How to Test Horizontal and Vertical Motion Independence

Two identical balls leave a bench at the same instant. One is dropped vertically. The other is launched horizontally at \(2.4\ \mathrm{m\,s^{-1}}\). After \(0.30\ \mathrm{s}\), which ball is closer to the ground?

Under the ideal projectile-motion model, neither. Both have fallen exactly the same vertical distance. The second ball has simply travelled further sideways.

But how could you demonstrate that experimentally, rather than just trusting an equation?

To test the independence of horizontal and vertical motion, change a ball’s horizontal launch speed while keeping its initial height and vertical velocity the same. Film the motion, measure vertical displacement at matching times, and check whether the vertical results agree within experimental uncertainty. Comparing landing times is useful, but tracking the entire fall provides stronger evidence.

01What are we actually testing?

Imagine the two balls just after release.

The dropped ball has no initial horizontal or vertical velocity. The launched ball has horizontal velocity, but its initial vertical velocity is also zero.

If air resistance is negligible, gravity acts vertically downwards on both. Neither ball needs a horizontal force to keep moving sideways.

Predict what happens if you double the launched ball’s horizontal speed. Does its vertical acceleration double, stay unchanged, or decrease?

It stays unchanged. Increasing horizontal velocity does not change the gravitational force on the ball, so both balls still accelerate downwards at approximately \(9.8\ \mathrm{m\,s^{-2}}\).

That is the independence we want to investigate.

For this experiment, use the following sign convention:

• \(x\) is horizontal displacement, positive to the right, measured in metres (m).
• \(d\) is vertical displacement from the release point, positive downwards, measured in metres (m).
• \(t\) is time since the ball actually left its launcher or support, measured in seconds (s).
• \(u_x\) is the initial horizontal velocity, measured in metres per second (\(\mathrm{m\,s^{-1}}\)).
• \(g\) is the magnitude of gravitational acceleration, approximately \(9.8\ \mathrm{m\,s^{-2}}\).

For a ball launched horizontally with zero initial vertical velocity, the ideal equations are

\[ x=u_xt \]

and

\[ d=\frac12gt^2 \]

The first equation tells us how far the ball travels sideways. The second tells us how far it falls.

Notice what is missing from the second equation: \(u_x\).

This is a prediction of the model, not experimental proof. Our job is to see whether measured motion supports it.

For the wider theory behind these equations, see Projectile Motion: Horizontal and Vertical Motion Explained. Here, we’ll concentrate on testing the model.

02The first experiment: drop one ball and launch another

The simplest apparatus uses two identical small balls released from the same height.

One ball is held stationary by an electromagnet. The other sits in a spring-powered launcher pointing horizontally.

A shared electronic trigger switches off the electromagnet and releases the spring mechanism. Ideally, both balls begin moving simultaneously.

The spring gives the launched ball horizontal velocity. The dropped ball begins falling without horizontal velocity.

What should happen?

Suppose both balls have a vertical distance of \(1.25\ \mathrm{m}\) to fall before first contact with the floor.

Strictly, this distance is measured from each ball’s starting centre position to its centre position at first contact. That accounts for the radius of the ball.

Using \(d=\frac12gt^2\), the predicted flight time is

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

Substituting:

\[ \begin{aligned} t&=\sqrt{\frac{2(1.25)}{9.8}}\\ &=0.505\ \mathrm{s} \end{aligned} \]

Both balls should reach the floor after approximately \(0.505\ \mathrm{s}\), provided their initial vertical velocities are zero.

The launched ball might travel a metre sideways before hitting the floor. That horizontal distance does not enter the flight-time calculation.

Why equal landing times aren’t enough

A first version of the experiment could simply record whether the two balls hit the floor together.

If they do, the result supports the prediction. But there are weaknesses.

The electromagnet might release its ball slightly later than the spring launcher. The spring might also give its ball a small upward or downward kick.

And even if both balls land simultaneously, they could have travelled through different vertical positions during the flight.

Equal landing times establish only that the balls reached the same final vertical level at approximately the same time. They do not establish that their vertical motion was identical throughout.

The improved experiment measures their positions between release and landing.

03The improved experiment: collect position data from video

Use a stationary slow-motion camera to record the two balls from the side.

The camera should face the vertical plane of motion directly rather than looking diagonally along the flight path. Place a calibration ruler in the same plane as the balls.

