How to Find g from a Projectile Video: HSC Physics Guide

A ball is thrown across a room, and you record its flight in slow motion. You can see it rise, reach its highest point, and fall. But how can you use that video to measure gravitational acceleration rather than simply describe the ball’s path?

To find \(g\) from a projectile video, measure the ball’s vertical position at known times, plot vertical position against time, and fit a quadratic equation:

\[ y=A+Bt+Ct^2 \]

With upwards defined as positive, the magnitude of gravitational acceleration is:

\[ g=-2C \]

You can also calculate \(g\) directly from three equally spaced video frames. The key is to measure how the ball’s vertical displacement changes between equal time intervals.

Here’s a prediction worth making first: if you throw the ball upwards twice as fast, should your measured value of \(g\) change?

It shouldn’t. The ball will rise higher and stay airborne longer, but its downward acceleration remains approximately \(9.8\ \mathrm{m\,s^{-2}}\) near Earth’s surface, assuming air resistance is negligible. Your experiment needs to separate that acceleration from the initial throwing speed.

01Set up the video so position means something

A camera doesn’t directly measure metres. It records pixels.

Before you can calculate acceleration, you need to turn those pixels into real distances and establish how much time passes between frames.

Arrange the apparatus

Imagine throwing a small, brightly coloured foam ball across a laboratory.

The ball travels from left to right through a vertical plane. A camera on a tripod faces that plane directly from the side.

A metre ruler is positioned vertically beside the flight path, at the same distance from the camera as the ball. It must not obstruct the ball.

For reliable measurements:

• Keep the camera stationary, level, and perpendicular to the plane of motion.
• Place the calibration ruler at the same depth as the projectile.
• Record the complete flight at a known capture frame rate.
• Use sufficient light and a short enough exposure to keep the ball’s centre identifiable.
• Track the centre of the ball, not whichever edge looks sharpest.
• Use the original frame timestamps or capture rate, not an altered slow-motion playback rate.
• Exclude frames where the ball is being thrown, caught, or colliding with another object.

Camera placement deserves attention because a ruler behind the ball can give the wrong scale even when the picture looks perfectly reasonable. See How to Choose a Camera Position for Projectile Motion for the geometry behind that error.

Calibrate the scale

Suppose a \(1.00\ \mathrm{m}\) ruler occupies 500 pixels vertically.

The image scale is:

\[ S=\frac{500}{1.00} =500\ \mathrm{pixels\,m^{-1}} \]

Here, \(S\) is the number of pixels representing one metre.

A measured vertical distance of 150 pixels therefore represents:

\[ \Delta y=\frac{150}{500} =0.300\ \mathrm{m} \]

We’ll take rightwards as positive \(x\) and upwards as positive \(y\). The position origin can be anywhere convenient, provided it stays fixed throughout the video.

Most image files number pixel rows downwards from the top. If your software does that, reverse the vertical direction or configure its coordinate axes so that upward movement gives increasing \(y\).

Get time from the recording

If the camera captures \(120\) frames per second, the interval between successive frames is:

\[ \Delta t=\frac{1}{120} \approx0.00833\ \mathrm{s} \]

Between frame 12 and frame 24:

\[ \Delta t=\frac{24-12}{120} =0.100\ \mathrm{s} \]

The numerator is the number of frame intervals, not the number of marked positions.

Why measuring just two positions fails

Suppose you know the ball’s height at \(t=0\) and \(t=0.20\ \mathrm{s}\).

You can calculate its average vertical velocity over that interval. But you cannot determine its acceleration without more information.

A ball launched upwards quickly and experiencing strong downward acceleration might reach the same final height as one launched more slowly with weaker downward acceleration.

Both explanations can fit two positions.

The original measurement has a problem: it cannot separate initial velocity from acceleration.

The simplest revision is to record at least three positions at known times. Better still, track many frames and examine the complete height-time relationship.

02Why three equally spaced frames reveal gravity

Picture a ball travelling upwards.

During one short time interval, it rises by 15 cm. During the next equally long interval, it rises by only 5 cm.

The ball is still moving upwards, but its upward velocity is decreasing. Gravity has been acting throughout.

If the intervals have the same duration, the difference between those two displacements tells you about acceleration.

