels forwards before hitting the floor.
We want to compare its predicted horizontal landing position with the position recorded experimentally.
Choose the following coordinate system:
- The origin, \(x=0\), is directly beneath the ball’s launch position, on the floor.
- Positive \(x\) points horizontally forwards, in the direction of launch.
- Positive \(y\) points vertically upwards.
- The floor is at \(y=0\).
- The ball leaves the launcher at height \(h\).
The landing position is the horizontal coordinate of the ball’s centre at its first contact with the floor.
This last detail matters. Measuring where the ball eventually comes to rest is not the same as measuring its landing position. The ball might bounce, roll, or slide after its first contact.
Calculate the predicted landing position
For a horizontal launch, the projectile has:
- Initial horizontal velocity \(v_x\).
- Initial vertical velocity \(v_{y0}=0\).
- Initial height \(h\).
- Vertical acceleration \(a_y=-g\).
We assume air resistance is negligible and gravitational acceleration is constant.
First, consider the vertical motion.
Using the constant-acceleration equation:
\[ y=y_0+v_{y0}t+\frac12a_yt^2 \]
At the instant the ball reaches the floor, \(y=0\). Substituting the initial conditions gives:
\[ 0=h-\frac12gt^2 \]
Rearranging:
\[ t=\sqrt{\frac{2h}{g}} \]
This is the predicted flight time.
Now consider the horizontal motion. With negligible air resistance, horizontal acceleration is zero, so horizontal velocity remains constant.
Therefore:
\[ x_{\text{pred}}=v_xt \]
Combining the two equations:
\[ \boxed{x_{\text{pred}}=v_x\sqrt{\frac{2h}{g}}} \]
Here:
- \(x_{\text{pred}}\) is the predicted horizontal landing position, in metres.
- \(v_x\) is the initial horizontal velocity, in \(\text{m s}^{-1}\).
- \(h\) is the vertical launch height, in metres.
- \(g\) is gravitational acceleration, approximately \(9.8\text{ m s}^{-2}\).
- \(t\) is the flight time, in seconds.
If the projectile is launched at an angle rather than horizontally, the same comparison method applies. You first resolve the initial velocity into horizontal and vertical components, then calculate the time when the projectile reaches the landing height.
For an initial vertical velocity \(v_{y0}\), with upward positive:
\[ t_{\text{land}}= \frac{v_{y0}+\sqrt{v_{y0}^2+2gh}}{g} \]
The predicted horizontal landing position is then:
\[ x_{\text{pred}}=v_{x0}t_{\text{land}} \]
The key is to calculate the position that the experiment is actually measuring.
02Collect an experimental landing position
A theoretical prediction is only useful if you can make a fair comparison with an actual measurement.
Consider a hypothetical school investigation using a horizontal projectile launcher.
The apparatus consists of:
- A launcher clamped to the edge of a bench.
- A small ball released horizontally.
- A measuring tape laid along the floor.
- A sheet of carbon paper or another suitable impact-marking surface.
- A fixed camera positioned beside the projectile’s plane of motion.
- A calibration ruler positioned in the same plane as the projectile.
The ball leaves the launcher, travels forwards, and strikes the floor. The impact surface records its first contact position.
The horizontal distance from the origin to this mark is the experimental landing position.
What should you measure?
You need three main measurements.
| Measurement | Symbol | How to obtain it |
|---|---|---|
| Launch height | \(h\) | Measure from the ball’s centre at release to the floor |
| Initial horizontal velocity | \(v_x\) | Use a photogate or calibrated video |
| Experimental landing position | \(x_{\text{exp}}\) | Measure from the floor origin to the first-contact mark |
Be careful with the launch height. Measuring from the bottom of the bench rather than the centre of the ball at release introduces an avoidable error.
Likewise, the horizontal velocity should represent the ball’s velocity immediately after leaving the launcher.
A video recording can help you extract both the initial velocity and the landing position. However, the camera must be positioned correctly to avoid distorting the measurements.
See How to Choose a Camera Position for Projectile Motion for a more detailed explanation of camera alignment and calibration.
Why repeat the experiment?
Imagine performing one launch and recording a landing position of \(1.20\text{ m}\).
That single measurement might be affected by:
- A slightly different launch speed.
- Small variations in the launch direction.
- Difficulty identifying the exact centre of the impact mark.
- An inconsistent release.
- Random measurement variation.
You cannot tell whether the result is representative of the experiment.
Instead, perform several launches under the same conditions.
An illustrative dataset might look like this:
| Trial | Experimental landing position (m) |
|---|---|
| 1 | 1.20 |
| 2 | 1.22 |
| 3 | 1.21 |
| 4 | 1.23 |
| 5 | 1.20 |
These measurements give you a better estimate of the experimental landing position.
The mean is:
\[ \bar{x}_{\text{exp}}= \frac{\sum x_i}{n} \]
where \(n\) is the number of trials.
Repeated measurements do not eliminate systematic errors, but they help identify random variation and make your comparison more reliable.