This allows you to convert distances in the video into metres without introducing unnecessary perspective errors.

A frame rate of \(240\) frames per second gives a time interval of

\[ \Delta t=\frac{1}{240}=0.00417\ \mathrm{s} \]

between consecutive frames.

The release frame matters. The electronic trigger is not necessarily the instant the ball begins moving.

Identify the first movement of each ball, and use that to establish its time origin.

What should you keep constant?

The independent variable is the horizontal launch speed.

The main dependent measurements are vertical displacement at specified times and the resulting vertical acceleration. Landing time provides another useful comparison.

Keep the following conditions as consistent as possible:

• The balls’ size, mass, shape, and surface material.
• Initial vertical position and initial vertical velocity.
• The release mechanism’s orientation and the floor level.
• Camera position, calibration, and timing procedure.
• Air conditions and any spin introduced during launch.

The most important control is initial vertical velocity. A launcher that points slightly downwards can make a ball fall faster even when gravity is behaving exactly as expected.

Use several spring settings to obtain different horizontal speeds. Include a vertically dropped ball as the control condition.

Repeat each setting, ideally at least five times, so one unusual launch doesn’t determine the conclusion. Five is a practical example, not a prescribed HSC minimum.

What data do you record?

For each recording:

1. Identify the ball’s centre in the release frame.
2. Record its horizontal and vertical centre positions at regular time intervals.
3. Convert image distances to metres using the calibration ruler.
4. Subtract the initial coordinates to obtain displacement.
5. Compare vertical displacements at the same elapsed times for different horizontal speeds.

For example, if the ball’s vertical centre coordinate is measured positive downwards, then

\[ d(t)=y(t)-y(0) \]

where \(y(t)\) is the measured vertical coordinate at time \(t\), and \(y(0)\) is its release coordinate.

Use the same coordinate convention for every trial.

04An illustrative dataset: does horizontal speed change the fall?

The following dataset is hypothetical. It is designed to demonstrate the analysis and is not a record of a performed experiment.

Three release conditions are tested:

• A ball dropped vertically, with \(u_x=0\).
• A slow horizontal launch, with \(u_x\approx1.2\ \mathrm{m\,s^{-1}}\).
• A fast horizontal launch, with \(u_x\approx2.4\ \mathrm{m\,s^{-1}}\).

All balls begin at the same vertical level with approximately zero initial vertical velocity.

The experiment estimates each vertical displacement with an uncertainty of \(\pm0.015\ \mathrm{m}\), excluding any additional uncertainty from aligning time origins between recordings.

Hypothetical vertical displacement data

The ideal predictions use \(g=9.8\ \mathrm{m\,s^{-2}}\) and zero initial vertical velocity.

The results aren’t numerically identical. Real measurements rarely are.

But the vertical displacements are very similar, even though the horizontal velocities differ substantially.

How much does timing uncertainty matter?

At \(240\) frames per second, one frame represents about \(0.0042\ \mathrm{s}\).

Suppose the release frame is identified one frame too early or late. Near \(t=0.40\ \mathrm{s}\), the resulting change in the predicted falling distance is approximately

\[ \begin{aligned} \Delta d&\approx gt\Delta t\\ &=9.8(0.40)(0.0042)\\ &\approx0.016\ \mathrm{m} \end{aligned} \]

Here, \(\Delta t\) is the timing error. The approximation assumes that the error is small compared with the elapsed time.

A one-frame timing error can therefore have an effect comparable to the position-measurement uncertainty.

If both balls are recorded in the same frame after simultaneous release, much of this timing uncertainty is shared. For separate trials, incorrectly aligned release frames can create an apparent difference between the vertical trajectories.

This is why an experimental conclusion should mention uncertainty in both position and timing.

05When the results seem to disagree: check the initial conditions

Suppose a horizontally launched ball consistently falls further than the dropped ball at the same time.

It would be tempting to conclude that its horizontal motion somehow increased its downward acceleration.

There’s another possibility: the launcher gave the ball a small downward velocity at release.

These two explanations make different predictions.

If the only difference is initial downward velocity, the extra displacement grows in proportion to time.

If the only difference is a constant extra downward acceleration, the extra displacement grows in proportion to time squared.

That difference gives us a diagnostic test.

06How to decide whether your experiment supports the model

A useful analysis separates three questions.

1. Were the starting conditions genuinely equivalent?

Check that the balls began at the same vertical level, at the same actual time, and with the same initial vertical velocity.