Under the standard projectile model, vertical position is:

\[ y=y_0+u_yt-\frac{1}{2}gt^2 \]

where:

• \(y\) is vertical position in metres.
• \(y_0\) is vertical position at \(t=0\), in metres.
• \(u_y\) is the initial vertical velocity in \(\mathrm{m\,s^{-1}}\), positive upwards.
• \(t\) is elapsed time in seconds.
• \(g\) is the positive magnitude of gravitational acceleration, in \(\mathrm{m\,s^{-2}}\).

The term \(u_yt\) describes the displacement that would occur at constant initial vertical velocity. The term \(-\frac12gt^2\) accounts for the growing effect of downward acceleration.

The important feature is that acceleration multiplies \(t^2\).

The three-frame formula

Take three heights \(y_1\), \(y_2\), and \(y_3\), measured at successive times separated by the same interval \(\Delta t\).

First calculate the change in vertical displacement:

\[ D=(y_3-y_2)-(y_2-y_1) \]

Rearranging:

\[ D=y_3-2y_2+y_1 \]

Here, \(D\) is the difference between two successive vertical displacements, measured in metres.

For constant vertical acceleration:

\[ D=-g(\Delta t)^2 \]

Therefore:

\[ \boxed{ g=-\frac{y_3-2y_2+y_1} {(\Delta t)^2} } \]

This works whether the ball is rising, falling, or passing its highest point.

The initial vertical velocity cancels out.

That is why the method can measure gravity without knowing how hard the ball was thrown.

03Use a quadratic fit when the video is noisy

Three points are enough mathematically, but three points are rarely the strongest experimental evidence.

A real ball might be blurred across several pixels. Its tracked centre could move slightly depending on which part of its outline is selected.

A one-pixel error seems tiny. Unfortunately, acceleration calculations amplify small position errors because they involve subtracting nearby measurements and dividing by a small time interval squared.

The more reliable approach is usually to measure many positions across the flight and fit a quadratic equation.

Read g from the fitted equation

Suppose your graphing software produces:

\[ y=A+Bt+Ct^2 \]

Compare this with:

\[ y=y_0+u_yt-\frac12gt^2 \]

The coefficients have physical meanings:

If the fitted equation is:

\[ y=1.20+2.40t-4.85t^2 \]

then:

\[ \begin{aligned} C&=-4.85\ \mathrm{m\,s^{-2}}\\ g&=-2C\\ &=9.70\ \mathrm{m\,s^{-2}} \end{aligned} \]

Notice that the initial height and initial upward velocity do not enter the final calculation of \(g\).

The quadratic coefficient is what you need.

A quadratic fit uses all the selected positions to estimate a single curve. That can reduce the influence of random tracking noise, provided the assumed constant-acceleration model is appropriate.

It cannot automatically correct a wrong calibration scale, incorrect timestamps, or acceleration caused by forces other than gravity.

04How to decide whether your result is trustworthy

A good experiment doesn’t end when a calculator produces \(9.7\ \mathrm{m\,s^{-2}}\).

You need to ask whether the video measured gravitational acceleration or some combination of gravity, other forces, and recording errors.

Start with the shape of the data

Under the simple projectile model:

• Horizontal position should be approximately linear in time.
• Vertical position should be approximately quadratic in time.
• The coefficient of \(t^2\) in the vertical fit should be negative when upwards is positive.
• Different sections of the flight should give compatible estimates of \(g\), within their uncertainty.

The horizontal data are a useful independent check. A changing horizontal velocity may indicate air resistance, another horizontal force, or a camera problem.

It does not automatically tell you which explanation is correct.

Distinguish measurement error from model failure

Contrast: a wrong clock can look like weaker gravity

Suppose the real time between selected frames is \(0.100\ \mathrm{s}\), but the slow-motion player makes you think it is \(0.200\ \mathrm{s}\).

Since:

\[ g\propto\frac{1}{(\Delta t)^2} \]

using twice the actual time produces:

\[ \frac{g_{\mathrm{wrong}}}{g_{\mathrm{true}}} =\left(\frac{0.100}{0.200}\right)^2 =\frac14 \]

A perfectly ordinary gravitational acceleration of \(9.8\ \mathrm{m\,s^{-2}}\) would be reported as:

\[ g_{\mathrm{wrong}}=2.45\ \mathrm{m\,s^{-2}} \]

The ball hasn’t moved to another planet. You’ve used the playback clock instead of the physical clock.

