Why a Ball Falls Vertically on a Train but Curves from the Ground

A passenger releases a ball inside a train travelling east at a constant speed. To the passenger, the ball drops straight down. To someone standing beside the railway track, the same ball moves forwards while falling, tracing a curved path.

Neither observer is mistaken. They are describing the ball’s motion relative to different reference frames.

The key is that the ball does not lose its horizontal velocity when it leaves the passenger’s hand. It retains the train’s forward velocity while gravity accelerates it downwards. The passenger moves forwards alongside the ball, so they see no horizontal movement. The roadside observer does not move with the train, so they see the ball travel forwards as it falls.

01The two observers are measuring different motion

Before calculating anything, identify the object and the observer.

For this problem, we will use three labels:

• \(B\): the ball, whose motion we want to describe.
• \(P\): the passenger, travelling with the train.
• \(G\): the ground, representing the stationary roadside observer.

A reference frame is the coordinate system relative to which an object’s position and motion are measured.

We therefore have two important reference frames.

The velocity of the ball relative to the passenger is written as:

\[
\vec v_{B/P}
\]

The velocity of the ball relative to the ground is:

\[
\vec v_{B/G}
\]

The velocity of the passenger relative to the ground is:

\[
\vec v_{P/G}
\]

These are vectors, meaning they have both magnitude and direction.

The relationship between them is:

\[
\boxed{
\vec v_{B/P}
=
\vec v_{B/G}-\vec v_{P/G}
}
\]

In words:

Object’s velocity relative to the ground minus observer’s velocity relative to the ground.

Notice that both velocities on the right-hand side are measured relative to the same reference frame.

This is important. You cannot correctly calculate relative velocity by comparing two measurements made in unrelated reference frames.

What happens at the instant of release?

Suppose the train travels east at a constant velocity of \(u\).

The passenger and the ball are initially travelling together. The ball is stationary relative to the passenger immediately before release.

Therefore:

\[
\vec v_{B/P}=\vec 0
\]

The passenger’s ground-frame velocity is:

\[
\vec v_{P/G}=(u,0)
\]

At the instant of release, the ball has exactly the same ground-frame velocity:

\[
\vec v_{B/G}=(u,0)
\]

This makes sense when you think about what the ball was doing before release. It was travelling forwards with the passenger.

Releasing it removes the contact force from the passenger’s hand. It does not suddenly remove its existing horizontal velocity.

02Why the roadside observer sees a curved path

We can now explain exactly why the two observers disagree about the shape of the ball’s trajectory.

Assume:

• The train travels in a straight line at constant velocity.
• The ball is released without an additional horizontal push.
• Air resistance is negligible.
• Gravity is the only force producing acceleration after release.

Choose east as the positive horizontal direction and upwards as the positive vertical direction.

Let:

• \(u\) = train’s horizontal velocity in \(\mathrm{m\,s^{-1}}\).
• \(t\) = time since release in seconds.
• \(g\) = gravitational acceleration, approximately \(9.8\,\mathrm{m\,s^{-2}}\).
• \(x_0\) = ball’s initial horizontal position.
• \(y_0\) = ball’s initial vertical position.

We will assume the ball is released from rest relative to the passenger, with no initial vertical velocity.

From the roadside observer’s frame

The ball has an initial horizontal velocity of \(u\).

There is no horizontal acceleration in our idealised model, so its horizontal velocity remains constant.

Its horizontal position is therefore:

\[
x_{B/G}=x_0+ut
\]

Vertically, gravity accelerates the ball downwards.

Using the standard displacement equation:

\[
y_{B/G}=y_0-\frac12gt^2
\]

The two equations describe the ball’s simultaneous horizontal and vertical motion.

The horizontal position changes linearly with time, while the vertical position changes quadratically with time.

This produces a parabolic trajectory.

From the passenger’s frame

The passenger is also travelling forwards at velocity \(u\).

The ball and passenger therefore have identical horizontal velocities.

Their horizontal separation does not change after release.

Consequently:

\[
x_{B/P}=0
\]

The ball still accelerates vertically under gravity, so its vertical displacement relative to the passenger is:

\[
y_{B/P}=-\frac12gt^2
\]

The passenger sees a straight vertical trajectory.

