How to Find Horizontal Distance in the Last Second of Flight

A projectile is airborne for 2.4 seconds and has a horizontal velocity of \(12.0\text{ m s}^{-1}\). How far does it travel horizontally during the final second?

The answer is \(12.0\text{ m}\).

You might expect the projectile to travel further horizontally near the end of its flight because it is falling faster. But that prediction confuses its total speed with its horizontal velocity.

Under the usual HSC projectile-motion model, horizontal velocity remains constant because gravity acts vertically. That means the projectile travels the same horizontal distance during every one-second interval.

The key method is:

\[ \boxed{d_{\text{last second}}=|u_x|\times 1.00} \]

Here, \(u_x\) is the horizontal velocity in metres per second, and the resulting distance is in metres.

The only complication is that you sometimes need to calculate the total flight time first, particularly when the projectile lands at a different height from where it was launched.

01Why the last second is a special interval

Imagine a ball launched at an angle. It rises, reaches its maximum height, and then falls towards the ground.

Its path is curved, but its motion can be separated into two independent directions.

• Horizontal motion: the ball maintains constant horizontal velocity.
• Vertical motion: gravity continuously changes its vertical velocity.

The important distinction is that the projectile’s vertical velocity changes throughout its flight, but its horizontal velocity does not change in the ideal model.

Deriving the final-second formula

Let:

• \(u_x\) = horizontal velocity in \(\text{m s}^{-1}\)
• \(T\) = total flight time in seconds
• \(x(t)\) = horizontal position at time \(t\)

With negligible air resistance, horizontal acceleration is zero.

Therefore:

\[ a_x=0 \]

The horizontal displacement after time \(t\) is:

\[ x(t)=u_xt \]

The final second of flight begins at \(T-1.00\) seconds and ends at \(T\).

Its horizontal displacement is:

\[ \begin{aligned} \Delta x &=x(T)-x(T-1.00)\\ &=u_xT-u_x(T-1.00)\\ &=u_x(1.00) \end{aligned} \]

Taking the magnitude gives the horizontal distance:

\[ \boxed{d_{\text{last second}}=|u_x|(1.00\text{ s})} \]

This is the central result.

Notice that the total flight time has disappeared from the final equation. Once you know the horizontal velocity, you can calculate the horizontal distance travelled during any complete one-second interval.

An important condition: The projectile must remain airborne for at least one second. If its entire flight lasts less than one second, there is no complete one-second interval within its flight.

What if the projectile is launched at an angle?

Suppose a projectile is launched with an initial speed \(u\) at an angle \(\theta\) above the horizontal.

Its initial velocity has two components:

\[ u_x=u\cos\theta \]

\[ u_y=u\sin\theta \]

Here:

• \(u\) is the initial speed in \(\text{m s}^{-1}\).
• \(\theta\) is the launch angle measured above the horizontal.
• \(u_x\) is the horizontal component.
• \(u_y\) is the vertical component.

Substitute the horizontal component into the final-second equation:

\[ \boxed{d_{\text{last second}}=|u\cos\theta|(1.00\text{ s})} \]

For a projectile travelling to the right, this simplifies to:

\[ d_{\text{last second}}=u\cos\theta \]

provided the flight lasts at least one second.

You do not need the vertical velocity at the end of the flight. You only need the horizontal component of the initial velocity.

02Finding the total flight time

Sometimes a question gives you the total flight time directly. If so, you can go straight to the horizontal calculation.

Other questions give you the launch speed, launch angle, and landing height. You must then calculate the flight time using vertical motion.

This is where students often lose marks by applying the wrong flight-time formula.

We need to distinguish two situations.

Case A: The projectile lands at the same height it was launched

Take upwards as the positive vertical direction.

The vertical acceleration is:

\[ a_y=-g \]

where \(g=9.8\text{ m s}^{-2}\) is the magnitude of gravitational acceleration near Earth’s surface.

The vertical displacement equation is:

\[ \Delta y=u_yt+\frac12a_yt^2 \]

Substituting the acceleration:

\[ \Delta y=u_yt-\frac12gt^2 \]

For a projectile that returns to its original height, its total vertical displacement is zero.

Therefore:

\[ 0=u_yT-\frac12gT^2 \]

Factorising:

\[ 0=T\left(u_y-\frac12gT\right) \]

There are two solutions:

• \(T=0\), which represents the instant of launch.
• \(T=\frac{2u_y}{g}\), which represents the later landing.

The total flight time is therefore:

\[ \boxed{T=\frac{2u_y}{g}} \]

Once you have this value, check that \(T\geq1.00\text{ s}\), then calculate the horizontal distance during the final second.

