How to Find Distance Travelled in the Last Second of a Fall

A ball is dropped from a height and hits the ground 3.0 seconds later. Did it travel one-third of its total distance during the last second?

Think about its speed just before impact. The ball is accelerating downwards, so it’s travelling faster in the third second than in the first. In fact, it travels 24.5 m in the last second, out of a total fall of 44.1 m.

To find the distance travelled during the last second of a fall, calculate how far the object has fallen when it hits the ground, then subtract how far it had fallen exactly one second earlier:

\[ d_{\text{last}}=s(T)-s(T-1.0\,\mathrm{s}) \]

Here, \(d_{\text{last}}\) is the distance travelled in the final second, \(T\) is the total time from release to impact, and \(s(t)\) is the downward displacement after time \(t\).

This subtraction gives the actual distance travelled as long as the object moves downwards throughout that final second. We’ll deal with a change of direction later.

01Why the last second covers more distance

Imagine recording a falling ball once every second. Gravity makes its downward velocity increase, so the gaps between successive positions get larger.

For a ball released from rest, ignoring air resistance, the downward displacement is:

\[ s(t)=\frac12 gt^2 \]

where:

• \(s(t)\) is downward displacement from the release point, in metres (m).
• \(g\) is gravitational acceleration, taken as \(9.8\,\mathrm{m\,s^{-2}}\).
• \(t\) is time since release, in seconds (s).

We’ll take downwards as positive. Because the ball starts from rest and keeps moving down, its displacement is also the distance it has fallen.

For our 3.0-second fall, the positions work out as follows.

These are calculated values for an idealised fall, not experimental measurements.

Notice that the last column isn’t the same as the middle column. The middle column gives the running total. The last column tells you how much distance was added during each individual second.

The calculation for the last second is:

\[ \begin{aligned} d_{\text{last}}&=s(3.0\,\mathrm{s})-s(2.0\,\mathrm{s})\\ &=44.1\,\mathrm{m}-19.6\,\mathrm{m}\\ &=24.5\,\mathrm{m} \end{aligned} \]

Dividing the total distance by three would give 14.7 m per second. That’s the average distance per second over the whole fall, not the distance travelled in each particular second.

The important distinction is between a cumulative measurement and the change in that measurement over the interval you actually want.

02How could you measure the distance travelled in the last second?

The calculation predicts that a falling ball covers more distance in each successive second. But how could you test that prediction without assuming the equation is correct?

You need two measurements: the ball’s position one second before impact, and its position at impact.

The apparatus and the measurement problem

Imagine a ball released remotely at the top of a safely enclosed vertical drop path. Reference marks at known heights run alongside its path.

A camera is fixed to the side, facing the path directly. It records the ball’s movement at a known frame rate. The reference marks and the ball’s path lie in the same plane, helping to minimise errors caused by viewing objects from an angle.

An initial design using a handheld stopwatch would have a serious weakness. Human reaction time makes it difficult to identify positions at precise one-second intervals.

Recording the fall allows us to revise the method:

1. Identify the frame where the ball is released and the frame immediately before impact.
2. Use the recording’s frame rate to determine the elapsed time.
3. Locate the frame exactly one second before impact.
4. Measure the ball’s vertical position in those two frames using the calibrated reference marks.
5. Subtract the positions to find the distance travelled.

A simple check is to compare distances in successive equal time intervals.

For example, consider an idealised 2.0-second fall from rest. Uniform acceleration predicts 4.9 m travelled during the first second and 14.7 m during the second.

If the ball instead moved at constant velocity, the two distances would be equal. Increasing separation between positions recorded at equal time intervals supports an accelerating-motion model.

The figures here are theoretical predictions, not recorded observations. Real video measurements would also need to account for frame timing, image resolution, parallax, and air resistance.

This is why the constant-acceleration equation is a model to test, rather than a guarantee that every real fall follows it perfectly.

03The method for any uniform-acceleration fall

For the simplest case, the ball is released from rest. But some HSC questions give an initial velocity, or ask you to calculate the time of impact from the drop height.

We can handle those using the same idea.

Step 1: Choose your positive direction

For a ball moving downwards, taking downwards as positive is usually convenient.

The displacement equation becomes:

\[ s(t)=ut+\frac12 gt^2 \]

Here:

• \(u\) is initial velocity, in metres per second (\(\mathrm{m\,s^{-1}}\)), with downward velocities positive.
• \(g\) is constant downward acceleration, in \(\mathrm{m\,s^{-2}}\).
• \(t\) is elapsed time, in seconds.
• \(s(t)\) is displacement from the release point, in metres.