Contrast: a clean parabola doesn’t prove gravity is the only force

Suppose a ball passes through an upward air current.

Its vertical acceleration depends on the combined force from gravity and the air.

The measured quadratic coefficient now describes the net vertical acceleration, not necessarily the gravitational acceleration alone.

Similarly, if the camera accelerates vertically, the ball’s position relative to the camera can curve differently from its position relative to the ground.

The underlying projectile assumptions are examined in Projectile Motion Assumptions in HSC Physics Explained.

The practical rule is to treat your calculated \(g\) as an estimate of gravitational acceleration only when the relevant force and measurement assumptions are justified.

05Questions and solutions

Question 1

A projectile is filmed using a stationary side-on camera.

The following vertical positions are measured relative to a fixed origin, with upwards positive:

Assume constant downward acceleration and negligible air resistance.

Calculate \(g\) from the three measurements. Explain why the ball’s increasing height does not contradict your result.

Solution 1

The gravitational acceleration is \(9.80\ \mathrm{m\,s^{-2}}\) downwards, even though the ball is moving upwards.

Using:

\[ g=-\frac{y_3-2y_2+y_1}{(\Delta t)^2} \]

Substitute the three measurements:

\[ \begin{aligned} D&=1.004-2(0.951)+0.800\\ &=-0.098\ \mathrm{m} \end{aligned} \]

Therefore:

\[ \begin{aligned} g&=-\frac{-0.098}{(0.10)^2}\\ &=\frac{0.098}{0.0100}\\ &=9.80\ \mathrm{m\,s^{-2}} \end{aligned} \]

The ball’s upward displacement is getting smaller over each equal time interval.

That is exactly what downward acceleration predicts for a ball travelling upwards. Gravity changes the vertical velocity; it does not require the ball to be moving downwards at every instant.

Question 2

A ball is filmed travelling from left to right. The recorded video is exported as a slow-motion file playing at 30 frames per second.

Three images taken from the original recording are displayed \(0.40\ \mathrm{s}\) apart during playback. The actual time between their capture is unknown.

The recording is calibrated at 500 pixels per metre. The measured positions are:

Pixel positions are measured relative to fixed axes, with rightwards and upwards positive.

An independent photogate measurement at release gives a horizontal velocity component of \(1.80\ \mathrm{m\,s^{-1}}\).

Assume the horizontal velocity remains constant during the recorded flight and that the camera is stationary.

(a) Determine the actual physical time interval between the three images.

(b) Calculate \(g\).

(c) A student uses the \(0.40\ \mathrm{s}\) playback interval and concludes that gravity is approximately \(0.61\ \mathrm{m\,s^{-2}}\). Explain precisely why the independent measurement contradicts this result.

Solution 2

The physical time interval is \(0.100\ \mathrm{s}\), and the measured gravitational acceleration is \(9.80\ \mathrm{m\,s^{-2}}\) downwards. The slow-motion playback clock is not the clock governing the actual flight.

(a) Find the physical time interval.

The horizontal displacement from A to B is:

\[ \begin{aligned} \Delta x&=\frac{200-110}{500}\\ &=0.180\ \mathrm{m} \end{aligned} \]

Since the horizontal velocity is assumed constant:

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

Rearranging:

\[ \begin{aligned} \Delta t&=\frac{0.180}{1.80}\\ &=0.100\ \mathrm{s} \end{aligned} \]

The next horizontal displacement is also 90 pixels, so the same time interval applies between B and C.

(b) Calculate gravitational acceleration.

Convert the change in vertical displacement directly from pixels to metres:

\[ \begin{aligned} D&=\frac{435-2(422)+360}{500}\\ &=\frac{-49}{500}\\ &=-0.098\ \mathrm{m} \end{aligned} \]

Therefore:

\[ \begin{aligned} g&=-\frac{-0.098}{(0.100)^2}\\ &=9.80\ \mathrm{m\,s^{-2}} \end{aligned} \]

(c) Identify the timing mistake.

Using the playback time gives:

\[ \begin{aligned} g_{\mathrm{wrong}} &=\frac{0.098}{(0.40)^2}\\ &=0.6125\ \mathrm{m\,s^{-2}} \end{aligned} \]

That calculation is mathematically consistent with the wrong time interval.