This is the crucial difference:

• In the ground frame, the ball moves horizontally and vertically.
• In the passenger frame, the ball moves vertically but has no horizontal displacement.

Both observers measure the same physical event. They simply use different coordinate systems.

Why doesn’t the ball lose its forward motion?

A common prediction is that the ball should fall behind the train because the passenger is no longer pushing it forwards.

That prediction feels reasonable if you think an object needs a continuous forward force to maintain its forward motion.

However, Newton’s first law tells us otherwise.

An object continues moving at constant velocity unless acted on by a resultant external force.

When the passenger releases the ball, its horizontal velocity is already \(u\). With no significant horizontal resultant force, that velocity remains unchanged.

Gravity changes the ball’s vertical velocity, not its horizontal velocity.

The ball therefore continues travelling forwards at the same speed as the train.

The vertical motion is ordinary free fall. If you need a refresher on why objects have the same ideal free-fall acceleration regardless of mass, see Why Mass Does Not Change Free-Fall Acceleration.

03Speed and velocity are not interchangeable

There is another important idea hiding in this problem.

Speed is a scalar. It describes how fast an object moves but does not include direction.

Velocity is a vector. It describes both speed and direction.

When calculating relative motion, you must subtract velocities, not simply subtract speedometer readings.

Consider a train travelling east at \(15\,\mathrm{m\,s^{-1}}\).

There are two possible situations.

In both situations, the ball and passenger have the same scalar speed when measured relative to the ground.

However, their relative velocities are completely different.

In the first situation, they travel in the same direction at the same speed. Their relative velocity is zero.

In the second situation, they travel in opposite directions. The ball moves west relative to the passenger at \(30\,\mathrm{m\,s^{-1}}\).

The relative speed is the magnitude of the relative velocity:

\[
\text{Relative speed}=|\vec v_{B/P}|
\]

It is not generally the difference between two scalar speeds.

This is why identifying direction before performing calculations matters. For more examples of this distinction, see Can Two Objects Have the Same Speed but Different Velocities?.

04When would the passenger see the ball move backwards?

So far, we have assumed that the train travels at constant velocity.

What happens if the train accelerates immediately after the ball is released?

Now the passenger and ball no longer have the same horizontal acceleration.

The ball has no horizontal acceleration in the ground frame, assuming negligible air resistance.

However, the passenger accelerates forwards with the train.

The ball therefore appears to move backwards relative to the passenger.

We can describe this using relative acceleration:

\[
\vec a_{B/P}
=
\vec a_{B/G}-\vec a_{P/G}
\]

Let the train accelerate eastwards at \(a\).

The ball’s horizontal acceleration relative to the ground is zero, so:

\[
a_{B/P,x}=0-a
\]

Therefore:

\[
\boxed{a_{B/P,x}=-a}
\]

If the ball initially has zero horizontal velocity relative to the passenger, its relative horizontal displacement after time \(t\) is:

\[
\boxed{
\Delta x_{B/P}=-\frac12at^2
}
\]

The negative sign means the ball moves backwards relative to the accelerating passenger.

If the vehicle is braking instead, the passenger decelerates while the ball retains its original horizontal velocity.

The passenger then sees the ball move forwards relative to them.

There is a useful distinction here:

Constant speed does not always mean constant velocity.

A train travelling around a bend at constant speed is still accelerating because its direction of motion is changing. A passenger on that train will not generally see a released ball fall vertically either.

05Testing the explanation: a two-camera experiment

How could we experimentally distinguish the two descriptions of the ball’s motion?

The important measurement problem is that a stationary camera only records the ball relative to the ground. It cannot directly show what a passenger moving with the vehicle measures.

A simple trolley experiment can address this.

The original apparatus

Imagine a low-friction trolley travelling along a straight, horizontal laboratory track.

A small ball is held above the trolley using an electromagnetic release mechanism. Switching off the electromagnet allows the ball to fall without deliberately giving it an additional horizontal push.

Initially, place a camera beside the track, pointing towards the moving trolley.