One important experimental improvement is to measure the first impact rather than the ball’s final resting position. An initial design that measures where the ball stops can accidentally include bouncing and rolling. Using an impact-marking surface fixes this problem by recording the event the theoretical model predicts.
03Account for uncertainty in the predicted position
Here is where many students make an important mistake.
They calculate a theoretical landing position to three decimal places and treat it as an exact answer.
It isn’t.
The measured launch height and horizontal velocity both have uncertainty. These uncertainties affect the predicted landing position.
For a horizontal launch:
\[ x_{\text{pred}}=v_x\sqrt{\frac{2h}{g}} \]
The fractional uncertainty in the predicted position can be estimated using:
\[ \boxed{ \frac{\delta x_{\text{pred}}}{x_{\text{pred}}} \approx \frac{\delta v_x}{v_x} + \frac12\frac{\delta h}{h} } \]
Here:
- \(\delta x_{\text{pred}}\) is the absolute uncertainty in predicted landing position.
- \(\delta v_x\) is the absolute uncertainty in horizontal velocity.
- \(\delta h\) is the absolute uncertainty in launch height.
This uses a conservative maximum-fractional-uncertainty approach, where the contributions are added.
We are treating \(g\) as known. If an investigation specifies uncertainty in \(g\), that contribution must also be considered.
Notice that the uncertainty in height contributes only half its fractional value.
This happens because the predicted range is proportional to the square root of height, rather than height itself.
Account for uncertainty in the experimental result
For repeated landing measurements, a useful school-level estimate of the spread is half the range:
\[ \delta x_{\text{spread}}= \frac{x_{\max}-x_{\min}}{2} \]
where:
- \(x_{\max}\) is the largest measured landing position.
- \(x_{\min}\) is the smallest measured landing position.
If there is a separate uncertainty associated with reading the measuring instrument, you can add it conservatively:
\[ \delta x_{\text{exp}}= \delta x_{\text{spread}}+ \delta x_{\text{reading}} \]
This gives a simple, conservative estimate rather than a formal statistical confidence interval.
You can then report the experimental result as:
\[ x_{\text{exp}}=\bar{x}_{\text{exp}}\pm\delta x_{\text{exp}} \]
For example, an experimental result might be:
\[ x_{\text{exp}}=1.21\pm0.02\text{ m} \]
This means the estimated experimental landing position is \(1.21\text{ m}\), with an uncertainty of approximately \(0.02\text{ m}\).
04Compare the two uncertainty intervals
Now you have two values:
\[ x_{\text{pred}}\pm\delta x_{\text{pred}} \]
and
\[ x_{\text{exp}}\pm\delta x_{\text{exp}} \]
The next step is to compare their uncertainty intervals.
The intervals are:
\[ [x_{\text{pred}}-\delta x_{\text{pred}}, x_{\text{pred}}+\delta x_{\text{pred}}] \]
and
\[ [x_{\text{exp}}-\delta x_{\text{exp}}, x_{\text{exp}}+\delta x_{\text{exp}}] \]
There are two main possibilities.
Case 1: The uncertainty intervals overlap
Consistent within uncertainty
If the intervals overlap, the experimental result is consistent with the theoretical prediction within the estimated uncertainty.
This does not prove that the model is perfectly correct.
It means that the difference between the two central values is small enough to be explained by the uncertainties you have accounted for.
Case 2: The uncertainty intervals do not overlap
Not consistent within uncertainty
If the intervals do not overlap, the experimental result is not consistent with the theoretical prediction within the stated uncertainties.
This is evidence that something needs investigating.
However, do not immediately conclude that the physics model is wrong.
The disagreement might come from:
- An incorrect calibration scale.
- A measurement origin that is incorrectly positioned.
- An inaccurate initial velocity measurement.
- Air resistance.
- A launch angle that differs from the assumed angle.
- An experimental condition that was not included in the model.
A useful mathematical test is to calculate the difference between the central values:
\[ d=x_{\text{exp}}-x_{\text{pred}} \]
For symmetric uncertainty intervals, they overlap when:
\[ \boxed{ |d|\leq \delta x_{\text{exp}}+\delta x_{\text{pred}} } \]
If the absolute difference is greater than the sum of the uncertainties, the intervals do not overlap.
This is a particularly useful decision rule for HSC experimental questions.
Importantly, there is no universal percentage difference that determines whether two results agree. A 5% difference might be acceptable for one investigation but significant for another. You must consider the actual uncertainties.
05Worked example 1: Does a 2 cm difference matter?
A student launches a small steel ball horizontally from a bench.
The measured launch height is:
\[ h=0.800\pm0.005\text{ m} \]
The initial horizontal velocity is:
\[ v_x=2.95\pm0.05\text{ m s}^{-1} \]
The student performs five launches and records the following experimental landing positions:
| Trial | Landing position (m) |
|---|---|
| 1 | 1.20 |
| 2 | 1.22 |
| 3 | 1.21 |
| 4 | 1.23 |
| 5 | 1.20 |
Each landing position has a reading uncertainty of \(\pm0.005\text{ m}\).