If the ball is dropped from rest, \(u=0\). If it is thrown downwards, \(u\) is positive. If it is thrown upwards, \(u\) is negative under this sign convention.

Step 2: Find the total flight time

Call the time of impact \(T\).

If the total time is supplied, you can use it immediately. If the drop height \(H\), measured vertically downwards from the release point, is supplied instead, solve:

\[ H=uT+\frac12gT^2 \]

A quadratic equation may produce two mathematical roots. Keep the root that describes impact after release, so \(T>0\). A negative time is not a second possible impact in the future.

You also need \(T\geq1.0\,\mathrm{s}\) for there to be a complete last second.

Step 3: Subtract the two positions

Let \(\tau=1.0\,\mathrm{s}\), the length of the interval we want.

Subtracting the two displacement equations gives:

\[ \begin{aligned} \Delta s_{\text{last}} &=s(T)-s(T-\tau)\\ &=u\tau+g\tau(T-\tfrac12\tau) \end{aligned} \]

The result is displacement during the final second.

For an object moving continuously downwards during that interval, this is also its distance travelled.

If you already know the velocity at impact, there’s another useful approach. Under constant acceleration, average velocity over an interval is the average of the initial and final velocities:

\[ \bar v=\frac{v_i+v_f}{2} \]

Here, \(\bar v\) is average velocity, while \(v_i\) and \(v_f\) are velocities at the beginning and end of the interval. Multiply \(\bar v\) by the one-second duration to obtain displacement. For continuous downward motion with downwards positive, this gives the distance as well.

04The important exception: When displacement is not distance

There’s one assumption you must check before using the subtraction method.

What happens if the object changes direction during the one-second interval?

Imagine a ball thrown vertically upwards from ground level at \(9.8\,\mathrm{m\,s^{-1}}\).

It reaches its highest point at \(t=1.0\,\mathrm{s}\), at a height of 4.9 m.

At \(t=0.5\,\mathrm{s}\), its height is 3.675 m. At \(t=1.5\,\mathrm{s}\), its height is also 3.675 m.

If we simply subtract the positions, we get:

\[ \Delta y=3.675-3.675=0\,\mathrm{m} \]

But the ball certainly travelled some distance between those times.

It moved up from 3.675 m to 4.9 m, then back down to 3.675 m:

\[ \begin{aligned} d&=(4.9-3.675)+(4.9-3.675)\\ &=2.45\,\mathrm{m} \end{aligned} \]

This particular interval is not the final second of the ball’s entire flight. It isolates the problem caused by a turning point.

The correct method is to split the interval at the instant when velocity is zero, calculate the distance on each side, and add the positive distances.

The general distinction is covered in Distance and Displacement in HSC Physics Explained Clearly. For more on why acceleration continues at the highest point, see Can an Object Have Zero Velocity and Nonzero Acceleration?.

Which method should you use?

These methods assume constant gravitational acceleration and negligible air resistance. They also end at the first impact. A bounce would introduce a new stage of motion and cannot be treated as one uninterrupted uniformly accelerated fall.

05Questions and solutions

The first question tests the basic subtraction method. The later questions require you to work backwards from measurements, compare competing explanations, and recognise when the final second doesn’t reveal the whole story of a fall.

Unless stated otherwise, use \(g=9.8\,\mathrm{m\,s^{-2}}\), ignore air resistance, and treat motion as vertical.

Question 1

A marble is released from rest and reaches the ground 2.5 seconds later.

Calculate the distance it travels during the final second before impact. State the time interval used.

Solution 1

The marble travels 19.6 m during its final second, from \(t=1.5\,\mathrm{s}\) to \(t=2.5\,\mathrm{s}\).

With downwards positive, the displacement from release is:

\[ s(t)=\frac12gt^2 \]

At impact:

\[ \begin{aligned} s(2.5\,\mathrm{s}) &=\frac12(9.8)(2.5)^2\,\mathrm{m}\\ &=30.625\,\mathrm{m} \end{aligned} \]

One second earlier:

\[ \begin{aligned} s(1.5\,\mathrm{s}) &=\frac12(9.8)(1.5)^2\,\mathrm{m}\\ &=11.025\,\mathrm{m} \end{aligned} \]

Therefore:

\[ \begin{aligned} d_{\text{last}} &=30.625-11.025\\ &=19.6\,\mathrm{m} \end{aligned} \]

The tempting mistake is to use the displacement during the first second, which would be 4.9 m. But the marble is moving much faster during the interval immediately before impact.

Question 2

Two stones, A and B, leave the same ledge, 44.1 m above the ground.

• Stone A is released from rest at \(t=0\).
• Stone B is thrown vertically downwards exactly 1.0 second later.
• Both stones strike the ground at the same instant.

