How to Choose Positive Direction in HSC Vertical Motion

Two students calculate the acceleration of the same falling ball. One gets \(-9.8\,\mathrm{m\,s^{-2}}\). The other gets \(+9.8\,\mathrm{m\,s^{-2}}\). Which student is wrong?

Possibly neither. In a vertical-motion calculation, you can choose either upwards or downwards as the positive direction. The rule is to choose once, draw an arrow showing your choice, and assign signs to displacement, initial velocity, final velocity, and acceleration consistently. If upwards is positive, gravitational acceleration is negative. If downwards is positive, gravitational acceleration is positive.

The important question isn’t which direction you choose. It’s whether every sign in your calculation agrees with that choice.

01What does choosing a positive direction actually mean?

Imagine standing beside a lift shaft with a measuring tape running vertically.

You could number the tape upwards, so positions higher up have larger coordinates. Or you could number it downwards, so positions lower down have larger coordinates.

Neither measuring system changes how the lift moves. You’re simply choosing how to describe that movement using numbers.

Vertical-motion calculations work the same way.

A positive sign means a quantity points in your chosen positive direction. A negative sign means it points in the opposite direction.

For motion near Earth’s surface, the magnitude of gravitational acceleration is approximately \(g=9.8\,\mathrm{m\,s^{-2}}\), assuming air resistance is negligible.

Gravity acts downwards regardless of which direction you call positive.

Here, \(u\) is initial vertical velocity, \(v\) is final vertical velocity, \(s\) is vertical displacement, and \(a\) is acceleration. The rows illustrate separate situations rather than measurements from one particular flight.

Notice that changing the positive direction reverses every signed vertical quantity. It doesn’t change any physical speed, distance, or elapsed time.

Displacement is not the same as height

Suppose a ball starts 15 m above the ground and finishes 4 m above the ground.

Its vertical displacement is 11 m downwards.

With upwards positive:

\[ s=4-15=-11\,\mathrm{m} \]

With downwards positive:

\[ s=+11\,\mathrm{m} \]

The ball’s final height is still 4 m above the ground. Height tells you where it is relative to the ground. Displacement tells you how its position has changed, including direction.

You can also choose where the coordinate zero is located. Putting zero at the ground or at the release point changes position coordinates, but not the physical displacement between two points.

02A falling-ball experiment: can the acceleration really have two signs?

How could we test whether two opposite acceleration signs describe the same physical motion?

Consider an idealised experiment using a falling steel ball and two light gates.

The apparatus and measurement problem

An electromagnetic release holds a small steel ball at the top of a vertical path.

Directly below the release are two light gates. Their light beams cross the ball’s path at fixed heights. The first beam is 0.196 m below the release point, and the second is 0.784 m below it.

A timer starts when the ball is released. As the ball falls, it passes through the first beam and then the second. Each interruption produces a time reading.

A soft collection tray sits below the second gate.

The original timing arrangement can tell us when the ball crosses each gate, but timing alone isn’t enough to calculate acceleration. We also need the separation of the measurement positions and a defined vertical coordinate.

Adding a calibrated ruler beside the falling path provides the distances. Marking the release position as zero and choosing a positive direction lets us assign signed displacements.

Suppose we model the ball as released from rest, with constant gravitational acceleration and no air resistance.

What should the gates record?

Predicted readings

Using \(g=9.8\,\mathrm{m\,s^{-2}}\), the idealised model predicts the following values.

These are illustrative theoretical data, not measurements from a performed experiment. For simplicity, each gate time represents the instant the centre of the ball reaches its marked height.

In either coordinate system, the ball reaches the same gates at exactly the same times.

The second gate is much further from the release point than the first, yet the ball reaches it only another 0.20 s later. That is consistent with the ball gaining downward speed.

Now calculate acceleration using the final reading.

For an object starting from rest under constant acceleration:

\[ s=\frac12at^2 \]

Rearranging:

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

With upwards positive:

\[ \begin{aligned} a&=\frac{2(-0.784)}{(0.40)^2}\\ &=-9.8\,\mathrm{m\,s^{-2}} \end{aligned} \]

With downwards positive:

\[ \begin{aligned} a&=\frac{2(+0.784)}{(0.40)^2}\\ &=+9.8\,\mathrm{m\,s^{-2}} \end{aligned} \]

Both results describe the same downward acceleration.