Using \(g=9.8\text{ m s}^{-2}\), determine whether the experimental results are consistent with the theoretical prediction.
Step 1
For a horizontal launch:
\[ t=\sqrt{\frac{2h}{g}} \]
Substitute:
\[ t=\sqrt{\frac{2(0.800)}{9.8}} \]
\[ t=0.4041\text{ s} \]
Step 2
Horizontal velocity is constant in the ideal projectile model.
\[ x_{\text{pred}}=v_xt \]
\[ x_{\text{pred}}=2.95(0.4041) \]
\[ \boxed{x_{\text{pred}}=1.192\text{ m}} \]
Now calculate its uncertainty.
\[ \frac{\delta x_{\text{pred}}}{x_{\text{pred}}} = \frac{0.05}{2.95} + \frac12\left(\frac{0.005}{0.800}\right) \]
\[ =0.02008 \]
Therefore:
\[ \delta x_{\text{pred}}=1.192(0.02008) \]
\[ \delta x_{\text{pred}}=0.024\text{ m} \]
The predicted result is:
\[ \boxed{x_{\text{pred}}=1.192\pm0.024\text{ m}} \]
Step 3
\[ \bar{x}_{\text{exp}}= \frac{1.20+1.22+1.21+1.23+1.20}{5} \]
\[ \boxed{\bar{x}_{\text{exp}}=1.212\text{ m}} \]
Now calculate the spread uncertainty:
\[ \delta x_{\text{spread}} = \frac{1.23-1.20}{2} \]
\[ =0.015\text{ m} \]
Add the reading uncertainty:
\[ \delta x_{\text{exp}}=0.015+0.005 \]
\[ \delta x_{\text{exp}}=0.020\text{ m} \]
Therefore:
\[ \boxed{x_{\text{exp}}=1.212\pm0.020\text{ m}} \]
Step 4
The predicted interval is:
\[ 1.192\pm0.024 \]
\[ [1.168,\ 1.216]\text{ m} \]
The experimental interval is:
\[ 1.212\pm0.020 \]
\[ [1.192,\ 1.232]\text{ m} \]
These intervals overlap.
The difference between the central values is:
\[ d=1.212-1.192 \]
\[ d=0.020\text{ m} \]
The combined uncertainty is:
\[ 0.024+0.020=0.044\text{ m} \]
Since:
\[ 0.020<0.044 \]
the results are consistent within uncertainty.
The experimental mean is 2 cm further than the predicted position, but this difference is not significant compared with the combined uncertainty.
05What if the experimental result disagrees?
A disagreement is not just a problem to report. It is information that can help you understand the limitations of your experiment or model.
There are two particularly important possibilities.
Experimental error
An experimental error occurs when the measurement process does not accurately represent the physical quantity being investigated.
For example, suppose the measuring tape’s zero mark is 5 cm away from the correct coordinate origin.
Every landing position will be shifted by approximately 5 cm.
Repeating the experiment will not fix this problem. The measurements might be very consistent, but they will still be systematically wrong.
This is called a systematic error.
Another example is using an incorrect camera calibration scale. If the camera underestimates every horizontal distance, all measured landing positions may be too short.
Limitations of the theoretical model
A theoretical model makes assumptions to simplify the physical situation.
For the standard projectile-motion model, we assume:
- Air resistance is negligible.
- Gravitational acceleration is constant.
- Horizontal acceleration is zero.
- The projectile follows a two-dimensional path.
- No additional forces act on the projectile after release.
These assumptions are useful, but they are not always perfectly true.
Consider two balls with the same diameter but very different masses: a steel ball and a ping-pong ball.
If both are launched horizontally at the same speed from the same height, the ideal projectile model predicts the same landing position.
Why?
Because the model does not include mass in its equations. Both balls have the same initial velocity, experience the same gravitational acceleration, and have the same flight time.
However, air resistance affects the two balls differently.
For similarly shaped balls travelling at similar speeds, the drag forces can be comparable, but the lighter ball experiences a greater acceleration due to that force.
Its horizontal velocity can decrease more significantly during flight.
This means the experimental landing positions might differ even though the theoretical model predicts identical results.
The disagreement is not necessarily a measurement mistake. It may reveal an effect that the simplified model has ignored.
The important task is to distinguish these possibilities.
07Worked example 2: Does air resistance explain the disagreement?
A student investigates two projectiles:
- A steel sphere.
- A ping-pong ball with the same diameter.
Both are launched horizontally from a height of \(0.900\pm0.010\text{ m}\).
Their initial horizontal velocities are independently measured as:
\[ v_x=3.20\pm0.03\text{ m s}^{-1} \]
The ideal projectile model predicts the same landing position for both balls.
The experimental results are:
| Projectile | Trial 1 (m) | Trial 2 (m) | Trial 3 (m) | Trial 4 (m) | Trial 5 (m) |
|---|---|---|---|---|---|
| Steel sphere | 1.36 | 1.37 | 1.38 | 1.36 | 1.37 |
| Ping-pong ball | 1.26 | 1.25 | 1.27 | 1.26 | 1.25 |
Each landing mark has a reading uncertainty of \(\pm0.005\text{ m}\).