Case B: The projectile lands at a different height

Now imagine launching a ball from the top of a cliff.

The ball does not land at its original height. It has additional vertical distance to fall.

You cannot use:

\[ T=\frac{2u_y}{g} \]

That equation assumes the launch and landing heights are equal.

Instead, return to the general vertical displacement equation:

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

Suppose the landing point is \(h\) metres below the launch point.

Taking upwards as positive:

\[ \Delta y=-h \]

Therefore:

\[ -h=u_yT-\frac12gT^2 \]

Rearranging:

\[ \frac12gT^2-u_yT-h=0 \]

This is a quadratic equation. Using the quadratic formula and selecting the positive root gives:

\[ \boxed{T=\frac{u_y+\sqrt{u_y^2+2gh}}{g}} \]

This formula applies when the landing point is below the launch point.

If the projectile lands above its launch point, use the original vertical displacement equation with the actual positive vertical displacement instead.

The key idea remains unchanged: the landing height determines the flight time, while the horizontal velocity determines the horizontal distance travelled during the final second.

A useful special case: horizontal launch

A horizontal launch is a particularly useful example because the initial vertical velocity is zero.

Therefore:

\[ u_y=0 \]

For a horizontal launch from a height \(h\):

\[ -h=-\frac12gT^2 \]

Rearranging:

\[ \boxed{T=\sqrt{\frac{2h}{g}}} \]

The horizontal velocity is simply the launch speed:

\[ u_x=u \]

If the flight lasts at least one second:

\[ d_{\text{last second}}=u(1.00\text{ s}) \]

Notice why the same-height formula would fail here. Substituting \(u_y=0\) into \(T=2u_y/g\) would incorrectly give zero flight time.

A horizontal launch from a raised position is a different situation from a projectile returning to its launch height.

03How an experiment can test the rule

The constant-horizontal-velocity model is not just a convenient equation. It makes a prediction that can be tested.

Consider a school investigation using a ball launcher and high-speed video.

A ball is launched across a flat testing area. A fixed camera records its flight from the side, while a measuring scale is positioned in the same vertical plane as the ball’s trajectory.

The camera records the ball’s position at equal time intervals.

What should the experiment show?

If horizontal velocity is constant, equal time intervals should produce equal horizontal displacements.

For example, suppose the ball’s total flight time is \(1.60\text{ s}\).

The final second therefore begins at:

\[ t=1.60-1.00=0.60\text{ s} \]

Imagine that video tracking produces the following illustrative data:

These are illustrative measurements, not results from an actual experiment.

Notice that the horizontal position increases by \(0.48\text{ m}\) every \(0.20\text{ s}\).

That is consistent with constant horizontal velocity.

The horizontal distance during the final second is:

\[ \Delta x=x(1.60)-x(0.60) \]

\[ \Delta x=3.84-1.44 \]

\[ \boxed{\Delta x=2.40\text{ m}} \]

The average horizontal velocity during this interval is:

\[ \bar v_x=\frac{2.40}{1.00} \]

\[ \bar v_x=2.40\text{ m s}^{-1} \]

The result supports the constant-horizontal-velocity model over the measured interval.

What could make the experiment unreliable?

A camera positioned at an angle to the plane of motion can introduce perspective errors. If the measuring scale is closer to the camera than the ball’s trajectory, the apparent distances may not match the actual distances.

A better design places the camera perpendicular to the plane of motion and the scale in that same plane. The camera should also remain fixed throughout the recording.

Air resistance is another limitation. A real ball experiences drag, which can reduce its horizontal velocity. Consequently, equal time intervals might not produce exactly equal horizontal displacements.

The experiment tests the model within its measurement accuracy. It does not prove that horizontal velocity is constant for every projectile in every real-world situation.

04Worked examples

05Common mistakes to avoid

A quick decision process

When you see a question asking for the horizontal distance during the last second of flight, follow this sequence:

1. Check whether total flight time is provided. If it is, use it. Otherwise, calculate it using vertical motion.
2. Check the launch and landing heights. Use the same-height formula only when the projectile returns to its original height.
3. Check that the flight lasts at least one second. Otherwise, there is no complete final-second interval.
4. Resolve the horizontal velocity. Use \(u_x=u\cos\theta\), unless the projectile is launched horizontally.
5. Calculate the horizontal distance. Multiply the horizontal velocity magnitude by \(1.00\text{ s}\).

There is also a useful shortcut when the question gives the total horizontal range \(R\) and total flight time \(T\).