Find the initial downward velocity of stone B and the distance each stone travels during its own final second.

Explain whether simultaneous impact implies that the stones have the same velocity at impact.

Solution 2

Stone B must be launched downwards at 12.25 m/s. Stone A travels 24.5 m during its final second, while B travels 26.95 m. Their impact velocities are different, even though they land together.

First, calculate stone A’s flight time using downwards as positive:

\[ \begin{aligned} 44.1&=\frac12(9.8)T_A^2\\ T_A^2&=9.0\,\mathrm{s^2}\\ T_A&=3.0\,\mathrm{s} \end{aligned} \]

Stone B is released one second later, so it has only 2.0 seconds to cover the same height.

Let its initial downward velocity be \(u_B\).

\[ \begin{aligned} 44.1&=u_B(2.0)+4.9(2.0)^2\\ 44.1&=2.0u_B+19.6\\ u_B&=12.25\,\mathrm{m\,s^{-1}} \end{aligned} \]

For stone A, the final second is from \(t=2.0\) to \(3.0\,\mathrm{s}\) after its release:

\[ \begin{aligned} d_A&=44.1-4.9(2.0)^2\\ &=24.5\,\mathrm{m} \end{aligned} \]

For stone B, the final second is from 1.0 to 2.0 seconds after B’s release.

Its downward displacement during its first second is:

\[ \begin{aligned} s_B(1.0\,\mathrm{s}) &=12.25(1.0)+4.9(1.0)^2\\ &=17.15\,\mathrm{m} \end{aligned} \]

Its remaining distance is:

\[ \begin{aligned} d_B&=44.1-17.15\\ &=26.95\,\mathrm{m} \end{aligned} \]

We can also compare impact velocities using \(v=u+gt\):

\[ \begin{aligned} v_A&=0+9.8(3.0)\\ &=29.4\,\mathrm{m\,s^{-1}}\\ v_B&=12.25+9.8(2.0)\\ &=31.85\,\mathrm{m\,s^{-1}} \end{aligned} \]

Both velocities are downwards.

It’s tempting to think that equal impact times and equal heights mean equal final-second distances. But B started later with a substantial initial velocity, and its impact speed is greater.

The relevant quantity is each stone’s velocity during its own final-second interval, not simply the clock time at which it lands.

Question 3

An object moves vertically downwards from the moment it leaves a ledge until it strikes the ground.

The ledge is 88.2 m above the ground. A video-tracking system gives the following idealised measurements:

Two explanations are proposed:

• Model A: The object was released from rest.
• Model B: The object was projected downwards at \(14.7\,\mathrm{m\,s^{-1}}\).

Assuming uniform acceleration, determine the acceleration from the video data. Then decide which model is consistent with both the measured distances and the height of the ledge. Find its total flight time.

Solution 3

The acceleration is \(9.8\,\mathrm{m\,s^{-2}}\) downwards, and Model B is consistent with the observations. The object was projected downwards at \(14.7\,\mathrm{m\,s^{-1}}\) and reached the ground after 3.0 seconds.

The key is that successive one-second distances reveal acceleration, but don’t by themselves reveal the initial velocity.

Under uniform acceleration, the distance travelled in one second increases by \(a(1.0\,\mathrm{s})^2\) in the next second, provided motion remains in the same direction.

Therefore:

\[ \begin{aligned} a&=\frac{39.2-29.4}{(1.0)^2}\\ &=9.8\,\mathrm{m\,s^{-2}} \end{aligned} \]

Now test Model A.

For a release from rest, the distance during the last second is:

\[ d_{\text{last}}=g\tau(T-\tfrac12\tau) \]

With \(\tau=1.0\,\mathrm{s}\):

\[ \begin{aligned} 39.2&=9.8(T-0.5)\\ T&=4.5\,\mathrm{s} \end{aligned} \]

This model correctly predicts 29.4 m during the preceding second as well.

But it requires a total fall of:

\[ \begin{aligned} H_A&=4.9(4.5)^2\\ &=99.225\,\mathrm{m} \end{aligned} \]

The measured height is only 88.2 m. Model A is inconsistent with that independent constraint.

Now test Model B.

With \(u=14.7\,\mathrm{m\,s^{-1}}\) downwards:

\[ 88.2=14.7T+4.9T^2 \]

Solving for the positive time gives \(T=3.0\,\mathrm{s}\).