The experiment’s physical observation would be a ball moving downwards and gaining speed. The signs come from the coordinates we use to describe that observation.

A real experiment would also require attention to release timing, gate alignment, the ball’s finite size, and measurement uncertainty. Air resistance could make the acceleration depart from the idealised constant value.

Those limitations affect measurement accuracy. They don’t make one positive-direction convention more scientifically correct than the other.

03How to choose the easiest positive direction

You can solve every constant-acceleration vertical-motion problem using either convention.

Still, some choices make the working easier to follow.

When upwards positive is convenient

Choose upwards positive when a ball is thrown upwards, when you’re calculating maximum height, or when most of the given displacements are upwards.

For example, a ball thrown vertically upwards at \(15\,\mathrm{m\,s^{-1}}\) starts with:

\[ u=+15\,\mathrm{m\,s^{-1}} \]

Its gravitational acceleration is:

\[ a=-9.8\,\mathrm{m\,s^{-2}} \]

As the ball rises, its positive velocity decreases. At the highest point, its vertical velocity is zero. As it falls, its velocity becomes negative.

Acceleration stays negative throughout the free flight.

A common mistake is to change acceleration from negative to positive when the ball starts falling.

That feels reasonable because the ball has changed direction. But gravity hasn’t changed direction. Only the velocity has.

The ball slows down on the way up because velocity and acceleration point in opposite directions. It speeds up on the way down because they point in the same direction.

If the instant of zero velocity is confusing, Can an Object Have Zero Velocity and Nonzero Acceleration? explains the turning point in more detail.

When downwards positive is convenient

Downwards positive often makes sense when an object is dropped, thrown downwards, or travelling mainly towards a lower point.

For a dropped object:

\[ u=0,\qquad a=+9.8\,\mathrm{m\,s^{-2}} \]

The displacement towards the ground is positive.

There are fewer negative numbers to manage, but the underlying calculation is identical.

A reliable decision process

Before substituting into an equation:

1. Draw a vertical arrow and label it positive. Choose up or down, whichever makes the given information easiest to organise.
2. Give each velocity and displacement a sign by comparing its direction with your arrow. Give gravitational acceleration its corresponding sign.
3. Use the signed values in the constant-acceleration equation. Do not change the axis halfway through a continuous flight.
4. Translate the mathematical result back into physical language. A negative velocity means motion opposite your positive arrow, not a negative speed.

Elapsed time after the chosen starting instant is non-negative. A negative mathematical time root therefore needs careful interpretation, rather than automatically being treated as a physical event after release.

06Three sign and root mistakes worth catching

Mistake 1: Making gravity negative automatically

Gravitational acceleration is not inherently a negative number.

Near Earth’s surface, gravity points downwards. Its signed value depends on your axis.

If downwards is positive, write \(a=+g\). If upwards is positive, write \(a=-g\).

Mistake 2: Changing positive direction when the ball turns around

A ball reaching maximum height doesn’t require a new coordinate system.

In an upward-positive system, the velocity simply passes from positive through zero to negative. Acceleration remains negative.

Changing your axis at the turning point introduces unnecessary sign conversions and makes it easier to use inconsistent quantities.

Mistake 3: Accepting every algebraic root without checking the motion

A quadratic equation may give two times, but neither should be accepted without checking what happens physically.

Suppose a ball would mathematically return to a particular height after 3.0 s, but a ceiling interrupts its flight after 1.0 s. The original free-flight equation no longer describes the ball after the collision.

The acceleration due to gravity may still be the same before and after the impact. However, the collision changes the velocity abruptly.

You must treat the motion as separate intervals, each with its own initial velocity and starting time.

This matters particularly when selecting roots. A root can be mathematically valid for an uninterrupted flight but physically impossible because the flight ended earlier.

07Questions and solutions

Use \(g=9.8\,\mathrm{m\,s^{-2}}\) throughout. Unless otherwise stated, ignore air resistance, treat objects as particles during free flight, and assume gravitational acceleration is uniform.

Question 1

A pebble is dropped from rest. It falls 4.90 m before passing a marked point.

Taking upwards as positive:

a) State the pebble’s initial velocity, signed displacement, and acceleration.

b) Calculate the time taken to reach the point.

c) Calculate its signed velocity and state its direction.