If not, different vertical displacements may simply reflect different starting conditions.

For a horizontal launch, the initial vertical velocity should be approximately zero, not merely assumed to be zero because the launcher looks level.

2. Do the vertical displacement-time results agree?

Compare \(d\) at equal elapsed times, preferably across several horizontal launch speeds.

If the measurements agree within uncertainty and follow approximately \(d=\frac12gt^2\), they support the gravity-only model.

A graph of \(d\) against \(t^2\) should be approximately a straight line with gradient \(g/2\) when the initial vertical velocity is zero.

However, a straight-line graph from one launch speed cannot establish independence. Independence requires a comparison between conditions with different horizontal velocities.

3. Could a measurement problem explain the difference?

Look for systematic errors before proposing a new physical effect.

The last case is especially important.

Two objects can reach the same destination at the same time without having followed identical motion between those events.

What the ideal model leaves out

In the usual HSC gravity-only projectile model, horizontal and vertical motion can be calculated independently.

Real balls also experience aerodynamic forces.

Air resistance depends on motion relative to the air. Once a ball is falling and travelling horizontally, the aerodynamic force may have both horizontal and vertical components. Spin can create additional forces.

Consequently, a large change in horizontal speed can affect the actual vertical motion through air resistance or lift.

That doesn’t make the gravity-only equations useless. It establishes their limits.

A well-written conclusion might be:

“The measured vertical displacements were consistent across the tested horizontal launch speeds within experimental uncertainty. The results support the independence of horizontal and vertical motion under approximately negligible air resistance. They do not establish that aerodynamic effects are absent or that the model applies at all speeds.”

07Questions and solutions

The questions below move from a standard HSC calculation to more demanding experimental reasoning. All datasets are hypothetical, and any measurement assumptions needed to obtain a unique answer are given.

Question 1

A ball is launched horizontally at \(2.0\ \mathrm{m\,s^{-1}}\) while an identical ball is dropped vertically from the same height at the same instant.

Both begin with zero vertical velocity, and each has a vertical distance of \(1.25\ \mathrm{m}\) to fall before first contact with the floor.

Assuming negligible air resistance and \(g=9.8\ \mathrm{m\,s^{-2}}\):

a. Calculate how far each ball falls in \(0.30\ \mathrm{s}\).

b. Calculate the launched ball’s horizontal displacement in that time.

c. Determine whether the balls reach the floor simultaneously.

Solution 1

Both balls fall \(0.441\ \mathrm{m}\) in \(0.30\ \mathrm{s}\), the launched ball travels \(0.600\ \mathrm{m}\) horizontally, and both reach the floor after approximately \(0.505\ \mathrm{s}\).

For part (a), use the vertical displacement equation:

\[ \begin{aligned} d&=\frac12gt^2\\ &=\frac12(9.8)(0.30)^2\\ &=0.441\ \mathrm{m} \end{aligned} \]

Both balls have the same initial vertical velocity and vertical acceleration, so they have equal vertical displacements at equal times.

For part (b):

\[ \begin{aligned} x&=u_xt\\ &=2.0(0.30)\\ &=0.600\ \mathrm{m} \end{aligned} \]

For part (c), the time needed to fall \(1.25\ \mathrm{m}\) is

\[ \begin{aligned} t&=\sqrt{\frac{2d}{g}}\\ &=\sqrt{\frac{2(1.25)}{9.8}}\\ &=0.505\ \mathrm{s} \end{aligned} \]

The tempting prediction is that the launched ball takes longer because it travels a greater distance along its curved path.

But the ground is reached when the vertical displacement equals \(1.25\ \mathrm{m}\). The length of the curved path does not determine that time.

Question 2

A laboratory group uses an electronic trigger to start a timer, launch a ball horizontally, and release a second ball from an electromagnet.

Both balls have a vertical distance of \(1.25\ \mathrm{m}\) to fall before floor contact.

The horizontally launched ball begins moving at the trigger instant. The electromagnet may take some time to release its ball.

The following displacement measurements are made relative to the electronic trigger:

Assume that both balls begin with zero vertical velocity when released, and that each accelerates downwards at \(9.8\ \mathrm{m\,s^{-2}}\). Treat the readings as sufficiently precise for estimating the release delay.

a. Estimate the electromagnet’s release delay using both rows of data.

b. Predict the two floor-contact times measured from the electronic trigger.

c. Explain why comparing displacements at equal trigger times is not a valid test of independence here, and describe a correction.

