How to Estimate Launch-Speed Uncertainty from Repeated Ranges

Five launches from the same projectile launcher land at slightly different distances. The measured ranges are 2.40 m, 2.43 m, 2.41 m, 2.46 m, and 2.42 m.

You can use these measurements to estimate the launch speed. But how certain should you be about your answer?

The key is that projectile range depends on the square of launch speed. This means the percentage uncertainty in launch speed is approximately half the percentage uncertainty in range, provided the launch angle is known and the usual ideal projectile assumptions apply.

Here’s how to collect the data, calculate the speed, estimate its uncertainty, and recognise when your answer looks precise but is actually wrong.

01Start with the relationship between range and launch speed

Imagine launching a ball at an angle above the horizontal. It travels forwards, rises, reaches a maximum height, and eventually lands.

Two things determine how far it travels horizontally:

• Its horizontal velocity determines how quickly it moves forwards.
• Its time in the air determines how long it can keep moving forwards.

For a projectile launched and landing at the same height, with negligible air resistance, these combine into the familiar range equation:

\[ R=\frac{u^2\sin(2\theta)}{g} \]

where:

• \(R\) is horizontal range, measured in metres (m).
• \(u\) is initial launch speed, measured in \(\mathrm{m\,s^{-1}}\).
• \(\theta\) is launch angle above the horizontal.
• \(g\) is gravitational acceleration, taken as \(9.8\ \mathrm{m\,s^{-2}}\).

You might already recognise this equation from studying horizontal range of a projectile.

However, we’re interested in using the equation in the opposite direction.

We have measured the range and want to work backwards to find the launch speed.

Rearranging:

\[ \boxed{u=\sqrt{\frac{Rg}{\sin(2\theta)}}} \]

This is the equation we’ll use throughout the article.

Why does range uncertainty produce a smaller speed uncertainty?

Before looking at the calculation, make a prediction.

Suppose your measured range has a percentage uncertainty of 4%. Would you expect your calculated launch speed to have a percentage uncertainty of 4% as well?

Not quite.

Look at the rearranged equation:

\[ u=\sqrt{\frac{Rg}{\sin(2\theta)}} \]

If the angle and gravitational acceleration are fixed, everything except \(R\) is constant.

Therefore:

\[ u\propto\sqrt R \]

Launch speed depends on the square root of range, not range itself.

For small measurement uncertainties, this gives:

\[ \boxed{ \frac{\Delta u}{u} \approx \frac{1}{2}\frac{\Delta R}{R} } \]

Here, \(\Delta u\) is the absolute uncertainty in launch speed, and \(\Delta R\) is the absolute uncertainty in range.

The important relationship is:

\[ \boxed{ \%\text{ uncertainty in }u \approx \frac{1}{2} (\%\text{ uncertainty in }R) } \]

So, if range has a 4% uncertainty, launch speed has approximately a 2% uncertainty.

This is the central idea behind estimating launch-speed uncertainty from repeated range measurements.

02Collect repeated range measurements

Before calculating anything, you need reliable measurements.

A single launch gives you one range value. However, small variations in the launcher, landing position, and measurement technique can produce slightly different results each time.

Repeating the experiment helps you estimate how consistent your measurements are.

Set up the experiment

For a reliable investigation:

1. Set the launcher to a fixed angle.
2. Keep the launch mechanism at the same setting for every trial.
3. Use the same projectile.
4. Keep the launch and landing surfaces at the same height.
5. Measure the horizontal distance from the launch point to the projectile’s first landing position.
6. Repeat the launch at least five times and record every result.

The range must be measured horizontally. Do not measure the curved distance travelled by the projectile.

A useful way to establish the starting position is to use a plumb line to identify the point directly beneath the launch exit.

If you’re extracting landing positions from video, camera position also matters. An oblique camera angle can distort the measured distances even when the projectile behaves consistently. See how to choose a camera position for projectile motion.

Calculate the mean range

Once you’ve collected your measurements, calculate their average.

The mean range is:

\[ \boxed{ \bar R=\frac{\sum R_i}{n} } \]

where:

• \(\bar R\) is the mean range.
• \(R_i\) represents each individual range measurement.
• \(n\) is the number of trials.