Solution 1

The pebble takes \(1.00\,\mathrm{s}\) to reach the point, with a signed velocity of \(-9.80\,\mathrm{m\,s^{-1}}\), meaning it is moving downwards.

Because upwards is positive, both the downward displacement and gravitational acceleration are negative.

For part (a):

\[ \begin{aligned} u&=0\,\mathrm{m\,s^{-1}}\\ s&=-4.90\,\mathrm{m}\\ a&=-9.8\,\mathrm{m\,s^{-2}} \end{aligned} \]

For part (b), use:

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

\[ \begin{aligned} -4.90&=0-4.9t^2\\ t^2&=1.00\,\mathrm{s^2}\\ t&=1.00\,\mathrm{s} \end{aligned} \]

We choose the positive time because the question asks about motion after release.

For part (c):

\[ \begin{aligned} v&=u+at\\ &=0-9.8(1.00)\\ &=-9.80\,\mathrm{m\,s^{-1}} \end{aligned} \]

The negative result indicates downward motion, not an impossible negative speed.

The tempting mistake is to use \(s=+4.90\,\mathrm{m}\) because the pebble travels a distance of 4.90 m. That confuses distance with signed displacement.

Question 2

A ball is launched vertically upwards from an unknown height above a courtyard.

A motion tracker records that the ball is 7.20 m above the courtyard at both \(t=0.40\,\mathrm{s}\) and \(t=2.00\,\mathrm{s}\), where time is measured from launch.

Assume the ball remains in uninterrupted free flight between these observations.

Using upwards as positive:

a) Determine the initial vertical velocity.

b) Determine the launch height above the courtyard.

c) Calculate the velocity at each recorded time and explain why the readings occur at the same height.

Solution 2

The ball was launched upwards at \(11.76\,\mathrm{m\,s^{-1}}\) from a height of 3.28 m. Its velocities at the two observations are \(+7.84\,\mathrm{m\,s^{-1}}\) and \(-7.84\,\mathrm{m\,s^{-1}}\), respectively.

The two readings are possible because the ball passes the same height once while rising and again while falling.

Let \(y_0\) be the launch height in metres, and let \(y\) be the height above the courtyard at time \(t\).

With upwards positive:

\[ y=y_0+ut-4.9t^2 \]

At \(t=0.40\,\mathrm{s}\):

\[ 7.20=y_0+0.40u-0.784 \]

At \(t=2.00\,\mathrm{s}\):

\[ 7.20=y_0+2.00u-19.6 \]

Subtract the first equation from the second:

\[ \begin{aligned} 0&=1.60u-18.816\\ u&=11.76\,\mathrm{m\,s^{-1}} \end{aligned} \]

Substitute into the first height equation:

\[ \begin{aligned} 7.20&=y_0+0.40(11.76)-0.784\\ y_0&=3.28\,\mathrm{m} \end{aligned} \]

Now calculate the velocities.

At \(t=0.40\,\mathrm{s}\):

\[ \begin{aligned} v&=u+at\\ &=11.76-9.8(0.40)\\ &=+7.84\,\mathrm{m\,s^{-1}} \end{aligned} \]

At \(t=2.00\,\mathrm{s}\):

\[ \begin{aligned} v&=11.76-9.8(2.00)\\ &=-7.84\,\mathrm{m\,s^{-1}} \end{aligned} \]

So the first observation catches the ball rising, and the second catches it falling.

The trap is assuming that two measurements at the same height mean the ball has not moved. Equal positions do not imply zero distance travelled between observations.

The ball reaches its highest point between the observations. Its velocity changes sign, but the downward acceleration does not.

Question 3

Two small balls move along the same vertical line.

Ball A is dropped from rest at a height of 18.0 m above the ground.

At the same instant, Ball B is thrown vertically upwards from ground level at \(12.0\,\mathrm{m\,s^{-1}}\).

Both experience the same constant downward gravitational acceleration.

a) Determine whether they meet before either reaches the ground.

b) If they meet, calculate the meeting time and height.

c) Determine the direction and velocity of each ball immediately before meeting.

Explain why the answer to part (c) might be surprising.