Determine whether the results agree with the theoretical prediction and explain what the difference suggests.
Step 1
\[ t=\sqrt{\frac{2h}{g}} \]
\[ t=\sqrt{\frac{2(0.900)}{9.8}} \]
\[ \boxed{t=0.4286\text{ s}} \]
Step 2
\[ x_{\text{pred}}=v_xt \]
\[ x_{\text{pred}}=3.20(0.4286) \]
\[ \boxed{x_{\text{pred}}=1.371\text{ m}} \]
Calculate the uncertainty:
\[ \frac{\delta x_{\text{pred}}}{x_{\text{pred}}} = \frac{0.03}{3.20} + \frac12\left(\frac{0.010}{0.900}\right) \]
\[ =0.01493 \]
Therefore:
\[ \delta x_{\text{pred}}=1.371(0.01493) \]
\[ \delta x_{\text{pred}}\approx0.020\text{ m} \]
The prediction is:
\[ \boxed{x_{\text{pred}}=1.371\pm0.020\text{ m}} \]
Step 3
\[ \bar{x}_{\text{steel}} = \frac{1.36+1.37+1.38+1.36+1.37}{5} \]
\[ \boxed{\bar{x}_{\text{steel}}=1.368\text{ m}} \]
Its spread uncertainty is:
\[ \delta x_{\text{spread}} = \frac{1.38-1.36}{2} \]
\[ =0.010\text{ m} \]
Including the reading uncertainty:
\[ \boxed{x_{\text{steel}}=1.368\pm0.015\text{ m}} \]
Step 4
\[ \bar{x}_{\text{ping-pong}} = \frac{1.26+1.25+1.27+1.26+1.25}{5} \]
\[ \boxed{\bar{x}_{\text{ping-pong}}=1.258\text{ m}} \]
Its uncertainty is:
\[ \delta x_{\text{exp}}= \frac{1.27-1.25}{2}+0.005 \]
\[ \boxed{\delta x_{\text{exp}}=0.015\text{ m}} \]
Therefore:
\[ \boxed{x_{\text{ping-pong}}=1.258\pm0.015\text{ m}} \]
Step 5
For the steel sphere:
\[ |1.368-1.371|=0.003\text{ m} \]
The combined uncertainty is:
\[ 0.020+0.015=0.035\text{ m} \]
The intervals overlap, so the steel sphere’s result is consistent with the ideal model.
For the ping-pong ball:
\[ |1.258-1.371|=0.113\text{ m} \]
The difference is much greater than the combined uncertainty of \(0.035\text{ m}\).
The intervals do not overlap.
The ping-pong ball lands approximately 8.3% shorter than the theoretical prediction.
Conclusion: The ideal projectile model describes the steel sphere reasonably well but does not adequately describe the ping-pong ball under these conditions.
Air resistance is a likely explanation. The lighter ball experiences a greater effect from drag relative to its mass, causing its horizontal motion to depart more significantly from the constant-velocity assumption.
However, the discrepancy alone does not prove that air resistance is responsible. The student should also check the launch speed measurements, camera calibration, and any other ball-dependent experimental effects.
06A practical decision process for experimental comparisons
When you are given an experimental investigation in an HSC Physics question, use the following sequence.
| Step | What to investigate | Why it matters |
|---|---|---|
| 1 | Check the coordinate origin | Both positions must use the same reference point |
| 2 | Calculate the theoretical position | Establishes the model’s prediction |
| 3 | Calculate its uncertainty | The prediction is not perfectly precise |
| 4 | Calculate the experimental mean | Reduces the influence of random variation |
| 5 | Estimate experimental uncertainty | Shows how reliable the measurements are |
| 6 | Compare uncertainty intervals | Determines whether the results are consistent |
| 7 | Investigate disagreement | Identifies experimental errors or model limitations |
There are also some useful patterns to recognise.
| Pattern in the results | Possible explanation |
|---|---|
| Large variation between repeated trials | Random experimental uncertainty |
| All results shifted by a similar distance | Incorrect coordinate origin |
| All results shifted by a similar percentage | Calibration or scale error |
| Disagreement increases for lighter projectiles | Air resistance may be significant |
| Disagreement increases at higher speeds | Air resistance or speed-dependent measurement errors may be important |
These patterns are clues, not automatic diagnoses. You should always explain what additional measurement or experimental check would help distinguish the possible causes.
09Questions and solutions
The following questions develop your ability to compare predicted and measured landing positions, progressing from basic uncertainty calculations to more demanding experimental reasoning.
Question 1
A ball is launched horizontally from a bench of height \(0.80\pm0.01\text{ m}\).
Its measured initial horizontal velocity is:
\[ v_x=2.50\pm0.05\text{ m s}^{-1} \]
Five experimental landing positions are recorded:
| Trial | Landing position (m) |
|---|---|
| 1 | 1.01 |
| 2 | 0.99 |
| 3 | 1.00 |
| 4 | 1.02 |
| 5 | 0.98 |
Each landing position has a reading uncertainty of \(\pm0.005\text{ m}\).