A measuring scale is fixed along the track so that the camera can record the ball’s horizontal and vertical positions over time.

What should the first design show?

If the ball retains the trolley’s horizontal velocity, the stationary camera should record two simultaneous motions:

• Horizontal movement at approximately constant velocity.
• Vertical movement with increasing downward velocity.

The resulting ground-frame trajectory should be parabolic.

This observation would support the idea that the ball retains its horizontal motion after release.

However, there is a limitation.

A stationary camera cannot directly establish that the ball follows a vertical path relative to the trolley. It only records ground-frame positions.

A curved trajectory in one reference frame does not disprove a straight trajectory in another.

Revising the experiment

Add a second camera, securely attached to the trolley.

Both cameras should be synchronised so that they record the same release event at the same times.

The stationary camera measures the ball relative to the laboratory floor.

The trolley-mounted camera measures the ball relative to the trolley.

A motion sensor or encoder should also record the trolley’s velocity, allowing us to check that it remains approximately constant during the experiment.

The revised apparatus now measures both reference frames directly.

Illustrative results

Suppose the trolley moves at a constant \(4.0\,\mathrm{m\,s^{-1}}\), and the ball is released from a height of \(1.00\,\mathrm{m}\).

The following table contains illustrative idealised position data, not measurements from an actual experiment.

Notice three things.

First, the ball moves \(0.40\,\mathrm{m}\) horizontally every \(0.10\,\mathrm{s}\) relative to the ground.

Its horizontal velocity is therefore:

\[
v_x=\frac{0.40}{0.10}
=4.0\,\mathrm{m\,s^{-1}}
\]

Second, the ball’s height decreases by increasingly large amounts during successive time intervals.

This is consistent with downward acceleration due to gravity.

Third, its horizontal position relative to the trolley remains zero.

The two cameras therefore record different horizontal positions for the same ball because they measure positions relative to different reference frames.

The experiment’s conclusion is not that one camera is correct and the other is wrong. Both measurements describe the same motion accurately.

The main experimental limitation is that the trolley must move at approximately constant velocity. If it accelerates, the trolley-mounted camera is measuring motion in an accelerating reference frame, and the ball will no longer appear to fall vertically.

07Questions and solutions

The following questions build from basic reference-frame identification to situations where the train’s motion changes during the ball’s flight.

Use \(g=9.8\,\mathrm{m\,s^{-2}}\), take east as positive horizontally, and ignore air resistance unless stated otherwise.

Question 1

A train travels east at a constant velocity of \(8.0\,\mathrm{m\,s^{-1}}\).

A passenger releases a ball from rest relative to the train.

After \(0.25\,\mathrm{s}\):

(a) What is the ball’s horizontal displacement relative to the train?

(b) What is its horizontal displacement relative to the ground?

(c) State the ball’s horizontal velocity relative to each observer.

Solution 1

The ball has zero horizontal displacement relative to the train and travels \(2.0\,\mathrm{m}\) east relative to the ground.

Part (a): Horizontal displacement relative to the train

The ball is released from rest relative to the passenger.

Since the train maintains constant velocity, the ball and passenger have identical horizontal velocities.

Therefore:

\[
\boxed{\Delta x_{B/P}=0\,\mathrm{m}}
\]

Part (b): Horizontal displacement relative to the ground

The ball’s ground-frame horizontal velocity is:

\[
v_{B/G,x}=8.0\,\mathrm{m\,s^{-1}}
\]

Using:

\[
\Delta x=v_xt
\]

\[
\begin{aligned}
\Delta x_{B/G}
&=8.0(0.25)\\
&=2.0\,\mathrm{m}
\end{aligned}
\]

The ball travels \(2.0\,\mathrm{m}\) east relative to the ground.

Part (c): Horizontal velocities

Relative to the ground:

\[
\boxed{v_{B/G,x}=+8.0\,\mathrm{m\,s^{-1}}}
\]

Relative to the passenger:

\[
\boxed{v_{B/P,x}=0\,\mathrm{m\,s^{-1}}}
\]

The passenger sees the ball fall vertically because the ball maintains the same horizontal velocity as the train.