Its displacement after the first second is:

\[ \begin{aligned} s(1.0\,\mathrm{s}) &=14.7(1.0)+4.9(1.0)^2\\ &=19.6\,\mathrm{m} \end{aligned} \]

After two seconds:

\[ \begin{aligned} s(2.0\,\mathrm{s}) &=14.7(2.0)+4.9(2.0)^2\\ &=49.0\,\mathrm{m} \end{aligned} \]

After three seconds it reaches the ground, 88.2 m below release.

The measured interval distances are therefore:

\[ \begin{aligned} d_{\text{second-last}}&=49.0-19.6\\ &=29.4\,\mathrm{m}\\ d_{\text{last}}&=88.2-49.0\\ &=39.2\,\mathrm{m} \end{aligned} \]

Model B matches all the stated evidence.

The subtle point is that Model A also matched the two interval distances. It failed only when we checked the total height.

You cannot always reconstruct the initial state of a falling object from its final-second movement alone. An independent measurement can be decisive.

Question 4

A ball is launched vertically upwards from level ground at \(7.35\,\mathrm{m\,s^{-1}}\). It rises, falls, and returns to its launch height.

An observer calculates the displacement during its final second and claims that the magnitude of this displacement must be the distance travelled.

Determine the total flight time, identify whether the ball changes direction during its final second, and calculate the actual distance travelled during that interval.

Explain precisely why the observer’s method succeeds or fails.

Solution 4

The ball is airborne for 1.50 s, and it travels 3.0625 m, or approximately 3.06 m, during its final second. It changes direction within that second, so subtracting its endpoint positions does not give its full distance travelled.

Choose upwards as positive.

The vertical position relative to launch height is:

\[ y(t)=7.35t-4.9t^2 \]

At ground level, \(y=0\):

\[ \begin{aligned} 0&=7.35T-4.9T^2\\ 0&=T(7.35-4.9T) \end{aligned} \]

The roots are \(T=0\) and \(T=1.50\,\mathrm{s}\).

The first root represents the launch itself. The second represents the return to ground.

Therefore, the final second runs from \(t=0.50\,\mathrm{s}\) to \(t=1.50\,\mathrm{s}\).

Now find the highest point.

At that instant, \(v=0\):

\[ \begin{aligned} 0&=7.35-9.8t_{\text{top}}\\ t_{\text{top}}&=0.75\,\mathrm{s} \end{aligned} \]

The ball reaches its highest point during the final second. We must split the interval there.

Its height at \(t=0.50\,\mathrm{s}\) is:

\[ \begin{aligned} y(0.50\,\mathrm{s}) &=7.35(0.50)-4.9(0.50)^2\\ &=2.45\,\mathrm{m} \end{aligned} \]

The highest position is:

\[ \begin{aligned} y(0.75\,\mathrm{s}) &=7.35(0.75)-4.9(0.75)^2\\ &=2.75625\,\mathrm{m} \end{aligned} \]

During the first part of the final second, the ball travels upwards:

\[ \begin{aligned} d_{\text{up}} &=2.75625-2.45\\ &=0.30625\,\mathrm{m} \end{aligned} \]

It then travels from its highest point all the way to ground level:

\[ d_{\text{down}}=2.75625\,\mathrm{m} \]

Add the two path lengths:

\[ \begin{aligned} d_{\text{last}} &=0.30625+2.75625\\ &=3.0625\,\mathrm{m} \end{aligned} \]

The observer’s proposed subtraction instead gives:

\[ \begin{aligned} \Delta y&=0-2.45\\ &=-2.45\,\mathrm{m} \end{aligned} \]

Its magnitude is 2.45 m, which is smaller than the actual 3.0625 m travelled.

The missing 0.6125 m comes from the extra upward-and-downward movement near the highest point. A net change in position ignores that reversal.

This is why you must check for a turning point before treating displacement as distance.

Question 5

Two identical balls are launched in separate trials from the same point, 44.1 m above level ground.

• In Trial A, the ball is launched vertically upwards at \(9.8\,\mathrm{m\,s^{-1}}\).
• In Trial B, it is launched vertically downwards at \(9.8\,\mathrm{m\,s^{-1}}\).

Both move under uniform gravitational acceleration with negligible air resistance.

A perfect camera records the height of each ball throughout its final 1.0 second before hitting the ground. The recordings are aligned so that impact occurs at the same displayed time.

A student claims that the camera can identify which recording belongs to the upward-launched ball, because that ball spent longer in the air.

Determine the total flight time in each trial, calculate each ball’s distance travelled in its final second, and decide whether the camera’s complete final-second position record can distinguish the two launches.

Give a mathematical explanation and identify an additional measurement that would distinguish the trials.

Solution 5

The two complete final-second camera recordings are identical, even though Trial A lasts approximately 4.162 seconds and Trial B approximately 2.162 seconds.