Solution 3

The balls meet after \(1.50\,\mathrm{s}\), at 6.98 m above the ground. Both balls are moving downwards immediately before meeting, even though Ball B was originally thrown upwards.

Take upwards as positive, and measure height from the ground.

For Ball A:

\[ y_A=18.0-4.9t^2 \]

For Ball B:

\[ y_B=12.0t-4.9t^2 \]

They meet when their heights are equal:

\[ 18.0-4.9t^2=12.0t-4.9t^2 \]

The acceleration terms cancel:

\[ \begin{aligned} 18.0&=12.0t\\ t&=1.50\,\mathrm{s} \end{aligned} \]

Why do the gravitational terms cancel?

Both balls accelerate downwards at the same rate. Gravity changes their individual velocities, but it doesn’t change the difference between those velocities while both remain in free flight.

Substitute the meeting time into either height equation:

\[ \begin{aligned} y_A&=18.0-4.9(1.50)^2\\ &=6.975\,\mathrm{m}\\ &\approx6.98\,\mathrm{m} \end{aligned} \]

For Ball A:

\[ \begin{aligned} v_A&=0-9.8(1.50)\\ &=-14.7\,\mathrm{m\,s^{-1}} \end{aligned} \]

For Ball B:

\[ \begin{aligned} v_B&=12.0-9.8(1.50)\\ &=-2.70\,\mathrm{m\,s^{-1}} \end{aligned} \]

Both velocities are negative, so both balls are descending.

Ball B actually reaches its highest point earlier:

\[ \begin{aligned} 0&=12.0-9.8t\\ t&=1.22\,\mathrm{s} \end{aligned} \]

By the meeting time of 1.50 s, Ball B has already turned around.

We should also check that neither ball has struck the ground first.

Ball A takes approximately 1.92 s to reach the ground. Ball B would return to ground level after approximately 2.45 s if uninterrupted.

The meeting at 1.50 s therefore occurs before either ground contact.

The tempting prediction is that the balls must meet while one rises and the other falls. But their meeting depends on relative position, not on their velocities pointing towards each other at that particular instant.

Question 4

An open lift platform is rising vertically at a constant speed of \(3.0\,\mathrm{m\,s^{-1}}\).

When the platform is 15.0 m above the ground, a ball is thrown vertically upwards at \(8.0\,\mathrm{m\,s^{-1}}\) relative to the platform.

After release, the ball experiences gravity only. The platform continues rising at the same constant speed.

Taking upwards as positive:

a) Determine the ball’s initial velocity relative to the ground.

b) Calculate when the ball next returns to the moving platform’s height.

c) Find the platform’s height at that instant and the ball’s velocity relative to the ground immediately before contact.

d) Explain why using \(u=+8.0\,\mathrm{m\,s^{-1}}\) in a ground-based calculation would be incorrect.

Solution 4

The ball initially moves upwards at \(11.0\,\mathrm{m\,s^{-1}}\) relative to the ground. It returns to the rising platform’s height after \(1.63\,\mathrm{s}\), when the platform is approximately 19.9 m above the ground. At that instant, the ball moves downwards at \(5.0\,\mathrm{m\,s^{-1}}\) relative to the ground.

The crucial constraint is that the ball is launched from a platform already moving upwards.

Its ground-frame initial velocity is:

\[ u=3.0+8.0 \]

\[ u=+11.0\,\mathrm{m\,s^{-1}} \]

Let \(y_b\) be the ball’s height above ground and \(y_p\) the platform’s height, both in metres.

For the ball:

\[ y_b=15.0+11.0t-4.9t^2 \]

For the platform, which moves at constant velocity:

\[ y_p=15.0+3.0t \]

The ball returns to the platform’s height when:

\[ 15.0+11.0t-4.9t^2=15.0+3.0t \]

Simplify:

\[ 8.0t-4.9t^2=0 \]

Factorise:

\[ t(8.0-4.9t)=0 \]

This gives:

\[ t=0 \]

or:

\[ t=\frac{8.0}{4.9}=1.63\,\mathrm{s} \]

The first solution is the release instant. The second is when the ball returns to the platform’s level.