The roadside observer sees horizontal motion because the observer is not travelling with the train.

Question 2

A laboratory trolley travels east at a constant \(4.0\,\mathrm{m\,s^{-1}}\). A ball is released from rest relative to the trolley at a height of \(1.00\,\mathrm{m}\).

Two synchronised cameras record the ball’s motion.

The following idealised data are obtained:

(a) Calculate the ball’s average horizontal velocity relative to the ground between \(0.10\,\mathrm{s}\) and \(0.30\,\mathrm{s}\).

(b) Determine its horizontal acceleration over this interval.

(c) Explain why the ball has a curved trajectory relative to the ground but a vertical trajectory relative to the trolley.

(d) A student concludes that the ball has no horizontal velocity because its horizontal position relative to the trolley remains zero. Explain the error.

Solution 2

The ball has a ground-frame horizontal velocity of \(4.0\,\mathrm{m\,s^{-1}}\) east and zero horizontal acceleration. Its horizontal velocity relative to the trolley is zero.

Part (a): Average horizontal velocity

Use the horizontal position measurements at \(0.10\,\mathrm{s}\) and \(0.30\,\mathrm{s}\).

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

Substitute:

\[
\begin{aligned}
v_{B/G,x}
&=\frac{1.20-0.40}{0.30-0.10}\\
&=\frac{0.80}{0.20}\\
&=4.0\,\mathrm{m\,s^{-1}}
\end{aligned}
\]

The ball travels east at an average horizontal velocity of \(4.0\,\mathrm{m\,s^{-1}}\).

Part (b): Horizontal acceleration

Consider the horizontal velocity during each \(0.10\,\mathrm{s}\) interval.

\[
v_{0.00-0.10}=\frac{0.40}{0.10}=4.0\,\mathrm{m\,s^{-1}}
\]

\[
v_{0.10-0.20}=\frac{0.40}{0.10}=4.0\,\mathrm{m\,s^{-1}}
\]

\[
v_{0.20-0.30}=\frac{0.40}{0.10}=4.0\,\mathrm{m\,s^{-1}}
\]

The horizontal velocity remains constant.

Therefore:

\[
\boxed{a_{B/G,x}=0\,\mathrm{m\,s^{-2}}}
\]

Part (c): Explain the different trajectories

The ground observer sees the ball move horizontally at constant velocity while accelerating vertically downwards.

The horizontal position increases linearly with time, while the vertical position changes quadratically.

This produces a parabolic trajectory.

The trolley observer moves forwards at the same horizontal velocity as the ball.

The ball therefore has zero horizontal displacement relative to the trolley and follows a vertical trajectory.

The different paths result from different reference frames, not different physical events.

Part (d): Evaluate the student’s conclusion

The student’s conclusion is incorrect.

The data show that the ball has zero horizontal velocity relative to the trolley, not zero horizontal velocity relative to the ground.

The trolley and ball both move east at \(4.0\,\mathrm{m\,s^{-1}}\).

Their equal horizontal velocities produce zero relative horizontal motion.

The student has confused zero relative velocity with zero absolute velocity in the ground frame.

Question 3

An airport shuttle travels east at \(15.0\,\mathrm{m\,s^{-1}}\).

At the instant a passenger releases a ball from rest relative to the shuttle, the shuttle begins braking with a constant acceleration of \(2.0\,\mathrm{m\,s^{-2}}\) westwards.

The ball is released \(1.80\,\mathrm{m}\) above the floor.

Assume the braking acceleration remains constant until the ball reaches the floor.

(a) Calculate the time taken for the ball to reach the floor.

(b) Calculate the ball’s horizontal displacement relative to the passenger at impact.

(c) Calculate the ball’s horizontal velocity relative to the passenger immediately before impact.

(d) Explain why the ball appears to move forwards relative to the passenger.

Solution 3

The ball reaches the floor after approximately \(0.606\,\mathrm{s}\). It moves \(0.367\,\mathrm{m}\) forwards relative to the passenger and has a forward relative velocity of approximately \(1.21\,\mathrm{m\,s^{-1}}\) at impact.