Both balls travel approximately 26.09 m during their final second.

The surprising result is that knowing every position during the final second still does not tell us whether a ball was originally launched upwards or downwards.

First, calculate the flight times.

Choose downwards as positive. The ground is 44.1 m below the launch position.

For Trial A, the initial velocity is \(-9.8\,\mathrm{m\,s^{-1}}\):

\[ 44.1=-9.8T_A+4.9T_A^2 \]

Rearranging and taking the positive root:

\[ \begin{aligned} T_A&=(1+\sqrt{10})\,\mathrm{s}\\ &\approx4.162\,\mathrm{s} \end{aligned} \]

For Trial B, the initial velocity is \(+9.8\,\mathrm{m\,s^{-1}}\):

\[ 44.1=9.8T_B+4.9T_B^2 \]

Its positive root is:

\[ \begin{aligned} T_B&=(\sqrt{10}-1)\,\mathrm{s}\\ &\approx2.162\,\mathrm{s} \end{aligned} \]

So the upward-launched ball does spend exactly 2.0 seconds longer in the air.

Now compare their impact velocities.

Under constant acceleration:

\[ v=u+gT \]

For Trial A:

\[ \begin{aligned} v_A&=-9.8+9.8(1+\sqrt{10})\\ &=9.8\sqrt{10}\,\mathrm{m\,s^{-1}} \end{aligned} \]

For Trial B:

\[ \begin{aligned} v_B&=9.8+9.8(\sqrt{10}-1)\\ &=9.8\sqrt{10}\,\mathrm{m\,s^{-1}} \end{aligned} \]

Both have the same impact velocity, approximately \(30.99\,\mathrm{m\,s^{-1}}\) downwards.

Why?

The upward-launched ball initially travels upwards, slows to zero velocity, then comes back through its release height. At that point, it is moving downwards at \(9.8\,\mathrm{m\,s^{-1}}\).

From that point onwards, its motion is exactly the same as that of the downward-launched ball, except shifted later in time.

This means their final-second distances are identical.

Let \(V\) represent their common downward impact speed, and let \(\tau=1.0\,\mathrm{s}\).

During the final second, both balls are moving downwards. Their common distance travelled is:

\[ d_{\text{last}}=V\tau-\frac12g\tau^2 \]

Substituting:

\[ \begin{aligned} d_{\text{last}} &=9.8\sqrt{10}(1.0)-4.9(1.0)^2\\ &\approx26.09\,\mathrm{m} \end{aligned} \]

The units of the two terms are metres.

But the camera records more than just the total distance. It records every position throughout the final second.

Could some subtle difference in the paths still reveal the initial launch direction?

No.

Let \(q\) be the number of seconds before impact, and \(h(q)\) the ball’s height above the ground at that instant.

For a ball falling towards the ground with final downward speed \(V\):

\[ h(q)=Vq-\frac12gq^2 \]

Both balls have exactly the same \(V\) and \(g\).

Therefore, for every time \(q\) between 0 and 1.0 seconds before impact, their heights are identical.

Their velocities at those corresponding times are identical too. One second before impact, for example, both have a downward speed of approximately \(21.19\,\mathrm{m\,s^{-1}}\), so neither changes direction within the recorded interval.

The two recordings contain the same physical information. They differ only in what happened before recording began.

To distinguish the trials, we could measure the elapsed time from launch to impact. That would reveal the 2.0-second difference immediately.

The tempting argument was that the longer flight must leave some trace in the final second. But under uniform acceleration, an object’s subsequent motion is determined by its position, velocity, and acceleration at the start of the interval. Its earlier path doesn’t create an extra effect.

This is a useful limitation of kinematic measurements: even perfect data over a restricted interval might not determine an object’s earlier motion uniquely.

06What to remember for the next kinematics question

When an HSC problem asks for the distance travelled during the last second of a fall, don’t begin by calculating the distance travelled in the first second.

Identify the instant of impact, move exactly 1.0 second backwards, and compare the object’s positions at those two times.

For continuous downward motion:

\[ d_{\text{last}}=s(T)-s(T-1.0\,\mathrm{s}) \]

If the flight time is unknown, find it first. If the equation gives two time roots, decide which describes the actual event after release. If the object turns around during the interval, split the motion and add the distances rather than the signed displacements.

The next useful connection is the velocity-time graph. Under uniform acceleration, velocity changes linearly with time, and the signed area beneath the graph gives displacement. Where the graph crosses zero velocity, calculating distance requires adding the magnitudes of the areas on either side.

That’s another way to see exactly why the last second of an accelerating fall can cover so much distance, and why a change of direction makes the subtraction method more delicate.