But the same incorrect clock would give a horizontal velocity:

\[ \begin{aligned} v_{x,\mathrm{wrong}} &=\frac{0.180}{0.40}\\ &=0.450\ \mathrm{m\,s^{-1}} \end{aligned} \]

This contradicts the independent horizontal measurement of \(1.80\ \mathrm{m\,s^{-1}}\), under the stated constant-horizontal-velocity assumption.

The photogate and measured horizontal displacement allow us to recover the physical time interval.

The important constraint is that both horizontal and vertical motion must use the same real clock. Without the independent velocity information or reliable original timestamps, the playback rate alone would not establish the physical timing.

Question 3

A student tracks a ball through five equally spaced frames in a stationary-camera video.

The recorded vertical positions are:

Each position is estimated to have a marking uncertainty of approximately \(\pm0.004\ \mathrm{m}\).

The ball remains airborne, and no additional force is known to act during the flight. Assume constant gravitational acceleration, and investigate whether the data support that model.

(a) Calculate the acceleration estimate using the three central measurements at \(0.10\), \(0.20\), and \(0.30\ \mathrm{s}\).

(b) Independently estimate \(g\) without using the measurement at \(0.20\ \mathrm{s}\). Use the average vertical velocities over \(0.00\)-\(0.10\ \mathrm{s}\) and \(0.30\)-\(0.40\ \mathrm{s}\). Explain which times those velocities represent under constant acceleration.

(c) Predict the height at \(0.20\ \mathrm{s}\) from your second estimate. Identify the most sensible next experimental action and explain why a brief additional force is not automatically the best explanation.

Solution 3

The central three-point calculation gives \(13.6\ \mathrm{m\,s^{-2}}\), but the surrounding measurements support a constant-acceleration estimate of \(9.60\ \mathrm{m\,s^{-2}}\). The height recorded at \(0.20\ \mathrm{s}\) is inconsistent with that otherwise smooth model and should be checked.

(a) Calculate the central estimate.

\[ \begin{aligned} D&=1.138-2(1.208)+1.142\\ &=-0.136\ \mathrm{m} \end{aligned} \]

Therefore:

\[ \begin{aligned} g&=-\frac{-0.136}{(0.10)^2}\\ &=13.6\ \mathrm{m\,s^{-2}} \end{aligned} \]

That is the apparent downward acceleration inferred from the central three positions.

It would be a mistake to assume this is automatically the real gravitational acceleration. The method is sensitive to a single incorrect middle position because that reading is multiplied by two.

(b) Estimate the acceleration without the middle measurement.

Calculate the average vertical velocity over the first interval:

\[ \begin{aligned} \bar v_1&=\frac{1.142-1.000}{0.10}\\ &=+1.42\ \mathrm{m\,s^{-1}} \end{aligned} \]

The second selected interval gives:

\[ \begin{aligned} \bar v_2&=\frac{0.992-1.138}{0.10}\\ &=-1.46\ \mathrm{m\,s^{-1}} \end{aligned} \]

For constant acceleration, the average velocity over an interval equals the instantaneous velocity at its midpoint.

So \(\bar v_1\) corresponds to \(t=0.05\ \mathrm{s}\), and \(\bar v_2\) corresponds to \(t=0.35\ \mathrm{s}\).

The time between those midpoints is \(0.30\ \mathrm{s}\).

\[ \begin{aligned} a_y&=\frac{-1.46-1.42}{0.30}\\ &=-9.60\ \mathrm{m\,s^{-2}} \end{aligned} \]

Therefore:

\[ g=9.60\ \mathrm{m\,s^{-2}} \]

This uses four measurements without relying on the potentially incorrect centre position.

(c) Predict the missing height.

First recover the initial vertical velocity.

The average velocity over the first \(0.10\ \mathrm{s}\) corresponds to its midpoint:

\[ \bar v_1=u_y+a_y(0.05) \]

Therefore:

\[ \begin{aligned} u_y&=1.42-(-9.60)(0.05)\\ &=1.90\ \mathrm{m\,s^{-1}} \end{aligned} \]

Now predict the position at \(t=0.20\ \mathrm{s}\):

\[ y=y_0+u_yt-\frac12gt^2 \]

\[ \begin{aligned} y&=1.000+1.90(0.20)\\ &\quad-\frac12(9.60)(0.20)^2\\ &=1.188\ \mathrm{m} \end{aligned} \]

The recorded position is \(1.208\ \mathrm{m}\), which is \(0.020\ \mathrm{m}\), or 20 mm, above this prediction.