Take east as positive.

The shuttle’s acceleration is:

\[
a_{P/G,x}=-2.0\,\mathrm{m\,s^{-2}}
\]

Part (a): Time taken to reach the floor

The vertical motion is unaffected by the shuttle’s horizontal braking.

\[
h=\frac12gt^2
\]

Rearrange:

\[
\begin{aligned}
t&=\sqrt{\frac{2h}{g}}\\
&=\sqrt{\frac{2(1.80)}{9.8}}\\
&=0.606\,\mathrm{s}
\end{aligned}
\]

Part (b): Horizontal displacement relative to the passenger

The ball retains its initial ground-frame horizontal velocity:

\[
v_{B/G,x}=15.0\,\mathrm{m\,s^{-1}}
\]

Its horizontal displacement is:

\[
\begin{aligned}
\Delta x_{B/G}
&=v_{B/G,x}t\\
&=15.0(0.606)\\
&=9.09\,\mathrm{m}
\end{aligned}
\]

The shuttle’s displacement is:

\[
\Delta x_{P/G}=ut+\frac12at^2
\]

Substitute:

\[
\begin{aligned}
\Delta x_{P/G}
&=15.0(0.606)+\frac12(-2.0)(0.606)^2\\
&=9.09-0.367\\
&=8.72\,\mathrm{m}
\end{aligned}
\]

The ball’s displacement relative to the passenger is:

\[
\begin{aligned}
\Delta x_{B/P}
&=\Delta x_{B/G}-\Delta x_{P/G}\\
&=9.09-8.72\\
&=+0.367\,\mathrm{m}
\end{aligned}
\]

The ball moves \(0.367\,\mathrm{m}\) forwards relative to the passenger.

Part (c): Relative horizontal velocity

The shuttle’s velocity immediately before impact is:

\[
\begin{aligned}
v_{P/G,x}&=u+at\\
&=15.0-2.0(0.606)\\
&=13.79\,\mathrm{m\,s^{-1}}
\end{aligned}
\]

The ball’s horizontal ground velocity remains \(15.0\,\mathrm{m\,s^{-1}}\).

Therefore:

\[
\begin{aligned}
v_{B/P,x}
&=v_{B/G,x}-v_{P/G,x}\\
&=15.0-13.79\\
&=+1.21\,\mathrm{m\,s^{-1}}
\end{aligned}
\]

The ball moves east relative to the passenger at approximately \(1.21\,\mathrm{m\,s^{-1}}\).

Part (d): Explain the direction

The shuttle is slowing down, but the ball maintains its original horizontal velocity.

The passenger therefore moves backwards relative to the ball’s original forward motion.

From the passenger’s perspective, the ball moves forwards.

This is the opposite of what happens when the vehicle accelerates forwards.

Question 4

A train travels east at a constant \(18.0\,\mathrm{m\,s^{-1}}\).

A passenger throws a ball horizontally westwards at \(18.0\,\mathrm{m\,s^{-1}}\) relative to the train.

The ball is thrown from a height of \(1.25\,\mathrm{m}\) above the track.

Ignoring air resistance, determine:

(a) The ball’s initial horizontal velocity relative to the ground.

(b) The time taken for the ball to reach the ground.

(c) The ball’s horizontal displacement relative to the ground and relative to the passenger before impact.

(d) Describe the trajectory observed by each observer.

Solution 4

The roadside observer sees the ball fall vertically, while the passenger sees it travel backwards and downwards along a curved trajectory.

The ball’s initial horizontal velocity relative to the ground is zero.

Part (a): Ground-frame horizontal velocity

Use the relative velocity relationship:

\[
\vec v_{B/P}=\vec v_{B/G}-\vec v_{P/G}
\]

Rearrange:

\[
\vec v_{B/G}=\vec v_{B/P}+\vec v_{P/G}
\]

Taking east as positive:

\[
\begin{aligned}
v_{B/G,x}
&=-18.0+18.0\\
&=0\,\mathrm{m\,s^{-1}}
\end{aligned}
\]

The ball has zero horizontal velocity relative to the ground.

Part (b): Time taken to reach the ground

The initial vertical velocity is zero.