Now calculate the platform height:

\[ \begin{aligned} y_p&=15.0+3.0(1.63)\\ &\approx19.9\,\mathrm{m} \end{aligned} \]

Finally, the ball’s ground-frame velocity:

\[ \begin{aligned} v_b&=11.0-9.8(1.63)\\ &\approx-5.0\,\mathrm{m\,s^{-1}} \end{aligned} \]

It is moving downwards at 5.0 m/s while the platform is still moving upwards at 3.0 m/s.

Relative to the platform, the ball is approaching downwards at \(8.0\,\mathrm{m\,s^{-1}}\).

The tempting mistake is using the \(8.0\,\mathrm{m\,s^{-1}}\) launch velocity directly in the ground-based equation. That is the velocity relative to the moving platform, not relative to the ground.

This question also shows why the position reference and positive direction are separate choices. Both the ball and the platform are described using an upward-positive ground coordinate. We don’t need to change the sign convention when the ball begins descending.

Question 5

A ball is launched vertically upwards from the floor of a shaft at \(14.7\,\mathrm{m\,s^{-1}}\).

There is a horizontal optical sensor 6.125 m above the floor and a rigid ceiling 9.80 m above the floor.

When the ball strikes the ceiling, it rebounds instantaneously with the same speed in the opposite direction. Treat this as an idealised collision: gravitational acceleration is constant during each free-flight interval, but the velocity changes abruptly at impact.

The ball is caught when it next reaches the floor.

A tracking program ignores the ceiling and models the ball as undergoing one uninterrupted flight. It predicts that the ball passes the sensor at:

\[ t_1=0.50\,\mathrm{s} \]

and:

\[ t_2=2.50\,\mathrm{s} \]

Using upwards as positive:

a) Determine whether the program’s two solutions are mathematically correct for its assumed uninterrupted flight.

b) Find the actual time when the ball first strikes the ceiling and its velocity immediately before and after impact.

c) Determine all actual times when the ball passes the sensor, including the signed velocities at those times.

d) Determine when the ball is next caught at floor level.

e) Another student suggests changing the positive direction to downwards when the ball hits the ceiling, without changing the original motion model. Explain whether changing the coordinate direction alone can correct the tracking program.

Solution 5

The tracking program’s two times are correct only for an imaginary flight without a ceiling. In the actual shaft, the ball passes the sensor at \(0.50\,\mathrm{s}\) while rising and \(1.50\,\mathrm{s}\) while falling. It strikes the ceiling at \(1.00\,\mathrm{s}\) and reaches the floor again at \(2.00\,\mathrm{s}\).

The central issue is not the sign convention. It’s that the collision interrupts the original constant-acceleration flight and changes the ball’s velocity.

Part (a): Test the uninterrupted-flight calculation.

Take the floor as zero height and upwards as positive.

Before any collision:

\[ \begin{aligned} u&=+14.7\,\mathrm{m\,s^{-1}}\\ a&=-9.8\,\mathrm{m\,s^{-2}} \end{aligned} \]

At the sensor:

\[ 6.125=14.7t-4.9t^2 \]

Rearrange:

\[ 4.9t^2-14.7t+6.125=0 \]

Divide by 4.9:

\[ t^2-3t+1.25=0 \]

Factorise:

\[ (t-0.50)(t-2.50)=0 \]

Therefore:

\[ t=0.50\,\mathrm{s} \]

or:

\[ t=2.50\,\mathrm{s} \]

Both are valid mathematical roots for a ball that moves freely without striking anything.

But the shaft has a ceiling. We need to check whether the ball reaches that ceiling before the second predicted passage.

Part (b): Find the ceiling collision.

At the ceiling, the ball is 9.80 m above its starting point.

\[ 9.80=14.7t-4.9t^2 \]

Rearrange and divide by 4.9:

\[ t^2-3t+2=0 \]

Factorise:

\[ (t-1)(t-2)=0 \]

The mathematical roots are \(t=1.00\,\mathrm{s}\) and \(t=2.00\,\mathrm{s}\).

The ball reaches the ceiling for the first time at \(t=1.00\,\mathrm{s}\). The later root describes a hypothetical return to ceiling height if the ball could pass through it, so it cannot describe the original free flight after the first collision.

Immediately before impact:

\[ \begin{aligned} v&=u+at\\ &=14.7-9.8(1.00)\\ &=+4.90\,\mathrm{m\,s^{-1}} \end{aligned} \]

The ball strikes the ceiling while moving upwards at 4.90 m/s.