Solution 2

The electromagnet releases its ball approximately \(0.040\ \mathrm{s}\) late. The launched ball reaches the floor about \(0.505\ \mathrm{s}\) after the trigger, and the dropped ball about \(0.545\ \mathrm{s}\) after the trigger.

For part (a), determine how long the dropped ball has actually been falling at each observation.

From \(d=\frac12gt^2\):

\[ t_{\mathrm{fall}}=\sqrt{\frac{2d}{g}} \]

For the first reading:

\[ \begin{aligned} t_{\mathrm{fall}}&=\sqrt{\frac{2(0.125)}{9.8}}\\ &\approx0.160\ \mathrm{s} \end{aligned} \]

But the timer reads \(0.200\ \mathrm{s}\).

The estimated delay is

\[ 0.200-0.160=0.040\ \mathrm{s} \]

For the second reading:

\[ \begin{aligned} t_{\mathrm{fall}}&=\sqrt{\frac{2(0.635)}{9.8}}\\ &\approx0.360\ \mathrm{s} \end{aligned} \]

The corresponding delay is

\[ 0.400-0.360=0.040\ \mathrm{s} \]

Both measurements are consistent with the same delay.

For part (b), each ball requires

\[ \begin{aligned} T&=\sqrt{\frac{2(1.25)}{9.8}}\\ &=0.505\ \mathrm{s} \end{aligned} \]

of actual falling time.

The launched ball starts at the trigger, so its predicted contact time is \(0.505\ \mathrm{s}\).

The dropped ball begins \(0.040\ \mathrm{s}\) later, giving

\[ 0.505+0.040=0.545\ \mathrm{s} \]

For part (c), the experiment initially compares unequal periods of falling.

At a trigger time of \(0.40\ \mathrm{s}\), one ball has fallen for \(0.40\ \mathrm{s}\), while the other has fallen for only \(0.36\ \mathrm{s}\).

The apparent difference therefore does not demonstrate that horizontal speed affects vertical motion.

The correction is to identify each ball’s actual first movement frame and calculate elapsed time separately from its release.

The model-based delay estimate should also be checked directly from the video, rather than simply assumed correct because it explains the measurements.

The trap is treating an electronic signal as proof that two mechanical actions occurred simultaneously.

Question 3

A student films two balls from a stationary camera.

The dropped ball moves in a vertical plane closer to the camera. The horizontally launched ball moves in a parallel plane further away.

Both balls begin moving simultaneously with zero initial vertical velocity.

The student records the following downward movements of the ball centres:

A calibration ruler in the dropped ball’s plane gives \(200\) pixels per metre.

A separate ruler in the launched ball’s plane gives \(180\) pixels per metre. Each ball remains in its respective plane throughout the flight.

A student uses the first scale for both balls and concludes that the horizontally launched ball has a smaller vertical acceleration.

a. Explain the problem with this conclusion.

b. Calculate both balls’ vertical displacements at \(0.40\ \mathrm{s}\), using the correct scales.

c. Estimate each ball’s vertical acceleration from the \(0.40\ \mathrm{s}\) measurements, assuming constant acceleration.

d. State what this corrected comparison does and does not establish.

Solution 3

The apparent difference is caused by applying the wrong spatial scale to the farther ball. With the correct calibration, both balls have approximately the same measured vertical acceleration, close to \(9.8\ \mathrm{m\,s^{-2}}\).

For part (a), the camera records image positions in pixels rather than physical distances.

An object further from the camera generally has a smaller image scale. The same real displacement can therefore occupy fewer pixels.

Using one conversion factor for both planes introduces a systematic error.

For part (b), convert pixels to metres.

Dropped ball:

\[ \begin{aligned} d_D&=\frac{157}{200}\\ &=0.785\ \mathrm{m} \end{aligned} \]

Launched ball:

\[ \begin{aligned} d_L&=\frac{141}{180}\\ &=0.7833\ \mathrm{m} \end{aligned} \]

The corrected displacements differ by only about

\[ 0.785-0.7833=0.0017\ \mathrm{m} \]

or \(1.7\ \mathrm{mm}\).