That discrepancy is five times the quoted uncertainty of an individual position measurement. The prediction also has uncertainty, because it was calculated from other measured positions, so a proper assessment should account for errors in all the readings.

Nevertheless, the sensible next action is to reinspect the frame at \(0.20\ \mathrm{s}\), check which point was marked as the ball’s centre, and repeat the tracking.

A genuine brief force would generally change the ball’s velocity and therefore affect subsequent positions. An isolated disagreement in one frame, followed by a return to an otherwise smooth trajectory, is a strong reason to suspect a tracking error first.

It is not proof that the measurement is wrong. The video and apparatus must decide that, rather than deleting points simply because they disagree with a preferred value.

Question 4

A \(0.100\ \mathrm{kg}\) ball travels from left to right across a laboratory.

During the first \(0.20\ \mathrm{s}\), gravity is the only significant vertical force acting on it.

Immediately after \(t=0.20\ \mathrm{s}\), the ball enters a region containing a steady upward air current from a fan. The fan exerts a constant upward force on the ball for the remainder of the recording.

The horizontal velocity remains constant, and the camera is stationary.

The illustrative height measurements are:

Take upwards as positive.

(a) Calculate gravitational acceleration using the first three measurements.

(b) Calculate the magnitude and direction of the net vertical acceleration while the ball is in the fan’s airflow.

(c) Determine the upward force exerted by the air current.

(d) A student uses the measurements at \(0.10\), \(0.20\), and \(0.30\ \mathrm{s}\) and obtains a third value. Calculate it and explain exactly what that value represents.

Solution 4

The gravitational acceleration is \(10.0\ \mathrm{m\,s^{-2}}\) downwards in this idealised dataset. The fan reduces the net downward acceleration to \(4.0\ \mathrm{m\,s^{-2}}\) by exerting an upward force of \(0.600\ \mathrm{N}\). Gravity itself has not changed.

(a) Calculate \(g\) before the ball enters the airflow.

Using the first three positions:

\[ \begin{aligned} D&=1.10-2(1.10)+1.00\\ &=-0.10\ \mathrm{m} \end{aligned} \]

Therefore:

\[ \begin{aligned} g&=-\frac{-0.10}{(0.10)^2}\\ &=10.0\ \mathrm{m\,s^{-2}} \end{aligned} \]

Because the problem states that gravity is the only significant vertical force in this first region, this acceleration measures \(g\).

(b) Calculate net acceleration in the airflow.

Use the positions at \(0.20\), \(0.30\), and \(0.40\ \mathrm{s}\).

\[ \begin{aligned} D&=0.92-2(1.03)+1.10\\ &=-0.04\ \mathrm{m} \end{aligned} \]

The net vertical acceleration is:

\[ \begin{aligned} a_y&=\frac{-0.04}{(0.10)^2}\\ &=-4.0\ \mathrm{m\,s^{-2}} \end{aligned} \]

The net acceleration is \(4.0\ \mathrm{m\,s^{-2}}\) downwards.

(c) Calculate the force from the fan.

Newton’s second law in the vertical direction gives:

\[ F_{\mathrm{air}}-mg=ma_y \]

Here, \(F_{\mathrm{air}}\) is the upward force from the air, \(m\) is the ball’s mass, and \(a_y\) is its signed net vertical acceleration.

Substitute:

\[ \begin{aligned} F_{\mathrm{air}}-0.100(10.0) &=0.100(-4.0)\\ F_{\mathrm{air}}-1.00&=-0.400\\ F_{\mathrm{air}}&=0.600\ \mathrm{N} \end{aligned} \]

The fan exerts an upward force of \(0.600\ \mathrm{N}\).

(d) Explain the measurement across the boundary.