Under the constant-horizontal-velocity model:

\[ u_x=\frac{R}{T} \]

Therefore:

\[ \boxed{d_{\text{last second}}=\frac{R}{T}(1.00\text{ s})} \]

This shortcut works because \(R/T\) equals the constant horizontal velocity. It is not a general rule for projectiles affected by significant air resistance.

06Questions and solutions

Use \(g=9.8\text{ m s}^{-2}\) unless otherwise stated. Ignore air resistance unless a question explicitly introduces experimental evidence of horizontal acceleration.

Question 1

A basketball is launched from ground level at an initial speed of \(14.0\text{ m s}^{-1}\), at an angle of \(30^\circ\) above the horizontal. It lands at the same height from which it was launched.

Calculate the horizontal distance travelled during the final second of flight.

Solution 1

The basketball travels approximately \(12.1\text{ m}\) horizontally during its final second.

First, resolve the initial velocity.

Horizontal component:

\[ u_x=14.0\cos30^\circ \]

\[ u_x=12.12\text{ m s}^{-1} \]

Vertical component:

\[ u_y=14.0\sin30^\circ \]

\[ u_y=7.00\text{ m s}^{-1} \]

Because the basketball lands at its original height, use:

\[ T=\frac{2u_y}{g} \]

Substitute:

\[ T=\frac{2(7.00)}{9.8} \]

\[ T=1.43\text{ s} \]

The flight lasts longer than one second, so a complete final-second interval exists.

The horizontal distance is:

\[ d_{\text{last second}}=u_x(1.00) \]

\[ d_{\text{last second}}=12.12(1.00) \]

\[ \boxed{d_{\text{last second}}=12.1\text{ m}} \]

The basketball covers the same horizontal distance during its first second. Its vertical velocity changes, but its horizontal velocity remains constant.

Question 2

A stone is launched from a ledge \(6.0\text{ m}\) above the ground. Its initial speed is \(16.0\text{ m s}^{-1}\), at an angle of \(35^\circ\) above the horizontal.

Ignoring air resistance:

a) Calculate the total flight time.

b) Calculate the horizontal distance travelled during the final second.

c) State the time interval corresponding to the final second.

Solution 2

The stone travels approximately \(13.1\text{ m}\) horizontally during its final second, from approximately \(1.39\text{ s}\) to \(2.39\text{ s}\).

Part A: Total flight time

First, resolve the initial velocity.

\[ u_x=16.0\cos35^\circ \]

\[ u_x=13.11\text{ m s}^{-1} \]

\[ u_y=16.0\sin35^\circ \]

\[ u_y=9.18\text{ m s}^{-1} \]

The landing point is \(6.0\text{ m}\) below the launch point.

Using upward as positive:

\[ -6.0=9.18T-4.90T^2 \]

Rearranging:

\[ 4.90T^2-9.18T-6.0=0 \]

Using the quadratic formula:

\[ T=\frac{9.18\pm\sqrt{9.18^2+4(4.90)(6.0)}}{9.80} \]

Selecting the positive root:

\[ \boxed{T=2.39\text{ s}} \]

Part B: Horizontal distance during the final second

Horizontal velocity remains constant.

\[ d_{\text{last second}}=u_x(1.00) \]

\[ d_{\text{last second}}=13.11(1.00) \]

\[ \boxed{d_{\text{last second}}=13.1\text{ m}} \]

Part C: Final-second interval

The final second starts one second before landing:

\[ t_{\text{start}}=2.39-1.00 \]

\[ t_{\text{start}}=1.39\text{ s} \]

Therefore, the final second runs from:

\[ \boxed{1.39\text{ s to }2.39\text{ s}} \]

The important detail is that the lower landing height changes the total flight time. It does not change the constant-horizontal-velocity rule.

Question 3

Two balls are launched horizontally at \(10.0\text{ m s}^{-1}\), but from platforms of different heights.

• Ball A is launched from a platform \(3.0\text{ m}\) above the ground.
• Ball B is launched from a platform \(19.6\text{ m}\) above the ground.

Both balls land on level ground. Ignore air resistance.

For each ball:

a) Calculate its total flight time.

b) Determine whether a complete final one-second interval exists.

c) Calculate the horizontal distance travelled during the final second, if a complete interval exists. Otherwise, calculate the horizontal distance travelled during the entire flight.

Explain why the two balls do not necessarily travel the same horizontal distance during their final portion of flight.

Solution 3

Ball A does not remain airborne for a complete second. It travels approximately \(7.82\text{ m}\) horizontally during its entire flight.

Ball B remains airborne for exactly \(2.00\text{ s}\), so it travels \(10.0\text{ m}\) horizontally during its final second.