For part (c), since both balls begin with zero vertical velocity,

\[ a=\frac{2d}{t^2} \]

For the dropped ball:

\[ \begin{aligned} a_D&=\frac{2(0.785)}{(0.40)^2}\\ &=9.81\ \mathrm{m\,s^{-2}} \end{aligned} \]

For the launched ball:

\[ \begin{aligned} a_L&=\frac{2(0.7833)}{(0.40)^2}\\ &=9.79\ \mathrm{m\,s^{-2}} \end{aligned} \]

Both estimates are consistent with approximately \(9.8\ \mathrm{m\,s^{-2}}\).

For comparison, incorrectly converting the farther ball’s \(141\) pixels using the nearer scale would give

\[ d_{\mathrm{wrong}}=\frac{141}{200}=0.705\ \mathrm{m} \]

This would produce an apparent acceleration of only

\[ a_{\mathrm{wrong}} =\frac{2(0.705)}{0.40^2} =8.81\ \mathrm{m\,s^{-2}} \]

That apparent reduction is an artefact of the measurement procedure.

For part (d), the corrected results support similar vertical motion in these two conditions.

They do not establish exact equality, because pixel-location, calibration, and timing uncertainties have not been quantified. Nor do they prove independence under different launch speeds or aerodynamic conditions.

The deeper problem is that a camera measures projections of motion. Correct physical reasoning cannot rescue incorrectly calibrated measurements.

Question 4

A dropped ball and a second ball from a nominally horizontal launcher begin moving at the same instant.

Both have a vertical distance of \(1.50\ \mathrm{m}\) to fall before floor contact.

The launched ball has a horizontal velocity component of \(2.5\ \mathrm{m\,s^{-1}}\), but its initial vertical velocity is not measured.

Hypothetical measurements show:

Assume that both balls undergo constant vertical acceleration during the flight. Each displacement reading has uncertainty \(\pm0.005\ \mathrm{m}\).

Consider two competing models:

• Model A: Both balls accelerate downwards at \(9.8\ \mathrm{m\,s^{-2}}\), but the launched ball has an unknown initial vertical velocity.
• Model B: The launched ball begins with zero initial vertical velocity but has a different constant downward acceleration.

a. Determine which model is consistent with both readings.

b. Find the launched ball’s initial vertical velocity under the supported model.

c. Calculate the predicted flight time of each ball and the launched ball’s horizontal range.

d. Explain whether their different flight times contradict independence.

Solution 4

Model A is consistent with both readings: the launched ball has an initial upward velocity of \(0.10\ \mathrm{m\,s^{-1}}\) while experiencing the same downward acceleration as the dropped ball. It consequently remains airborne slightly longer.

For part (a), first examine Model B.

If the launched ball starts with zero vertical velocity, the \(0.20\ \mathrm{s}\) measurement implies

\[ \begin{aligned} a&=\frac{2d}{t^2}\\ &=\frac{2(0.176)}{(0.20)^2}\\ &=8.8\ \mathrm{m\,s^{-2}} \end{aligned} \]

This acceleration predicts a displacement at \(0.40\ \mathrm{s}\) of

\[ \begin{aligned} d&=\frac12(8.8)(0.40)^2\\ &=0.704\ \mathrm{m} \end{aligned} \]

But the measured displacement is \(0.744\ \mathrm{m}\).

The discrepancy of \(0.040\ \mathrm{m}\) is much larger than the stated measurement uncertainty. Model B cannot explain both readings with one constant acceleration.

For part (b), apply Model A.

The vertical motion is

\[ d=u_dt+\frac12gt^2 \]

Using the \(0.20\ \mathrm{s}\) measurement:

\[ \begin{aligned} 0.176&=0.20u_d+0.196\\ 0.20u_d&=-0.020\\ u_d&=-0.10\ \mathrm{m\,s^{-1}} \end{aligned} \]

The negative sign means the initial vertical velocity is upwards.

At \(0.40\ \mathrm{s}\), the model predicts

\[ \begin{aligned} d&=(-0.10)(0.40)+4.9(0.40)^2\\ &=-0.040+0.784\\ &=0.744\ \mathrm{m} \end{aligned} \]

It matches the second measurement.