The mean gives you a representative range for the repeated experiment.

Estimate the range uncertainty

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

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

Here:

• \(R_{\max}\) is the largest measured range.
• \(R_{\min}\) is the smallest measured range.

You can then express your measured range as:

\[ R=\bar R\pm\Delta R \]

This is a simple estimate of the observed spread. It is not a formal 95% confidence interval.

If your investigation specifies standard deviation or standard error instead, use the required statistical method. Do not automatically treat half the range as the uncertainty in the mean.

03Convert range uncertainty into launch-speed uncertainty

Now we can combine the two parts of the investigation.

You have:

1. A mean range.
2. An estimated uncertainty in that range.
3. A known launch angle.
4. A value for gravitational acceleration.

First, calculate the launch speed:

\[ u=\sqrt{\frac{\bar Rg}{\sin(2\theta)}} \]

Then calculate the percentage uncertainty in range:

\[ \%\Delta R=\frac{\Delta R}{\bar R}\times100 \]

Because speed depends on the square root of range, divide this percentage by two:

\[ \%\Delta u= \frac{1}{2}\times\%\Delta R \]

Finally, convert the percentage uncertainty back into an absolute uncertainty:

\[ \boxed{ \Delta u=u\left(\frac{\Delta R}{2\bar R}\right) } \]

Your final result is:

\[ \boxed{u=\bar u\pm\Delta u} \]

The uncertainty is expressed in the same units as speed, \(\mathrm{m\,s^{-1}}\).

04What if the launch angle also has uncertainty?

So far, we’ve treated the launch angle as exact.

Real measurements aren’t quite that convenient.

Suppose you set a launcher to \(40.0^\circ\), but your protractor only allows you to determine the angle within \(0.5^\circ\).

Your angle is now:

\[ \theta=40.0^\circ\pm0.5^\circ \]

This matters because the launch speed equation also contains \(\sin(2\theta)\).

If the angle is uncertain, you cannot account for the entire speed uncertainty using range measurements alone.

For a small uncertainty in range and a known angle uncertainty, a useful approach is to calculate the extreme possible speeds.

The method is:

• Calculate the maximum speed using the largest possible range and the angle that produces the smallest \(\sin(2\theta)\).
• Calculate the minimum speed using the smallest possible range and the angle that produces the largest \(\sin(2\theta)\).

You must consider how the sine function changes over the particular angle interval. The direction of the effect changes around \(45^\circ\).

05Understand what your uncertainty actually tells you

A small uncertainty does not automatically mean your calculated speed is accurate.

You need to understand what causes the variation in your measurements and whether the projectile model actually describes the experiment.

There are three important types of limitations to consider.

Random uncertainty

Random uncertainty causes measurements to vary unpredictably between trials.

Examples include:

• Small differences in identifying the projectile’s landing position.
• Slight variations in the launch mechanism.
• Difficulty reading the measuring scale precisely.
• Small changes in the projectile’s release conditions.

Repeating the experiment helps reveal this variation.

A larger number of measurements can also improve an estimate of the mean when the appropriate statistical method is used.

Systematic uncertainty

Systematic uncertainty shifts measurements consistently in one direction.

For example, imagine that your measuring tape starts 5 cm to the right of the actual launch origin.

Every recorded range will be 5 cm too short.

Repeating the experiment 20 times won’t fix this problem. You’ll simply obtain 20 consistently biased measurements.

Other examples include:

• An incorrectly calibrated measuring scale.
• A camera positioned at an angle that distorts horizontal distances.
• An incorrectly measured launch angle.
• A consistent error in identifying the launch point.

Model limitations

The projectile range equation assumes:

• Negligible air resistance.
• Constant gravitational acceleration.
• No significant sideways motion.
• The projectile launches and lands at the same vertical height.

The final assumption is especially important.

A ball launched from a tabletop and landing on the floor does not satisfy the same-height condition.

You cannot automatically use:

\[ R=\frac{u^2\sin(2\theta)}{g} \]

for that experiment.