Ball A

For a horizontal launch:

\[ u_y=0 \]

Use the vertical displacement equation:

\[ -h=-\frac12gT^2 \]

Rearranging:

\[ T=\sqrt{\frac{2h}{g}} \]

Substitute the height:

\[ T=\sqrt{\frac{2(3.0)}{9.8}} \]

\[ \boxed{T=0.782\text{ s}} \]

The flight lasts less than one second.

Therefore, there is no complete one-second interval within the flight.

The ball’s entire horizontal distance is:

\[ d=u_xT \]

\[ d=10.0(0.782) \]

\[ \boxed{d=7.82\text{ m}} \]

Ball B

Again, use:

\[ T=\sqrt{\frac{2h}{g}} \]

Substituting:

\[ T=\sqrt{\frac{2(19.6)}{9.8}} \]

\[ \boxed{T=2.00\text{ s}} \]

A complete final-second interval exists, from \(1.00\text{ s}\) to \(2.00\text{ s}\).

Since the horizontal velocity is constant:

\[ d=u_x(1.00) \]

\[ d=10.0(1.00) \]

\[ \boxed{d=10.0\text{ m}} \]

The difference comes from the available flight time.

Both balls have the same horizontal velocity, but Ball A does not stay airborne long enough to complete a full second. Ball B does.

This is why you should always check the total flight time before applying the final-second formula.

Question 4

Two projectiles are launched from the same level at the same initial speed of \(20.0\text{ m s}^{-1}\).

• Projectile A is launched at \(30^\circ\) above the horizontal.
• Projectile B is launched at \(60^\circ\) above the horizontal.

Both land at their original launch height. Ignore air resistance.

a) Calculate the total flight time of each projectile.

b) Calculate the horizontal distance travelled during the final second for each projectile.

c) Both projectiles have the same total horizontal range. Explain why their final-second horizontal distances are different.

Solution 4

Projectile A travels approximately \(17.3\text{ m}\) horizontally during its final second, while projectile B travels \(10.0\text{ m}\).

Both have the same total range, but they have different horizontal velocities and flight times.

Part A: Calculate the flight times

For projectile A:

\[ u_{y,A}=20.0\sin30^\circ \]

\[ u_{y,A}=10.0\text{ m s}^{-1} \]

Using the same-height flight-time equation:

\[ T_A=\frac{2u_{y,A}}{g} \]

\[ T_A=\frac{2(10.0)}{9.8} \]

\[ \boxed{T_A=2.04\text{ s}} \]

For projectile B:

\[ u_{y,B}=20.0\sin60^\circ \]

\[ u_{y,B}=17.32\text{ m s}^{-1} \]

Therefore:

\[ T_B=\frac{2(17.32)}{9.8} \]

\[ \boxed{T_B=3.53\text{ s}} \]

Projectile B remains airborne for longer because it has a larger initial vertical velocity.

Part B: Calculate the final-second distances

For projectile A:

\[ u_{x,A}=20.0\cos30^\circ \]

\[ u_{x,A}=17.32\text{ m s}^{-1} \]

Therefore:

\[ d_A=17.32(1.00) \]

\[ \boxed{d_A=17.3\text{ m}} \]

For projectile B:

\[ u_{x,B}=20.0\cos60^\circ \]

\[ u_{x,B}=10.0\text{ m s}^{-1} \]

Therefore:

\[ d_B=10.0(1.00) \]

\[ \boxed{d_B=10.0\text{ m}} \]

Part C: Explain the difference

Both projectiles have the same total range because the higher launch angle gives projectile B more time in the air, compensating for its smaller horizontal velocity.

We can check their ranges.

For projectile A:

\[ R_A=17.32(2.04) \]

\[ R_A\approx35.3\text{ m} \]

For projectile B:

\[ R_B=10.0(3.53) \]

\[ R_B\approx35.3\text{ m} \]

Their ranges are equal.

However, the final-second distance depends on horizontal velocity, not total range.

Projectile A has a greater horizontal velocity, so it covers more horizontal ground during that one-second interval.

This is an important distinction: two projectiles can have identical ranges but different distances travelled during their final second.

Question 5

A ball is launched from an elevated platform and lands on the ground \(3.00\text{ s}\) later.

A calibrated side-on video recording gives the following horizontal position measurements:

A student calculates the average horizontal velocity over the entire flight:

\[ \bar v_x=\frac{42.0}{3.00} \]

They obtain \(14.0\text{ m s}^{-1}\) and conclude that the ball must have travelled \(14.0\text{ m}\) horizontally during its final second.