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

Part (c): Horizontal displacement in both frames

In the ground frame:

\[
\Delta x_{B/G}=v_{B/G,x}t
\]

\[
\boxed{\Delta x_{B/G}=0\,\mathrm{m}}
\]

In the passenger frame:

\[
\begin{aligned}
\Delta x_{B/P}
&=v_{B/P,x}t\\
&=-18.0(0.505)\\
&=-9.09\,\mathrm{m}
\end{aligned}
\]

The ball travels approximately \(9.09\,\mathrm{m}\) west relative to the passenger.

Part (d): Compare the trajectories

The roadside observer sees a vertical trajectory because the ball has no horizontal velocity relative to the ground.

The passenger sees a backward-moving ball with a downward vertical acceleration.

The passenger therefore observes a parabolic trajectory.

This is an important reversal of the original situation.

A roadside observer does not always see a curved trajectory simply because an object is released from a moving vehicle. The ball’s ground-frame horizontal velocity determines whether it moves forwards, backwards, or not at all.

Question 5

An electric train travels east at \(16.0\,\mathrm{m\,s^{-1}}\).

At \(t=0\), a passenger throws a ball vertically upwards at \(9.8\,\mathrm{m\,s^{-1}}\) relative to the train.

At the same instant, the train begins accelerating:

• During the first \(1.0\,\mathrm{s}\), it accelerates eastwards at \(2.0\,\mathrm{m\,s^{-2}}\).
• During the next \(1.0\,\mathrm{s}\), it accelerates westwards at \(2.0\,\mathrm{m\,s^{-2}}\).

Assume the train maintains these constant accelerations during the two intervals. Ignore air resistance and assume the ball experiences no horizontal force after release.

The ball returns to its original vertical height after \(2.0\,\mathrm{s}\).

(a) Calculate the ball’s horizontal displacement relative to the ground after \(2.0\,\mathrm{s}\).

(b) Calculate the train’s total horizontal displacement over the same period.

(c) Determine the ball’s horizontal displacement relative to the passenger when it returns to its original height.

(d) Calculate the ball’s horizontal velocity relative to the passenger at that instant.

(e) Does the ball return to the passenger’s hand? Explain why returning to the train’s original velocity does not necessarily mean the ball returns to its original position.

Solution 5

No, the ball does not return to the passenger’s hand. It returns to its original height \(2.0\,\mathrm{m}\) behind the passenger, even though the ball and train have the same horizontal velocity at that instant.

The important distinction is between relative velocity and relative displacement.

Part (a): Ball’s horizontal displacement relative to the ground

The ball initially shares the train’s horizontal velocity.

Therefore:

\[
v_{B/G,x}=+16.0\,\mathrm{m\,s^{-1}}
\]

Since there is no horizontal force on the ball, its horizontal velocity remains constant.

Over \(2.0\,\mathrm{s}\):

\[
\begin{aligned}
\Delta x_{B/G}
&=v_{B/G,x}t\\
&=16.0(2.0)\\
&=32.0\,\mathrm{m}
\end{aligned}
\]

The ball travels \(32.0\,\mathrm{m}\) east relative to the ground.

Part (b): Train’s total horizontal displacement

We must calculate the displacement during each acceleration interval separately.

During the first second:

\[
u=16.0\,\mathrm{m\,s^{-1}}
\]

\[
a=+2.0\,\mathrm{m\,s^{-2}}
\]

Using:

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

\[
\begin{aligned}
s_1
&=16.0(1.0)+\frac12(2.0)(1.0)^2\\
&=17.0\,\mathrm{m}
\end{aligned}
\]

At the end of the first second, the train’s velocity is:

\[
v=16.0+2.0(1.0)
=18.0\,\mathrm{m\,s^{-1}}
\]

During the second second, the train accelerates westwards.