You must account for the vertical displacement and calculate the flight time using vertical motion.

We’ll test this distinction in the final practice question.

A subtle issue: Does the launcher produce exactly the same speed every time?

Repeated measurements are often treated as though the projectile has exactly the same launch speed in every trial.

That might not be true.

Suppose the launcher produces slightly different speeds, \(u_1,u_2,u_3,\ldots\), across successive shots.

Because range depends on speed squared, averaging the ranges and then converting the mean range into speed gives the root-mean-square launch speed:

\[ u_{\mathrm{RMS}}= \sqrt{\frac{u_1^2+u_2^2+\cdots+u_n^2}{n}} \]

This is not exactly the same as the ordinary arithmetic mean of the individual speeds.

For small variations, the difference is usually small. However, if the launcher produces substantially different speeds, you should recognise that the spread may represent genuine variation in launch performance rather than just measurement uncertainty.

A useful alternative is to calculate an individual speed from each measured range, then calculate the mean and spread of those inferred speeds.

06Questions and solutions

The following questions build from a standard launch-speed calculation to more demanding experimental reasoning.

Question 1

A projectile is launched at \(30.0^\circ\) above the horizontal. It lands at the same height from which it was launched.

Five repeated measurements give the following horizontal ranges:

\[ 2.42,\ 2.46,\ 2.44,\ 2.40,\ 2.43\ \mathrm m \]

Using \(g=9.8\ \mathrm{m\,s^{-2}}\):

(a) Calculate the mean range.

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

(c) Calculate the launch speed.

(d) Determine the absolute uncertainty in launch speed.

Solution 1

The launch speed is approximately:

\[ \boxed{u=(5.24\pm0.03)\ \mathrm{m\,s^{-1}}} \]

The calculation requires the mean range, its half-range uncertainty, and the square-root relationship between range and speed.

(a) Calculate the mean range.

\[ \bar R= \frac{2.42+2.46+2.44+2.40+2.43}{5} \]

\[ \boxed{\bar R=2.43\ \mathrm m} \]

(b) Calculate the absolute range uncertainty.

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

\[ =\frac{2.46-2.40}{2} \]

\[ \boxed{\Delta R=0.03\ \mathrm m} \]

(c) Calculate the launch speed.

Use:

\[ u=\sqrt{\frac{Rg}{\sin(2\theta)}} \]

Substitute:

\[ u=\sqrt{ \frac{2.43\times9.8}{\sin60^\circ} } \]

\[ u=\sqrt{27.50} \]

\[ \boxed{u=5.24\ \mathrm{m\,s^{-1}}} \]

(d) Calculate the absolute speed uncertainty.

First, calculate the percentage uncertainty in range:

\[ \%\Delta R=\frac{0.03}{2.43}\times100 \]

\[ \%\Delta R=1.23\% \]

The percentage uncertainty in speed is half this value:

\[ \%\Delta u=\frac{1.23}{2}=0.62\% \]

Convert this into an absolute uncertainty:

\[ \Delta u=5.24\times0.0062 \]

\[ \Delta u\approx0.03\ \mathrm{m\,s^{-1}} \]

Therefore:

\[ \boxed{ u=(5.24\pm0.03)\ \mathrm{m\,s^{-1}} } \]

The speed uncertainty is smaller than the range uncertainty as a percentage because the speed depends on the square root of range.

Question 2

A student performs a projectile investigation using a fixed launch angle of \(35.0^\circ\).

Their repeated measurements produce:

\[ R=3.20\pm0.08\ \mathrm m \]

The student calculates the launch speed and makes the following statement:

“The range has a percentage uncertainty of 2.5%, so the launch speed must also have a percentage uncertainty of 2.5%.”

(a) Calculate the launch speed.

(b) Determine the correct percentage uncertainty in launch speed.

(c) Calculate the absolute uncertainty in launch speed.

(d) Explain why the student’s statement is incorrect.

Use \(g=9.8\ \mathrm{m\,s^{-2}}\).