Another student argues that the video measurements show otherwise.

Assume the measurements are accurate.

a) Calculate the actual horizontal distance travelled during the final second.

b) Explain why the first student’s calculation does not establish the final-second distance.

c) Determine whether the measured positions are consistent with constant horizontal acceleration. If so, calculate the horizontal acceleration and initial horizontal velocity.

d) Explain why these measurements challenge the usual ideal projectile model. Do they prove that air resistance is responsible?

Solution 5

The ball travels \(12.0\text{ m}\) horizontally during its final second, not \(14.0\text{ m}\).

The measurements also show that its horizontal velocity is decreasing. They are consistent with constant horizontal acceleration of \(-2.0\text{ m s}^{-2}\), although they do not establish the physical cause of that acceleration.

Part A: Actual final-second horizontal distance

The final second runs from \(t=2.00\text{ s}\) to \(t=3.00\text{ s}\).

Therefore:

\[ \Delta x=x(3.00)-x(2.00) \]

Substituting the measured positions:

\[ \Delta x=42.0-30.0 \]

\[ \boxed{\Delta x=12.0\text{ m}} \]

Part B: Why the average velocity calculation is insufficient

The first student calculated:

\[ \bar v_x=\frac{42.0}{3.00} \]

\[ \bar v_x=14.0\text{ m s}^{-1} \]

This is the average horizontal velocity over the entire three-second flight.

It does not tell us the average velocity during the final second if horizontal velocity changes.

We can see the difference by examining the displacement during each one-second interval:

The horizontal distance covered in each successive second decreases.

The final-second distance is therefore \(12.0\text{ m}\), not the full-flight average of \(14.0\text{ m}\).

Part C: Determine the horizontal acceleration

The measured data are consistent with constant horizontal acceleration.

Use the horizontal displacement equation:

\[ \Delta x=u_xt+\frac12a_xt^2 \]

At \(t=1.00\text{ s}\):

\[ 16.0=u_x+\frac12a_x \]

At \(t=2.00\text{ s}\):

\[ 30.0=2u_x+2a_x \]

Doubling the first equation gives:

\[ 32.0=2u_x+a_x \]

Subtracting the second equation:

\[ 32.0-30.0=a_x-2a_x \]

Therefore:

\[ \boxed{a_x=-2.0\text{ m s}^{-2}} \]

Substitute this value into the first equation:

\[ 16.0=u_x+\frac12(-2.0) \]

\[ 16.0=u_x-1.0 \]

\[ \boxed{u_x=17.0\text{ m s}^{-1}} \]

We can check this model at \(t=3.00\text{ s}\):

\[ x=17.0(3.00)+\frac12(-2.0)(3.00)^2 \]

\[ x=51.0-9.0 \]

\[ x=42.0\text{ m} \]

This matches the final measured position.

The constant-acceleration model is therefore consistent with all four measurements.

Part D: Evaluate the ideal projectile model

The usual ideal projectile model assumes:

\[ a_x=0 \]

This means horizontal velocity remains constant.

If that assumption were correct, the ball would cover equal horizontal distances during each one-second interval.

However, the measured distances are \(16.0\text{ m}\), \(14.0\text{ m}\), and \(12.0\text{ m}\).

The data show that horizontal velocity is changing, so the standard ideal model cannot explain the measurements.

Air resistance could explain this behaviour. As the ball moves forwards through the air, drag acts against its motion and can reduce its horizontal velocity.

However, the measurements alone do not prove that air resistance is responsible. A horizontal wind, another external force, or an unaccounted measurement issue could also produce horizontal acceleration.

The important conclusion is that the assumption of zero horizontal acceleration is inconsistent with these measurements.

This question exposes a limitation of the simple projectile model: constant horizontal velocity is an assumption about the forces acting on the projectile, not an unconditional rule for every real-world flight.

06Where to go next

Finding the horizontal distance travelled during the last second is really an application of two ideas working together:

• Vertical motion determines when the projectile lands.
• Horizontal motion determines how far it travels during a particular time interval.

Once you understand this separation, you can approach more complicated projectile questions without searching for a different formula every time.

For a broader explanation of how horizontal and vertical components work together, see Projectile Motion: Horizontal and Vertical Motion Explained.

If you want to extend the calculation to the entire flight, including projectiles launched from cliffs or other elevated positions, read Horizontal Range of a Projectile: HSC Physics Guide.

The useful habit to carry into your next question is simple: calculate the flight time using vertical motion, then calculate the horizontal distance using horizontal motion. Always check that the time interval you’re calculating actually exists.