For part (c), the dropped ball’s flight time is

\[ \begin{aligned} T_D&=\sqrt{\frac{2(1.50)}{9.8}}\\ &=0.553\ \mathrm{s} \end{aligned} \]

For the launched ball:

\[ 1.50=-0.10T_L+4.9T_L^2 \]

Rearranging:

\[ 4.9T_L^2-0.10T_L-1.50=0 \]

Using the positive solution of the quadratic:

\[ \begin{aligned} T_L&=\frac{0.10+\sqrt{29.41}}{9.8}\\ &=0.564\ \mathrm{s} \end{aligned} \]

Its horizontal range is

\[ \begin{aligned} R&=u_xT_L\\ &=2.5(0.564)\\ &\approx1.41\ \mathrm{m} \end{aligned} \]

For part (d), the launched ball remains airborne approximately \(0.011\ \mathrm{s}\) longer because it began with an upward vertical velocity.

The two balls are not starting with identical vertical conditions.

Their different flight times therefore do not contradict independent horizontal and vertical motion.

The tempting explanation is that the launched ball falls more slowly because of its sideways speed. The two-time comparison instead identifies a different initial vertical velocity.

A useful revision would be to inspect the first video frames for upward motion, check the launcher’s orientation, and repeat after correcting it.

Question 5

A research group constructs an improved drop-and-launch experiment using two identical balls.

Ball D is dropped vertically from rest. Ball F is launched with a much greater horizontal velocity.

Both begin at the same centre height and at the same instant.

The vertical fall distance to first floor contact is \(1.225\ \mathrm{m}\).

The video produces two surprising observations:

• Both balls reach the floor after exactly \(0.500\ \mathrm{s}\).
• At \(t=0.250\ \mathrm{s}\), ball F is \(0.060\ \mathrm{m}\) higher than ball D.

The uncertainty in the measured vertical separation at \(0.250\ \mathrm{s}\) is \(\pm0.012\ \mathrm{m}\).

One student concludes that equal landing times prove independence.

Another concludes that the intermediate separation proves that greater horizontal speed increases gravitational acceleration.

For this question, assume that:

• Ball D has downward acceleration \(g=9.8\ \mathrm{m\,s^{-2}}\).
• Each ball has constant vertical acceleration during the flight, although those accelerations may differ.
• Ball F’s initial vertical velocity is unknown.
• Timing uncertainty is negligible compared with the stated separation uncertainty.
• Both balls start and finish at equal vertical levels.

a. Assess both students’ conclusions.

b. Determine Ball F’s initial vertical velocity and vertical acceleration from the measurements.

c. Show when the vertical separation between the balls is greatest, under the stated constant-acceleration assumptions.

d. Explain why the result still cannot establish that horizontal speed caused the difference in vertical acceleration. Propose a suitable experimental improvement.

Solution 5

Neither student’s conclusion is justified.

Equal landing times do not establish identical vertical motion, and a difference in vertical acceleration does not, by itself, show that horizontal speed caused that difference.

Under the stated assumptions, Ball F must begin with an upward vertical velocity of \(0.48\ \mathrm{m\,s^{-1}}\) and experience a downward acceleration of \(11.72\ \mathrm{m\,s^{-2}}\).

The key is that its initial upward motion and larger downward acceleration can compensate over the complete flight, producing the same landing time.

Step 1: Express the vertical motion of both balls.

Use downward displacement as positive.

Ball D is dropped from rest:

\[ d_D=\frac12gt^2 \]

Ball F has unknown initial downward velocity \(u\) and unknown downward acceleration \(a\):

\[ d_F=ut+\frac12at^2 \]

If \(u\) is negative, Ball F initially moves upwards.

Define the difference in downward displacement as

\[ \Delta d=d_F-d_D \]

Therefore,

\[ \Delta d=ut+\frac12(a-g)t^2 \]

Step 2: Use the equal landing times.

At \(T=0.500\ \mathrm{s}\), both balls have fallen the same vertical distance.

Thus

\[ \Delta d(T)=0 \]

Substituting:

\[ 0=uT+\frac12(a-g)T^2 \]

Since \(T\neq0\), divide by \(T\):

\[ u=-\frac12(a-g)T \]

With \(T=0.500\ \mathrm{s}\):

\[ u=-0.250(a-g) \]

This is the first constraint.

It says that if Ball F has greater downward acceleration, it must also start with an upward velocity to reach the floor at the same time.

Step 3: Use the midpoint measurement.

At \(t=0.250\ \mathrm{s}\), Ball F is \(0.060\ \mathrm{m}\) higher.