Using \(g=9.8\text{ m s}^{-2}\):
(a) Calculate the theoretical landing position.
(b) Calculate its uncertainty using the conservative fractional-uncertainty method.
(c) Calculate the mean experimental landing position and its uncertainty, using half the range and the reading uncertainty.
(d) Determine whether the experimental result is consistent with the theoretical prediction.
Solution 1
The experimental result is consistent with the theoretical prediction because the two uncertainty intervals overlap.
Step 1: Calculate the theoretical flight time.
\[ t=\sqrt{\frac{2h}{g}} \]
\[ t=\sqrt{\frac{2(0.80)}{9.8}} \]
\[ t=0.4041\text{ s} \]
Step 2: Calculate the predicted landing position.
\[ x_{\text{pred}}=v_xt \]
\[ x_{\text{pred}}=2.50(0.4041) \]
\[ \boxed{x_{\text{pred}}=1.010\text{ m}} \]
Step 3: Calculate the uncertainty in the prediction.
\[ \frac{\delta x_{\text{pred}}}{x_{\text{pred}}} = \frac{0.05}{2.50} + \frac12\left(\frac{0.01}{0.80}\right) \]
\[ =0.02625 \]
Therefore:
\[ \delta x_{\text{pred}}=1.010(0.02625) \]
\[ \boxed{\delta x_{\text{pred}}=0.027\text{ m}} \]
The theoretical result is:
\[ x_{\text{pred}}=1.010\pm0.027\text{ m} \]
Step 4: Calculate the experimental mean.
\[ \bar{x}_{\text{exp}}= \frac{1.01+0.99+1.00+1.02+0.98}{5} \]
\[ \boxed{\bar{x}_{\text{exp}}=1.000\text{ m}} \]
Calculate the half-range uncertainty:
\[ \delta x_{\text{spread}} = \frac{1.02-0.98}{2} \]
\[ =0.020\text{ m} \]
Including the reading uncertainty:
\[ \delta x_{\text{exp}}=0.020+0.005 \]
\[ \boxed{\delta x_{\text{exp}}=0.025\text{ m}} \]
Step 5: Compare the results.
The predicted interval is:
\[ [0.983,\ 1.037]\text{ m} \]
The experimental interval is:
\[ [0.975,\ 1.025]\text{ m} \]
The intervals overlap.
Therefore, the experimental result is consistent with the theoretical prediction within the estimated uncertainty.
The experimental mean is slightly smaller than the prediction, but the difference is not significant compared with the combined uncertainty.
Question 2
A ball is launched horizontally from a height of \(0.800\pm0.005\text{ m}\).
Its initial horizontal velocity is measured independently as:
\[ v_x=3.00\pm0.03\text{ m s}^{-1} \]
Five experimental landing positions are recorded:
| Trial | Landing position (m) |
|---|---|
| 1 | 1.14 |
| 2 | 1.16 |
| 3 | 1.15 |
| 4 | 1.15 |
| 5 | 1.14 |
Each measurement has a reading uncertainty of \(\pm0.005\text{ m}\).
Using \(g=9.8\text{ m s}^{-2}\):
(a) Calculate the predicted landing position and its uncertainty.
(b) Calculate the mean experimental landing position and its uncertainty.
(c) Determine whether the results are consistent.
(d) Assuming the measured flight time is correct, calculate the average horizontal velocity implied by the experimental landing position.
(e) Explain why the disagreement does not, by itself, prove that air resistance is responsible.
Solution 2
The experimental result is not consistent with the theoretical prediction within the stated uncertainties. The measured average horizontal velocity is approximately \(2.84\text{ m s}^{-1}\), suggesting that the ideal constant-horizontal-velocity assumption may not describe the actual motion.
Step 1: Calculate the flight time.
\[ t=\sqrt{\frac{2(0.800)}{9.8}} \]
\[ \boxed{t=0.4041\text{ s}} \]
Step 2: Calculate the predicted landing position.
\[ x_{\text{pred}}=3.00(0.4041) \]
\[ \boxed{x_{\text{pred}}=1.212\text{ m}} \]
Calculate its uncertainty:
\[ \frac{\delta x_{\text{pred}}}{x_{\text{pred}}} = \frac{0.03}{3.00} + \frac12\left(\frac{0.005}{0.800}\right) \]
\[ =0.013125 \]
Therefore:
\[ \delta x_{\text{pred}}=1.212(0.013125) \]
\[ \boxed{\delta x_{\text{pred}}=0.016\text{ m}} \]
Step 3: Calculate the experimental mean.
\[ \bar{x}_{\text{exp}} = \frac{1.14+1.16+1.15+1.15+1.14}{5} \]
\[ \boxed{\bar{x}_{\text{exp}}=1.148\text{ m}} \]
The half-range uncertainty is:
\[ \delta x_{\text{spread}} = \frac{1.16-1.14}{2} \]
\[ =0.010\text{ m} \]
Adding the reading uncertainty:
\[ \boxed{\delta x_{\text{exp}}=0.015\text{ m}} \]
Step 4: Compare the uncertainty intervals.