Therefore:

\[
u=18.0\,\mathrm{m\,s^{-1}}
\]

\[
a=-2.0\,\mathrm{m\,s^{-2}}
\]

Its displacement during this interval is:

\[
\begin{aligned}
s_2
&=18.0(1.0)+\frac12(-2.0)(1.0)^2\\
&=17.0\,\mathrm{m}
\end{aligned}
\]

The total train displacement is:

\[
\begin{aligned}
\Delta x_{P/G}
&=s_1+s_2\\
&=17.0+17.0\\
&=34.0\,\mathrm{m}
\end{aligned}
\]

The train travels \(34.0\,\mathrm{m}\) east.

Notice that its final velocity is:

\[
v=18.0-2.0(1.0)
\]

\[
\boxed{v_{P/G}=16.0\,\mathrm{m\,s^{-1}}}
\]

It has returned to its original velocity.

Part (c): Ball’s horizontal displacement relative to the passenger

Subtract the train’s displacement from the ball’s displacement:

\[
\begin{aligned}
\Delta x_{B/P}
&=\Delta x_{B/G}-\Delta x_{P/G}\\
&=32.0-34.0\\
&=-2.0\,\mathrm{m}
\end{aligned}
\]

The ball is \(2.0\,\mathrm{m}\) west of the passenger.

It has not returned to the passenger’s hand.

Part (d): Relative horizontal velocity at \(t=2.0\,\mathrm{s}\)

At this instant:

\[
v_{B/G,x}=16.0\,\mathrm{m\,s^{-1}}
\]

The train has also returned to:

\[
v_{P/G,x}=16.0\,\mathrm{m\,s^{-1}}
\]

Therefore:

\[
\begin{aligned}
v_{B/P,x}
&=v_{B/G,x}-v_{P/G,x}\\
&=16.0-16.0\\
&=0\,\mathrm{m\,s^{-1}}
\end{aligned}
\]

The ball and passenger have zero relative horizontal velocity at the instant the ball returns to its original height.

Part (e): Why doesn’t the ball return to the passenger’s hand?

This is the most important part of the problem.

A tempting argument is:

1. The train starts at \(16.0\,\mathrm{m\,s^{-1}}\).
2. The train finishes at \(16.0\,\mathrm{m\,s^{-1}}\).
3. The ball also travels horizontally at \(16.0\,\mathrm{m\,s^{-1}}\).
4. Therefore, the ball should return to the passenger’s hand.

The problem is in the final conclusion.

Having the same velocity at the end does not mean two objects have the same position.

We can see this by tracking their relative velocities.

During the first second, the train accelerates ahead of the ball.

The ball’s relative velocity changes from \(0\) to \(-2.0\,\mathrm{m\,s^{-1}}\).

During the second second, the train slows down again. The ball’s relative velocity changes from \(-2.0\) back to zero.

However, it remains negative throughout that second.

We can calculate the relative displacement using the area under the relative velocity-time graph.

During the first second, the average relative velocity is:

\[
\bar v_1=\frac{0+(-2.0)}{2}
=-1.0\,\mathrm{m\,s^{-1}}
\]

Therefore:

\[
\Delta x_1=-1.0(1.0)=-1.0\,\mathrm{m}
\]

During the second second:

\[
\bar v_2=\frac{-2.0+0}{2}
=-1.0\,\mathrm{m\,s^{-1}}
\]

Therefore:

\[
\Delta x_2=-1.0(1.0)=-1.0\,\mathrm{m}
\]

The total relative displacement is:

\[
\begin{aligned}
\Delta x_{B/P}
&=\Delta x_1+\Delta x_2\\
&=-1.0-1.0\\
&=-2.0\,\mathrm{m}
\end{aligned}
\]

The ball finishes \(2.0\,\mathrm{m}\) behind the passenger.

The train has returned to its original velocity, but it has travelled further than the ball.

This is the distinction to remember:

• Relative velocity describes how quickly the separation between two objects is changing.
• Relative displacement describes how much their separation has changed.

A zero relative velocity at one instant does not mean the relative displacement is zero.

The next time you encounter a relative-motion problem, identify the object, identify the observer, and write both velocities in a common reference frame before doing any calculations.

That same approach extends naturally to two-dimensional problems involving boats in currents, aircraft in wind, and projectiles observed from moving platforms. The key is always to establish whose motion you are describing before deciding what the trajectory should look like.