Solution 2

The correct result is:

\[ \boxed{ u=(5.78\pm0.07)\ \mathrm{m\,s^{-1}} } \]

The student’s mistake is treating the relationship between range and speed as linear when it is actually quadratic.

(a) Calculate the launch speed.

Use:

\[ u=\sqrt{\frac{Rg}{\sin(2\theta)}} \]

Substitute:

\[ u=\sqrt{ \frac{3.20\times9.8}{\sin70^\circ} } \]

\[ u=\sqrt{33.37} \]

\[ \boxed{u=5.78\ \mathrm{m\,s^{-1}}} \]

(b) Calculate the percentage uncertainty in range.

\[ \%\Delta R= \frac{0.08}{3.20}\times100 \]

\[ \boxed{\%\Delta R=2.5\%} \]

Since speed depends on the square root of range:

\[ \%\Delta u=\frac{2.5}{2} \]

\[ \boxed{\%\Delta u=1.25\%} \]

(c) Calculate the absolute uncertainty in speed.

\[ \Delta u=5.78\times\frac{1.25}{100} \]

\[ \Delta u=0.0723\ \mathrm{m\,s^{-1}} \]

Therefore:

\[ \boxed{\Delta u=0.07\ \mathrm{m\,s^{-1}}} \]

(d) Explain the student’s mistake.

The student’s prediction would be correct if launch speed were directly proportional to range.

However:

\[ R\propto u^2 \]

Rearranging gives:

\[ u\propto\sqrt R \]

A change in range therefore produces a smaller fractional change in speed.

There is also a units issue worth noticing. The range uncertainty is \(0.08\ \mathrm m\), whereas the speed uncertainty must be measured in \(\mathrm{m\,s^{-1}}\). You cannot simply copy the numerical uncertainty from one physical quantity to another.

Question 3

A student performs six projectile launches at a fixed angle of \(30.0^\circ\).

Their measured ranges are:

\[ 2.88,\ 2.90,\ 2.89,\ 2.91,\ 2.88,\ 2.90\ \mathrm m \]

They calculate a small uncertainty and conclude that their launch-speed estimate must be highly accurate.

However, they discover that the zero mark on their measuring tape was positioned 0.060 m to the right of the true vertical projection of the launch point.

Assume the 0.060 m offset is known exactly.

(a) Calculate the mean measured range.

(b) Determine the half-range uncertainty.

(c) Calculate the corrected mean range.

(d) Calculate the corrected launch speed and its uncertainty.

(e) Explain why repeating the experiment would not remove the original error.

Solution 3

The corrected launch-speed estimate is:

\[ \boxed{ u=(5.78\pm0.02)\ \mathrm{m\,s^{-1}} } \]

The measuring tape introduced a systematic error, so the recorded ranges must be corrected before calculating the speed.

(a) Calculate the mean measured range.

\[ \bar R_{\mathrm{measured}}= \frac{2.88+2.90+2.89+2.91+2.88+2.90}{6} \]

\[ \boxed{ \bar R_{\mathrm{measured}}=2.893\ \mathrm m } \]

(b) Calculate the half-range uncertainty.

\[ \Delta R=\frac{2.91-2.88}{2} \]

\[ \boxed{\Delta R=0.015\ \mathrm m} \]

(c) Correct the mean range.

The measuring tape begins 0.060 m to the right of the actual launch origin.

Consequently, every recorded range is 0.060 m shorter than the true range.

Therefore:

\[ R_{\mathrm{corrected}}=2.893+0.060 \]

\[ \boxed{R_{\mathrm{corrected}}=2.953\ \mathrm m} \]

(d) Calculate the corrected launch speed and uncertainty.

Using the corrected range:

\[ u=\sqrt{ \frac{2.953\times9.8}{\sin60^\circ} } \]

\[ \boxed{u=5.78\ \mathrm{m\,s^{-1}}} \]

The known systematic offset has been corrected. Assuming it has no remaining uncertainty, the repeatability uncertainty is still estimated from the measured spread.

\[ \%\Delta R= \frac{0.015}{2.953}\times100 \]

\[ \%\Delta R=0.508\% \]

Therefore:

\[ \%\Delta u=\frac{0.508}{2} \]

\[ \%\Delta u=0.254\% \]

Convert to absolute uncertainty:

\[ \Delta u=5.78\times0.00254 \]

\[ \boxed{\Delta u\approx0.02\ \mathrm{m\,s^{-1}}} \]

The corrected result is:

\[ \boxed{ u=(5.78\pm0.02)\ \mathrm{m\,s^{-1}} } \]

(e) Explain why repeating the experiment does not remove the error.