Using the positions at \(0.10\), \(0.20\), and \(0.30\ \mathrm{s}\):

\[ \begin{aligned} D&=1.03-2(1.10)+1.10\\ &=-0.07\ \mathrm{m} \end{aligned} \]

Therefore:

\[ \begin{aligned} a_y&=\frac{-0.07}{(0.10)^2}\\ &=-7.0\ \mathrm{m\,s^{-2}} \end{aligned} \]

The apparent acceleration magnitude is \(7.0\ \mathrm{m\,s^{-2}}\).

This three-point calculation crosses the instant when the net force changes. It combines information from the earlier interval with downward acceleration \(10.0\ \mathrm{m\,s^{-2}}\) and the later interval with downward acceleration \(4.0\ \mathrm{m\,s^{-2}}\).

Because the two adjacent intervals are equal and each has constant acceleration, the central calculation gives their equally weighted average:

\[ \frac{10.0+4.0}{2} =7.0\ \mathrm{m\,s^{-2}} \]

That value is neither the gravitational acceleration nor the net acceleration throughout either region.

The trap is treating a quadratic calculation as a measurement of gravity even after the forces acting on the ball have changed.

Question 5

A camera records two balls, A and B, moving through the same vertical plane. Both balls are already airborne and experience only gravity during the recording.

However, the camera itself is mounted on a platform that can move vertically with an unknown, possibly changing velocity.

The camera maintains a constant image scale, remains correctly aligned, and records all positions relative to its own moving coordinate system.

A fixed marker on a laboratory wall is also visible at the same calibrated distance from the camera as the balls.

The measured vertical coordinates, converted to metres with upwards positive, are:

Assume the timestamps are accurate, both balls have constant downward gravitational acceleration, and no other significant forces act on them.

A student calculates the curvature of ball A’s recorded path and concludes that:

\[ g=6.0\ \mathrm{m\,s^{-2}} \]

Another student calculates the curvature for ball B and obtains exactly the same result.

They argue that two independent balls confirming the same value proves the measurement is reliable.

(a) Explain why this conclusion is not justified.

(b) Use the fixed marker to determine the actual gravitational acceleration.

(c) Calculate the camera’s vertical acceleration.

(d) Suppose the fixed marker had not been recorded. Could perfect position measurements of the two balls alone determine the actual value of \(g\)? Give a reasoned answer, considering that their initial velocities may be different.

Solution 5

The actual gravitational acceleration is \(10.0\ \mathrm{m\,s^{-2}}\) downwards, and the camera accelerates downwards at \(4.0\ \mathrm{m\,s^{-2}}\). The agreement between the two ball-only measurements does not establish the true value of \(g\), because both share the same accelerating camera reference frame.

The fixed marker is the crucial additional measurement.

(a) Explain why the two matching results are misleading.

For ball A:

\[ \begin{aligned} D_A&=1.180-2(1.120)+1.000\\ &=-0.060\ \mathrm{m} \end{aligned} \]

The observed acceleration is:

\[ \begin{aligned} a_{A,\mathrm{obs}} &=\frac{-0.060}{(0.10)^2}\\ &=-6.0\ \mathrm{m\,s^{-2}} \end{aligned} \]

For ball B:

\[ \begin{aligned} D_B&=0.800-2(0.780)+0.700\\ &=-0.060\ \mathrm{m} \end{aligned} \]

So:

\[ a_{B,\mathrm{obs}} =-6.0\ \mathrm{m\,s^{-2}} \]

Both measurements really do show the same acceleration relative to the moving camera.

But a moving reference frame can change the observed acceleration.

In a vertical coordinate system where the camera maintains its orientation and scale:

\[ y_{\mathrm{obs}}=y_{\mathrm{world}}-y_{\mathrm{camera}} \]

Consequently:

\[ a_{\mathrm{obs}} =a_{\mathrm{world}}-a_{\mathrm{camera}} \]

The observed acceleration is not necessarily the ball’s acceleration relative to the ground.

Both balls share the same camera motion, so agreement between their observed accelerations cannot eliminate that common systematic effect.

(b) Use the fixed marker to recover \(g\).

The wall marker is stationary in the laboratory.

Its position in the camera’s image can change only because of camera movement, under the stated assumptions.

A useful method is to subtract the marker’s observed position from the ball’s observed position in every frame.

For ball A:

The camera’s movement cancels when the two simultaneous positions are subtracted.