Because downward displacement is positive,

\[ \Delta d=-0.060\ \mathrm{m} \]

Substitute the expression for \(u\) into the difference equation:

\[ \Delta d=\frac12(a-g)t(t-T) \]

At \(t=0.250\ \mathrm{s}\) and \(T=0.500\ \mathrm{s}\):

\[ -0.060 =\frac12(a-g)(0.250)(-0.250) \]

So

\[ -0.060=-0.03125(a-g) \]

Therefore,

\[ a-g=1.92\ \mathrm{m\,s^{-2}} \]

and

\[ \begin{aligned} a&=9.8+1.92\\ &=11.72\ \mathrm{m\,s^{-2}} \end{aligned} \]

Now recover the initial vertical velocity:

\[ \begin{aligned} u&=-0.250(1.92)\\ &=-0.48\ \mathrm{m\,s^{-1}} \end{aligned} \]

Ball F begins with a vertical velocity of \(0.48\ \mathrm{m\,s^{-1}}\) upwards.

Step 4: Check that both observations are satisfied.

At the midpoint, Ball D has fallen

\[ \begin{aligned} d_D&=4.9(0.250)^2\\ &=0.30625\ \mathrm{m} \end{aligned} \]

Ball F has fallen

\[ \begin{aligned} d_F&=(-0.48)(0.250)\\ &\quad+\frac12(11.72)(0.250)^2\\ &=0.24625\ \mathrm{m} \end{aligned} \]

Their separation is

\[ 0.30625-0.24625=0.060\ \mathrm{m} \]

At landing, Ball F’s downward displacement is

\[ \begin{aligned} d_F&=(-0.48)(0.500)\\ &\quad+\frac12(11.72)(0.500)^2\\ &=-0.240+1.465\\ &=1.225\ \mathrm{m} \end{aligned} \]

Both constraints are satisfied.

The midpoint discrepancy is also larger than its stated uncertainty. In the constant-acceleration model, the inferred extra acceleration is approximately \(1.92\pm0.38\ \mathrm{m\,s^{-2}}\), considering only the uncertainty in the measured gap.

Step 5: Determine when the separation is greatest.

From the equal-start and equal-finish conditions, we established

\[ \Delta d=\frac12(a-g)t(t-T) \]

The magnitude of the separation is therefore proportional to

\[ t(T-t) \]

For times between \(0\) and \(T\),

\[ t(T-t)=\left(\frac T2\right)^2 -\left(t-\frac T2\right)^2 \]

The squared term is never negative. Subtracting it gives the largest possible value when it equals zero.

That occurs at

\[ t=\frac T2 \]

Therefore, under these assumptions, the maximum vertical separation occurs halfway through the flight:

\[ t=0.250\ \mathrm{s} \]

This is the surprising result.

The balls are together at release and together again at landing, but their vertical separation reaches its maximum between those times.

A landing-time-only experiment would completely miss it.

Step 6: Decide what the experiment actually establishes.

The results are inconsistent with both balls having identical initial vertical velocities and identical constant vertical accelerations.

They can be explained by the combination of an initial upward kick and a larger downward acceleration for Ball F.

But what caused those differences?

Possibilities include a launcher that initially directs Ball F upwards, additional vertical forces associated with its spin, or another force acting on Ball F during flight.

An additional vertical force that does not depend on horizontal velocity would produce a different downward acceleration without demonstrating a coupling between the horizontal and vertical components of motion.

The measurements therefore identify a difference in motion, not its physical cause.

To test whether horizontal speed itself is responsible, the group should repeat the experiment at several horizontal speeds while controlling initial vertical velocity, spin, ball properties, air conditions, and release timing.

It should also measure complete displacement-time curves to test whether acceleration is approximately constant.

Only then can changes in vertical motion be meaningfully associated with changes in horizontal launch speed.

The general lesson is that matching endpoints can hide different motion histories. A strong experimental test measures the behaviour between those endpoints and controls the conditions that could produce the same observations for different reasons.

08What to investigate next

Once your measurements support the same vertical displacement-time relationship across different horizontal speeds, you can extract another useful result: an experimental estimate of \(g\).

Instead of assuming \(9.8\ \mathrm{m\,s^{-2}}\), fit your vertical displacement measurements to the constant-acceleration model and determine the acceleration from the data.

That leads naturally to How to Find g from a Projectile Video: HSC Physics Guide.

The important decision is whether your video is good enough to distinguish a small physical effect from timing error, launch conditions, and measurement uncertainty. That’s what turns a demonstration of projectile motion into a useful physics experiment.