The measuring tape’s incorrect starting position shifts every measurement by the same amount.

This is a systematic error, not random variation.

Repeating the experiment might make the mean measured range more consistent, but it will not correct the incorrectly positioned zero mark.

This question illustrates an important experimental distinction: a small spread of measurements demonstrates good repeatability, not necessarily good accuracy.

Question 4

A physics class investigates whether complementary launch angles produce the same horizontal range when launch speed is constant.

The projectile is launched and lands at the same height in both investigations. Air resistance is negligible.

Two sets of hypothetical measurements are collected.

Experiment A: Launch angle \(30.0^\circ\)

Experiment B: Launch angle \(60.0^\circ\)

(a) Calculate the mean range and half-range uncertainty for each experiment.

(b) Calculate the launch speed and its uncertainty for each experiment.

(c) Determine whether the two results are consistent with the prediction that complementary launch angles produce equal ranges for the same launch speed.

(d) Suggest two possible experimental reasons for the difference between the results.

Use \(g=9.8\ \mathrm{m\,s^{-2}}\).

Solution 4

The estimated launch speeds are:

\[ \boxed{u_A=(5.21\pm0.02)\ \mathrm{m\,s^{-1}}} \]

\[ \boxed{u_B=(5.29\pm0.03)\ \mathrm{m\,s^{-1}}} \]

These results are not consistent within their estimated repeatability uncertainties. This suggests an experimental difference or a failure of one or more assumptions.

(a) Calculate the mean range and uncertainty for each experiment.

For Experiment A:

\[ \bar R_A= \frac{2.38+2.40+2.42+2.39+2.41}{5} \]

\[ \boxed{\bar R_A=2.40\ \mathrm m} \]

The half-range uncertainty is:

\[ \Delta R_A=\frac{2.42-2.38}{2} \]

\[ \boxed{\Delta R_A=0.02\ \mathrm m} \]

For Experiment B:

\[ \bar R_B= \frac{2.44+2.48+2.46+2.50+2.47}{5} \]

\[ \boxed{\bar R_B=2.47\ \mathrm m} \]

The half-range uncertainty is:

\[ \Delta R_B=\frac{2.50-2.44}{2} \]

\[ \boxed{\Delta R_B=0.03\ \mathrm m} \]

(b) Calculate the launch speed and uncertainty for each experiment.

For Experiment A:

\[ u_A=\sqrt{ \frac{2.40\times9.8}{\sin60^\circ} } \]

\[ \boxed{u_A=5.21\ \mathrm{m\,s^{-1}}} \]

Its percentage range uncertainty is:

\[ \%\Delta R_A= \frac{0.02}{2.40}\times100 \]

\[ =0.833\% \]

Therefore:

\[ \%\Delta u_A=0.417\% \]

The absolute uncertainty is:

\[ \Delta u_A=5.21\times0.00417 \]

\[ \boxed{\Delta u_A=0.02\ \mathrm{m\,s^{-1}}} \]

For Experiment B:

\[ u_B=\sqrt{ \frac{2.47\times9.8}{\sin120^\circ} } \]

Remember:

\[ \sin120^\circ=\sin60^\circ \]

Therefore:

\[ \boxed{u_B=5.29\ \mathrm{m\,s^{-1}}} \]

Its percentage range uncertainty is:

\[ \%\Delta R_B= \frac{0.03}{2.47}\times100 \]

\[ =1.215\% \]

Therefore:

\[ \%\Delta u_B=0.608\% \]

The absolute uncertainty is:

\[ \Delta u_B=5.29\times0.00608 \]

\[ \boxed{\Delta u_B=0.03\ \mathrm{m\,s^{-1}}} \]

(c) Determine whether the results are consistent.