The predicted interval is:
\[ [1.196,\ 1.228]\text{ m} \]
The experimental interval is:
\[ [1.133,\ 1.163]\text{ m} \]
There is no overlap.
The difference between the central values is:
\[ d=1.148-1.212 \]
\[ \boxed{d=-0.064\text{ m}} \]
The experimental result is approximately 6.4 cm shorter than predicted.
Step 5: Calculate the average horizontal velocity implied by the experiment.
Average horizontal velocity is:
\[ v_{x,\text{avg}}=\frac{\Delta x}{\Delta t} \]
Substituting:
\[ v_{x,\text{avg}}=\frac{1.148}{0.4041} \]
\[ \boxed{v_{x,\text{avg}}=2.84\text{ m s}^{-1}} \]
This is approximately 5.3% lower than the independently measured initial velocity.
The difference is consistent with the possibility that horizontal velocity decreases during flight.
Step 6: Explain the limitation of the conclusion.
Air resistance could cause the horizontal velocity to decrease.
However, an incorrect distance calibration or an inaccurate initial velocity measurement could produce a similar disagreement.
To distinguish these possibilities, the student could track horizontal position at several points throughout the flight and check whether horizontal velocity actually decreases.
An independent calibration check would also help identify systematic measurement errors.
Question 3
A student investigates the relationship between launch height and horizontal landing position.
The same ball is launched horizontally at a constant measured velocity of \(2.64\text{ m s}^{-1}\).
The theoretical predictions and experimental means are:
| Launch height (m) | Predicted landing position (m) | Experimental mean (m) |
|---|---|---|
| 0.45 | 0.800 | 0.760 |
| 0.80 | 1.067 | 1.027 |
| 1.25 | 1.333 | 1.293 |
Each experimental mean has an uncertainty of \(\pm0.010\text{ m}\).
The student suspects that air resistance is causing the disagreement.
However, a separate inspection reveals that the zero mark of the floor measuring tape is positioned \(0.040\text{ m}\) forwards from the correct coordinate origin.
(a) Calculate the difference between the predicted and experimental positions for each launch height.
(b) Explain the pattern in these differences.
(c) Correct the experimental landing positions using the actual coordinate origin.
(d) Explain why this is a systematic measurement error rather than convincing evidence of a limitation in the projectile-motion model.
Solution 3
The disagreement is caused by a systematic coordinate-origin error. Correcting the tape’s zero position makes all three experimental results agree with the theoretical predictions.
Step 1: Calculate the differences.
Use:
\[ d=x_{\text{exp}}-x_{\text{pred}} \]
For the first height:
\[ d=0.760-0.800 \]
\[ d=-0.040\text{ m} \]
For the second height:
\[ d=1.027-1.067 \]
\[ d=-0.040\text{ m} \]
For the third height:
\[ d=1.293-1.333 \]
\[ d=-0.040\text{ m} \]
All three measurements are 0.040 m shorter than predicted.
Step 2: Interpret the pattern.
The difference is approximately the same absolute distance for every launch height.
This suggests an additive measurement error rather than an error that changes with the projectile’s range.
The measuring tape’s zero mark is positioned forwards of the true origin, so every measured distance is underestimated by the same amount.
Step 3: Correct the experimental measurements.
Because the tape zero is 0.040 m forwards from the correct origin, add 0.040 m to each experimental reading.
First height:
\[ x_{\text{corrected}}=0.760+0.040 \]
\[ \boxed{x_{\text{corrected}}=0.800\text{ m}} \]
Second height:
\[ x_{\text{corrected}}=1.027+0.040 \]
\[ \boxed{x_{\text{corrected}}=1.067\text{ m}} \]
Third height:
\[ x_{\text{corrected}}=1.293+0.040 \]
\[ \boxed{x_{\text{corrected}}=1.333\text{ m}} \]
All three corrected measurements match the theoretical predictions.
Step 4: Explain why the model was not the problem.
Air resistance is a physical effect that changes the projectile’s motion.
An incorrectly positioned measuring-tape zero does not change the projectile’s motion at all. It changes only the coordinate assigned to the landing position.
The fact that all three results have the same absolute offset, together with the independent inspection of the tape, identifies the measurement problem.
This is why systematic errors must be investigated before changing the theoretical model.
Question 4
A student uses video analysis to investigate the landing position of the same compact projectile at three different launch speeds.
The theoretical predictions and experimental results are:
| Trial | Predicted position (m) | Experimental mean (m) |
|---|---|---|
| A | \(0.80\pm0.01\) | \(0.74\pm0.01\) |
| B | \(1.20\pm0.01\) | \(1.11\pm0.01\) |
| C | \(1.60\pm0.01\) | \(1.48\pm0.01\) |
(a) Calculate the absolute difference between the theoretical and experimental positions for each trial.