The collision reverses its velocity without changing its speed.

Immediately afterwards:

\[ v=-4.90\,\mathrm{m\,s^{-1}} \]

This reversal is caused by the collision, not by the choice of positive direction.

Part (c): Find the actual sensor crossings.

The first crossing at \(t=0.50\,\mathrm{s}\) occurs before the ceiling collision and remains valid.

Its velocity is:

\[ \begin{aligned} v&=14.7-9.8(0.50)\\ &=+9.80\,\mathrm{m\,s^{-1}} \end{aligned} \]

For the motion after impact, start a new constant-acceleration interval.

Let \(\tau\) be the elapsed time since the ceiling collision.

At the start of this new interval:

\[ \begin{aligned} u'&=-4.90\,\mathrm{m\,s^{-1}}\\ a&=-9.8\,\mathrm{m\,s^{-2}} \end{aligned} \]

The sensor is 3.675 m below the ceiling.

Because upwards is positive:

\[ s=6.125-9.80 \]

\[ s=-3.675\,\mathrm{m} \]

Use the displacement equation for this new interval:

\[ s=u'\tau+\frac12a\tau^2 \]

Substitute:

\[ -3.675=-4.9\tau-4.9\tau^2 \]

Divide by \(-4.9\):

\[ 0.75=\tau+\tau^2 \]

Rearrange:

\[ \tau^2+\tau-0.75=0 \]

Factorise:

\[ (\tau-0.50)(\tau+1.50)=0 \]

So:

\[ \tau=0.50\,\mathrm{s} \]

or:

\[ \tau=-1.50\,\mathrm{s} \]

We discard the negative elapsed time because it refers to a time before the collision that started this interval.

The ball reaches the sensor 0.50 s after striking the ceiling.

Measured from the original launch:

\[ t=1.00+0.50=1.50\,\mathrm{s} \]

Its velocity at this second crossing is:

\[ \begin{aligned} v&=-4.90-9.8(0.50)\\ &=-9.80\,\mathrm{m\,s^{-1}} \end{aligned} \]

The actual crossings are therefore:

Part (d): Check the floor arrival.

The ball starts the post-collision flight at height 9.80 m with downward velocity \(-4.90\,\mathrm{m\,s^{-1}}\).

At the floor, its height is zero:

\[ 0=9.80-4.90\tau-4.9\tau^2 \]

Divide by 4.9 and rearrange:

\[ \tau^2+\tau-2=0 \]

Factorise:

\[ (\tau-1)(\tau+2)=0 \]

The physically relevant time since collision is:

\[ \tau=1.00\,\mathrm{s} \]

Therefore, the ball reaches the floor at:

\[ t=1.00+1.00=2.00\,\mathrm{s} \]

This confirms that the ball passes the sensor on the way down at 1.50 s, before reaching the floor.

The uninterrupted-flight prediction of a second sensor crossing at 2.50 s is physically impossible in the actual experiment: the ball has already been caught at the floor.

Part (e): Can a different positive direction fix the original calculation?

No. Changing the positive coordinate direction alone cannot change the physical time of an event.

If we switch to downwards positive after the ceiling collision, the correct post-collision values become:

\[ \begin{aligned} u'&=+4.90\,\mathrm{m\,s^{-1}}\\ a&=+9.8\,\mathrm{m\,s^{-2}}\\ s&=+3.675\,\mathrm{m} \end{aligned} \]

We would calculate:

\[ 3.675=4.90\tau+4.9\tau^2 \]

This produces the same physical result:

\[ \tau=0.50\,\mathrm{s} \]

The calculation works because we have correctly restarted the motion after the collision, not because we changed the sign convention.

That’s the deeper distinction.

A change of positive direction changes how quantities are represented. A collision changes the actual motion.

A correct vertical-motion calculation must account for both, but only the collision requires a new physical initial condition.

Once you can keep displacement, velocity, and acceleration signs consistent, the same method carries into two-dimensional motion. In Projectile Motion: Horizontal and Vertical Motion Explained, the vertical component still follows these constant-acceleration equations; you simply analyse it alongside the horizontal component.

The habit to carry forward is simple: choose an axis, keep it fixed, and check whether each mathematical solution can actually occur during the physical motion you’re modelling.