The uncertainty intervals are approximately:

For Experiment A:

\[ 5.19\leq u_A\leq5.23 \]

For Experiment B:

\[ 5.26\leq u_B\leq5.32 \]

These intervals do not overlap.

The difference between the estimated speeds is:

\[ 5.29-5.21=0.08\ \mathrm{m\,s^{-1}} \]

This is larger than the combined half-range uncertainty estimates.

Therefore, the results are not consistent with the same launch speed within the estimated repeatability.

However, half-range estimates are not formal confidence intervals. The result does not prove that the underlying launch speeds were different.

(d) Suggest two possible experimental reasons.

One possibility is that the launcher produced different initial speeds when its angle was changed. Using the same launcher setting does not automatically guarantee identical launch conditions.

Another possibility is that the launch angle was measured incorrectly or the launch and landing heights were not actually equal.

Other possibilities include systematic range measurement errors or differences in camera calibration.

The tempting conclusion is that complementary angles must always produce equal ranges. However, this prediction only applies when the launch speeds are equal and the other projectile assumptions are satisfied.

Question 5

A projectile is launched from a platform 0.80 m above the landing surface.

Its launch angle is \(30.0^\circ\), and five repeated horizontal range measurements are:

\[ 3.79,\ 3.83,\ 3.81,\ 3.80,\ 3.82\ \mathrm m \]

A student calculates the launch speed using:

\[ R=\frac{u^2\sin(2\theta)}{g} \]

They obtain a small uncertainty and conclude that their answer is reliable.

Another student argues that the calculation is invalid because the projectile does not land at the same height from which it was launched.

(a) Calculate the mean range and its half-range uncertainty.

(b) Explain why the standard same-height range equation cannot be used.

(c) Derive an equation for launch speed using the measured range, launch angle, and vertical height difference.

(d) Calculate the correct launch speed.

(e) Estimate the uncertainty in launch speed using the extreme range measurements.

(f) Explain why the original calculation can produce a substantially different answer despite the small spread in repeated measurements.

Use \(g=9.8\ \mathrm{m\,s^{-2}}\), and assume negligible air resistance.

Solution 5

The correct launch-speed estimate is:

\[ \boxed{ u=(5.62\pm0.02)\ \mathrm{m\,s^{-1}} } \]

The original calculation is invalid because the projectile lands below its launch point. The extra falling distance increases its time in the air, so the same-height range equation does not describe the experiment.

(a) Calculate the mean range and its uncertainty.

The mean range is:

\[ \bar R= \frac{3.79+3.83+3.81+3.80+3.82}{5} \]

\[ \boxed{\bar R=3.81\ \mathrm m} \]

The half-range uncertainty is:

\[ \Delta R=\frac{3.83-3.79}{2} \]

\[ \boxed{\Delta R=0.02\ \mathrm m} \]

Therefore:

\[ R=3.81\pm0.02\ \mathrm m \]

(b) Explain why the standard range equation cannot be used.

The equation:

\[ R=\frac{u^2\sin(2\theta)}{g} \]

is derived for a projectile that lands at the same vertical height from which it was launched.

Here, the projectile lands 0.80 m below its launch point.

The projectile therefore spends additional time falling after passing the launch height.

This changes the total flight time and consequently changes the horizontal range.

We need to return to the horizontal and vertical equations of motion.

(c) Derive the correct launch-speed equation.

Take the launch point as the vertical origin.

Choose upwards as the positive vertical direction.

The vertical displacement at landing is:

\[ \Delta y=-h \]

where \(h=0.80\ \mathrm m\).