(b) Calculate the percentage shortfall for each trial, using the theoretical prediction as the reference value.
(c) Explain why the pattern suggests a multiplicative systematic error rather than a constant coordinate-origin error.
(d) Identify one independent experimental check that would help determine whether the discrepancy comes from calibration or from a limitation of the projectile-motion model.
Solution 4
All three experimental results underestimate the theoretical predictions by 7.5%. This suggests a common multiplicative systematic error, such as an incorrect distance or velocity calibration, rather than a fixed coordinate-origin offset.
Step 1: Calculate the absolute differences.
Trial A:
\[ 0.80-0.74=0.06\text{ m} \]
Trial B:
\[ 1.20-1.11=0.09\text{ m} \]
Trial C:
\[ 1.60-1.48=0.12\text{ m} \]
The absolute discrepancy increases with the predicted landing distance.
Step 2: Calculate the percentage shortfall.
Use:
\[ \%\text{ shortfall} = \frac{x_{\text{pred}}-x_{\text{exp}}} {x_{\text{pred}}}\times100 \]
For trial A:
\[ \frac{0.80-0.74}{0.80}\times100 \]
\[ \boxed{7.5\%} \]
For trial B:
\[ \frac{1.20-1.11}{1.20}\times100 \]
\[ \boxed{7.5\%} \]
For trial C:
\[ \frac{1.60-1.48}{1.60}\times100 \]
\[ \boxed{7.5\%} \]
The percentage shortfall is identical for all three trials.
Step 3: Interpret the pattern.
A fixed coordinate-origin error would produce approximately the same absolute difference for every measurement.
For example, an incorrectly positioned origin might make all landing positions 0.05 m too short.
Here, the absolute differences increase as the predicted range increases, but the percentage difference remains constant.
This is characteristic of a multiplicative error.
The measurements satisfy:
\[ x_{\text{exp}}\approx0.925x_{\text{pred}} \]
The experimental measurement process is effectively reporting only 92.5% of the predicted distance.
An incorrect camera calibration scale could produce this pattern.
A systematic overestimate of initial horizontal velocity could also make the theoretical predictions consistently too large.
Step 4: Identify an independent check.
One useful test is to place a known-length ruler in the projectile’s plane of motion and independently check the camera’s pixel-to-metre calibration.
The student could also measure the initial horizontal velocity using a separate photogate.
If these independent measurements confirm the original calibration, a limitation in the physical model becomes more plausible.
The data alone cannot uniquely identify the cause. A systematic calibration error and a systematic error in the theoretical inputs must both be considered.
Question 5
A student investigates projectile motion using a moving landing platform.
A ball is launched horizontally from a height of \(1.25\text{ m}\), with an initial horizontal velocity of \(3.00\text{ m s}^{-1}\).
The coordinate origin is on the floor directly beneath the launch point. Positive \(x\) is forwards.
A horizontal trolley carries a wide landing tray. At the instant of launch, the centre of the tray is at:
\[ x_{\text{tray},0}=1.50\text{ m} \]
Two experiments are performed using identical projectile launch conditions.
In experiment A, the trolley moves forwards at:
\[ u_A=0.40\text{ m s}^{-1} \]
In experiment B, the trolley moves forwards at:
\[ u_B=0.70\text{ m s}^{-1} \]
Both trolley speeds remain constant.
The ball’s first-contact position is measured relative to the centre of the moving tray, rather than relative to the laboratory floor.
The measured offsets are:
| Experiment | Measured impact position relative to tray centre |
|---|---|
| A | \(-0.187\pm0.003\text{ m}\) |
| B | \(-0.338\pm0.003\text{ m}\) |
For this question, treat the launch conditions and trolley velocities as exact. Only the measured impact offsets have the stated uncertainty.
A student compares the theoretical landing position with the two negative experimental offsets and concludes that the projectile model is completely wrong.
(a) Calculate the theoretical flight time and the predicted landing position relative to the laboratory floor.
(b) Calculate the position of the trolley’s centre at the instant of impact in each experiment.
(c) Calculate the predicted impact position relative to the centre of the trolley in each experiment.
(d) Use the difference between the two experimental impact offsets to calculate an independent estimate of the projectile’s flight time. Compare this with the theoretical flight time.
(e) Reconstruct the experimental landing position relative to the laboratory floor for both experiments. Determine whether the results agree with the theoretical prediction.
(f) Explain precisely why comparing the negative trolley-relative measurements directly with the positive laboratory-frame prediction is incorrect.
Solution 5
Both experimental results are consistent with the theoretical prediction. The negative measurements describe positions relative to a moving trolley, not positions relative to the laboratory floor.
This question tests an important distinction: the position of an object depends on the reference frame from which it is measured.
Step 1: Calculate the theoretical flight time.
For a horizontal launch:
\[ t=\sqrt{\frac{2h}{g}} \]
Substitute:
\[ t=\sqrt{\frac{2(1.25)}{9.8}} \]
\[ \boxed{t=0.5051\text{ s}} \]
Step 2: Calculate the predicted laboratory landing position.