Now calculate the acceleration from the relative heights:

\[ \begin{aligned} D&=0.620-2(0.610)+0.500\\ &=-0.100\ \mathrm{m} \end{aligned} \]

Therefore:

\[ \begin{aligned} a_y&=\frac{-0.100}{(0.10)^2}\\ &=-10.0\ \mathrm{m\,s^{-2}} \end{aligned} \]

So the actual gravitational acceleration is:

\[ g=10.0\ \mathrm{m\,s^{-2}} \]

This calculation works even though the camera accelerates because the ball and fixed marker are measured at the same moments in the same calibrated frame.

(c) Find the camera’s acceleration.

Calculate the observed acceleration of the fixed marker:

\[ \begin{aligned} D_M&=0.560-2(0.510)+0.500\\ &=+0.040\ \mathrm{m} \end{aligned} \]

Therefore:

\[ \begin{aligned} a_{M,\mathrm{obs}} &=\frac{0.040}{(0.10)^2}\\ &=+4.0\ \mathrm{m\,s^{-2}} \end{aligned} \]

The stationary marker appears to accelerate upwards at \(4.0\ \mathrm{m\,s^{-2}}\).

Since the marker has zero acceleration in the laboratory frame, the camera must accelerate in the opposite direction:

\[ a_{\mathrm{camera}} =-4.0\ \mathrm{m\,s^{-2}} \]

Check that this agrees with the ball measurement:

\[ \begin{aligned} a_{A,\mathrm{obs}} &=a_{A,\mathrm{world}}-a_{\mathrm{camera}}\\ &=-10.0-(-4.0)\\ &=-6.0\ \mathrm{m\,s^{-2}} \end{aligned} \]

It matches exactly.

(d) Decide whether two balls alone can determine \(g\).

No. If the camera has unknown vertical acceleration, the two ball trajectories alone cannot determine the actual gravitational acceleration.

The reason is subtle.

Consider the difference between the observed positions of balls A and B.

Their relative vertical position increases by the same amount in each equal time interval.

Calculate its acceleration:

\[ \begin{aligned} D_{A-B}&=0.380-2(0.340)+0.300\\ &=0\ \mathrm{m} \end{aligned} \]

Thus:

\[ a_{A-B}=0\ \mathrm{m\,s^{-2}} \]

Both balls experience the same gravitational acceleration, so gravity cancels when their motions are compared.

The camera’s acceleration also cancels.

What remains is their relative motion due to different initial positions and velocities. Their separation contains no information about the common gravitational acceleration.

There are therefore multiple possible physical explanations for the same two observed trajectories.

For example:

• If the camera had zero vertical acceleration and the balls accelerated downwards at \(6.0\ \mathrm{m\,s^{-2}}\), both recorded accelerations would be \(-6.0\ \mathrm{m\,s^{-2}}\).
• If the camera accelerated downwards at \(4.0\ \mathrm{m\,s^{-2}}\) and the balls accelerated downwards at \(10.0\ \mathrm{m\,s^{-2}}\), both recorded accelerations would still be \(-6.0\ \mathrm{m\,s^{-2}}\).

The two-ball observations cannot distinguish these explanations, even with perfect position measurements.

The fixed marker resolves the ambiguity because it provides an object whose acceleration in the laboratory frame is already known to be zero.

The deeper lesson is that a mathematical fit can be exact without identifying the physical cause of the fitted acceleration.

To infer gravity, you need not only position and time but also a trustworthy reference frame and an appropriate force model.

05What to do after finding g

Once you have a reliable quadratic fit, use it to make a prediction rather than just reporting a number.

If:

\[ y=A+Bt+Ct^2 \]

then the constant-acceleration model predicts the vertical velocity at time \(t\) to be:

\[ v_y=B+2Ct \]

Here, \(v_y\) is in \(\mathrm{m\,s^{-1}}\), \(B\) is the fitted initial vertical velocity, and \(C\) is the quadratic coefficient.

You can estimate vertical velocities from neighbouring video positions and compare them with that prediction. You can also combine the vertical component with the measured horizontal velocity to predict the projectile’s speed and direction later in its flight.

That leads directly to Final Velocity of a Projectile: HSC Physics Guide.

The strongest result from your video is not simply a number close to \(9.8\). It is a model that consistently explains the measured positions, survives reasonable uncertainty checks, and makes testable predictions about what the projectile will do next.