The initial horizontal and vertical velocity components are:

\[ u_x=u\cos\theta \]

\[ u_y=u\sin\theta \]

From horizontal motion:

\[ R=u\cos\theta\,T \]

Rearrange to find flight time:

\[ T=\frac{R}{u\cos\theta} \]

Now use the vertical displacement equation:

\[ \Delta y=u_yT-\frac12gT^2 \]

Substitute the known vertical displacement and initial vertical velocity:

\[ -h=u\sin\theta\,T-\frac12gT^2 \]

Replace \(T\) with the expression from horizontal motion:

\[ -h= u\sin\theta \left(\frac{R}{u\cos\theta}\right) - \frac12g \left(\frac{R}{u\cos\theta}\right)^2 \]

Simplify:

\[ -h=R\tan\theta- \frac{gR^2}{2u^2\cos^2\theta} \]

Rearrange to isolate \(u^2\):

\[ R\tan\theta+h= \frac{gR^2}{2u^2\cos^2\theta} \]

Therefore:

\[ \boxed{ u= \sqrt{ \frac{gR^2} {2\cos^2\theta(R\tan\theta+h)} } } \]

This equation accounts for the lower landing height.

(d) Calculate the correct launch speed.

Substitute:

• \(R=3.81\ \mathrm m\)
• \(h=0.80\ \mathrm m\)
• \(\theta=30.0^\circ\)
• \(g=9.8\ \mathrm{m\,s^{-2}}\)

\[ u= \sqrt{ \frac{9.8(3.81)^2} {2\cos^2 30^\circ (3.81\tan30^\circ+0.80)} } \]

Using:

\[ \cos^230^\circ=0.75 \]

and:

\[ \tan30^\circ=0.577 \]

gives:

\[ u=\sqrt{31.62} \]

\[ \boxed{u=5.62\ \mathrm{m\,s^{-1}}} \]

(e) Estimate the uncertainty in launch speed.

Because the speed equation is more complicated than the same-height equation, we should not automatically use the rule that percentage uncertainty in speed is half the percentage uncertainty in range.

Instead, calculate the speeds at the extreme measured ranges.

For the minimum range, \(R=3.79\ \mathrm m\):

\[ u_{\min}= \sqrt{ \frac{9.8(3.79)^2} {2(0.75)(3.79\tan30^\circ+0.80)} } \]

\[ \boxed{u_{\min}=5.60\ \mathrm{m\,s^{-1}}} \]

For the maximum range, \(R=3.83\ \mathrm m\):

\[ u_{\max}= \sqrt{ \frac{9.8(3.83)^2} {2(0.75)(3.83\tan30^\circ+0.80)} } \]

\[ \boxed{u_{\max}=5.64\ \mathrm{m\,s^{-1}}} \]

Estimate the uncertainty using half the difference:

\[ \Delta u=\frac{5.64-5.60}{2} \]

\[ \boxed{\Delta u=0.02\ \mathrm{m\,s^{-1}}} \]

The final result is:

\[ \boxed{ u=(5.62\pm0.02)\ \mathrm{m\,s^{-1}} } \]

(f) Explain why the original calculation gives a different answer.

The student used the same-height equation:

\[ u=\sqrt{\frac{Rg}{\sin(2\theta)}} \]

Substituting the measured mean range:

\[ u= \sqrt{ \frac{3.81\times9.8}{\sin60^\circ} } \]

\[ \boxed{u=6.57\ \mathrm{m\,s^{-1}}} \]

This is substantially higher than the correct result of 5.62 m/s.

Why?

The same-height model assumes the projectile returns to its original height before landing. It therefore predicts a shorter flight time than the actual experiment allows.

To explain the measured horizontal range using this incorrectly shortened flight time, the student calculates a larger launch speed.

The repeated measurements are still consistent with one another. The problem is not the spread of the data. It is the model used to interpret them.

This is the important lesson: a small uncertainty does not rescue an incorrect physical model.

07What to do with your result next

Once you have estimated launch speed and its uncertainty, the next useful step is to test how well your experimental measurements agree with the projectile-motion model.

You can compare the measured launch speed with an independently calculated value, investigate whether the launcher produces consistent speeds, or examine how changing launch angle affects range.

The key is to keep three things separate:

• Measurement uncertainty tells you how much your repeated readings vary.
• Systematic uncertainty tells you whether your measurements are consistently shifted.
• Model limitations tell you whether your equations accurately describe the physical experiment.

Getting all three right is what turns a collection of projectile range measurements into a defensible physics result.