The horizontal velocity remains constant in the ideal model.
\[ x_{\text{pred}}=v_xt \]
\[ x_{\text{pred}}=3.00(0.5051) \]
\[ \boxed{x_{\text{pred}}=1.515\text{ m}} \]
This position is measured from the laboratory origin on the floor.
Step 3: Calculate the trolley positions at impact.
The trolley moves at constant velocity, so:
\[ x_{\text{tray}}=x_{\text{tray},0}+ut \]
For experiment A:
\[ x_{\text{tray,A}}=1.50+0.40(0.5051) \]
\[ \boxed{x_{\text{tray,A}}=1.702\text{ m}} \]
For experiment B:
\[ x_{\text{tray,B}}=1.50+0.70(0.5051) \]
\[ \boxed{x_{\text{tray,B}}=1.854\text{ m}} \]
The trolley moves further forwards in experiment B because its velocity is greater.
Step 4: Calculate the predicted positions relative to the trolley.
The relative position of the ball with respect to the trolley is:
\[ x_{\text{ball/tray}} = x_{\text{ball}}-x_{\text{tray}} \]
For experiment A:
\[ x_{\text{ball/tray,A}} = 1.515-1.702 \]
\[ \boxed{x_{\text{ball/tray,A}}=-0.187\text{ m}} \]
For experiment B:
\[ x_{\text{ball/tray,B}} = 1.515-1.854 \]
\[ \boxed{x_{\text{ball/tray,B}}=-0.338\text{ m}} \]
These predictions match the experimental measurements.
The negative signs simply mean that the ball strikes behind the centre of the moving tray.
They do not mean the ball has travelled backwards relative to the laboratory.
Step 5: Use the difference between the experimental offsets to estimate flight time.
This is the particularly interesting part of the investigation.
Both experiments have the same initial projectile position, the same initial horizontal velocity, and the same flight time.
The difference between the two trolley-relative impact positions is therefore caused by the difference in trolley displacement.
We can write:
\[ \Delta x_{\text{relative}} = -(u_B-u_A)t \]
Rearranging:
\[ t= -\frac{\Delta x_{\text{relative}}} {u_B-u_A} \]
Calculate the difference between the measured offsets:
\[ \Delta x_{\text{relative}} = -0.338-(-0.187) \]
\[ \Delta x_{\text{relative}}=-0.151\text{ m} \]
The difference in trolley velocities is:
\[ u_B-u_A=0.70-0.40 \]
\[ =0.30\text{ m s}^{-1} \]
Substitute:
\[ t=-\frac{-0.151}{0.30} \]
\[ \boxed{t_{\text{exp}}=0.503\text{ s}} \]
The conservative uncertainty in the difference between the two measured offsets is:
\[ 0.003+0.003=0.006\text{ m} \]
Therefore, the uncertainty in this estimated flight time is:
\[ \delta t=\frac{0.006}{0.30} \]
\[ \delta t=0.020\text{ s} \]
The experimental estimate is:
\[ \boxed{t_{\text{exp}}=0.503\pm0.020\text{ s}} \]
This is consistent with the theoretical flight time of \(0.5051\text{ s}\).
The two experiments have therefore provided an independent check of the predicted time of flight.
Step 6: Reconstruct the laboratory landing positions.
We know:
\[ x_{\text{ball}}= x_{\text{tray}}+x_{\text{ball/tray}} \]
For experiment A:
\[ x_{\text{ball,A}} = 1.50+0.40(0.5051)-0.187 \]
\[ \boxed{x_{\text{ball,A}}=1.515\text{ m}} \]
For experiment B:
\[ x_{\text{ball,B}} = 1.50+0.70(0.5051)-0.338 \]
\[ \boxed{x_{\text{ball,B}}=1.516\text{ m}} \]
The small difference between these reconstructed positions is within the stated measurement uncertainty and rounding.
Both results agree with the predicted laboratory landing position of approximately \(1.515\text{ m}\).
Step 7: Explain the original mistake.
The student compared two different types of position.
The theoretical prediction was measured relative to the laboratory floor.
The experimental measurements were measured relative to the centre of a moving trolley.
These are not directly comparable.
The trolley moves during the projectile’s flight, so its position must be included when converting the experimental measurements into laboratory coordinates.
This is an example of relative position. The same principle applies to moving observers, boats, aircraft, and other situations involving two-dimensional motion. See Relative Position and Motion in Two Dimensions for HSC Physics for further work with reference frames.
The deeper lesson is that a disagreement between two numerical values is meaningless until you establish that they represent the same physical quantity, measured from the same reference frame.
10Where to take this next
Once the predicted and experimental landing positions have been compared, the next useful step is to investigate the projectile’s entire trajectory rather than just its final position.
A landing position can agree with a prediction by coincidence. Tracking horizontal and vertical position at several times provides a stronger test of the model.
You can then investigate whether horizontal velocity remains constant and whether vertical acceleration remains approximately equal to gravitational acceleration.
That moves the investigation from comparing one endpoint to testing the physics of the entire motion.