How to Solve Horizontal Launches from Moving Vehicles

A passenger releases a ball from the tray of a ute travelling east at \(15.0\text{ m s}^{-1}\). Does the ball fall straight down, or does it move forwards while falling?

Think about what the ball is doing just before it leaves the ute. It is already travelling forwards at \(15.0\text{ m s}^{-1}\). Releasing it does not suddenly remove that horizontal velocity.

From the ground, the ball therefore travels forwards while gravity pulls it downwards.

This is the key to solving a horizontal launch from a moving vehicle: add the vehicle’s velocity to the object’s launch velocity relative to the vehicle, then solve the horizontal and vertical motion separately.

You need three things:

1. The object’s initial horizontal velocity relative to the ground.
2. The time it takes to fall, calculated using vertical motion.
3. The horizontal displacement, calculated using horizontal motion.

Let’s work through the method, including what changes when the object is thrown backwards or the vehicle accelerates.

01First, find the object’s horizontal velocity

The most common mistake in these questions is confusing velocity relative to the vehicle with velocity relative to the ground.

These are not necessarily the same.

Imagine someone standing in the back of a moving ute and holding a ball. Both the person and the ball are already moving forwards with the ute.

When the person releases the ball without pushing it, the ball keeps that forwards velocity.

This is an example of the principle of inertia: an object maintains its velocity unless a resultant force changes its motion.

The two reference frames

A reference frame is simply the viewpoint from which we describe motion.

For this problem, we usually need two:

• Ground frame: An observer standing beside the road measures the ball’s motion relative to the ground.
• Vehicle frame: An observer travelling inside the vehicle measures the ball’s motion relative to the vehicle.

For a vehicle travelling at constant velocity, these observers describe the same motion differently.

If a ball is simply released, the ground observer sees it travel forwards while falling. The observer inside the vehicle sees it fall vertically.

Both descriptions are correct.

Adding the velocities

Take the vehicle’s forward direction as positive.

Define:

• \(V\) = vehicle’s horizontal velocity relative to the ground, in \(\text{m s}^{-1}\).
• \(u_{\mathrm{rel}}\) = object’s initial horizontal velocity relative to the vehicle, in \(\text{m s}^{-1}\).
• \(v_{0x}\) = object’s initial horizontal velocity relative to the ground, in \(\text{m s}^{-1}\).

The relationship is:

\[ \boxed{v_{0x}=V+u_{\mathrm{rel}}} \]

The sign of \(u_{\mathrm{rel}}\) depends on the launch direction.

Notice something important about the final row.

Throwing an object backwards relative to the vehicle does not necessarily mean it travels backwards relative to the ground.

For example, suppose a truck travels forwards at \(20\text{ m s}^{-1}\), and a ball is thrown backwards at \(6\text{ m s}^{-1}\) relative to the truck.

Its ground-frame velocity is:

\[ v_{0x}=20-6 \]

\[ \boxed{v_{0x}=14\text{ m s}^{-1}} \]

The ball still travels forwards relative to the ground.

It simply travels more slowly than the truck.

If the ball were thrown backwards at \(25\text{ m s}^{-1}\), its ground-frame velocity would instead be \(5\text{ m s}^{-1}\) backwards.

The direction of motion depends on the combined velocity, not just the direction of the throw.

02Separate horizontal and vertical motion

Once you know the initial horizontal velocity relative to the ground, the projectile calculation becomes much easier.

We make the usual HSC projectile-motion assumptions:

• Air resistance is negligible.
• The vehicle travels in a straight horizontal direction.
• The object is launched horizontally.
• Gravitational acceleration is constant.
• The landing surface is horizontal.

Choose:

• Positive \(x\): forwards along the road.
• Positive \(y\): upwards.
• The launch point: origin \((0,0)\).

Let \(h\) represent the vertical height between the launch point and the landing surface.

The object’s initial vertical velocity is zero because the launch is horizontal.

Gravity acts downwards, so:

\[ a_y=-g \]

where \(g=9.8\text{ m s}^{-2}\).

Horizontal motion

There is no horizontal acceleration in our model.

Therefore:

\[ a_x=0 \]

The horizontal velocity remains constant:

\[ v_x=v_{0x} \]

Using the constant-velocity equation:

\[ \boxed{\Delta x=v_{0x}t} \]

Remember that \(v_{0x}\) includes the vehicle’s velocity.

Vertical motion

The object starts with zero vertical velocity and accelerates downwards under gravity.

Using the SUVAT equation:

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

Since \(u_y=0\) and \(a_y=-g\):

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

The landing surface is \(h\) metres below the launch point, so:

\[ \Delta y=-h \]

Substitute:

\[ -h=-\frac12gt^2 \]

Rearrange:

\[ h=\frac12gt^2 \]

Therefore:

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

This gives the time taken for the object to reach the landing surface.

Combine the two equations

We now have everything needed.

The horizontal velocity is:

\[ v_{0x}=V+u_{\mathrm{rel}} \]

The time of flight is:

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

Substitute both into the horizontal displacement equation:

\[ \Delta x=v_{0x}t \]

Giving:

\[ \boxed{ \Delta x=(V+u_{\mathrm{rel}}) \sqrt{\frac{2h}{g}} } \]

This is the useful combined equation for a horizontal launch from a moving vehicle, under the assumptions above.

It calculates horizontal displacement relative to the ground.

03Don’t confuse a horizontal launch with a level-ground launch

This is an important distinction in HSC projectile questions.

You might remember the standard horizontal range equation:

\[ R=\frac{u^2\sin(2\theta)}{g} \]

However, this equation applies when a projectile lands at the same height from which it was launched, assuming negligible air resistance.

A horizontal launch from a vehicle’s tray onto the road is a different situation.

The object starts above the landing surface.

Its initial vertical velocity is zero, but gravity still gives it time to fall.

For a horizontal launch:

\[ \theta=0^\circ \]

If you substitute this into the standard level-ground range equation, you get:

\[ R=\frac{u^2\sin(0^\circ)}{g}=0 \]

That result does not describe a ball falling from a vehicle’s elevated tray.

It describes the limiting same-height situation: a projectile launched horizontally from a point that is already at the landing height has no non-zero flight time before reaching that level.

For a vehicle launch from height \(h\), use:

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

Then:

\[ \Delta x=v_{0x}t \]

For more detail about the conditions behind projectile range equations, see Horizontal Range of a Projectile: HSC Physics Guide.

If the object is launched at an angle from a vehicle rather than horizontally, the vertical equation needs an initial vertical velocity component. See Projectile Motion from a Launch Height: HSC Physics Guide.

04Where does the object land relative to the vehicle?

So far, we’ve calculated the object’s horizontal displacement relative to the ground.

Sometimes an exam question asks something different:

How far ahead of or behind the vehicle does the object land?

You need to pay attention to the reference frame.

Case 1: The vehicle maintains constant velocity

Suppose an object is launched with horizontal velocity \(u_{\mathrm{rel}}\) relative to a vehicle travelling at constant velocity \(V\).

The object’s ground displacement is:

\[ \Delta x_{\mathrm{object}}=(V+u_{\mathrm{rel}})t \]

The vehicle’s displacement is:

\[ \Delta x_{\mathrm{vehicle}}=Vt \]

Subtract the vehicle’s displacement from the object’s displacement:

\[ \Delta x_{\mathrm{relative}} =(V+u_{\mathrm{rel}})t-Vt \]

Therefore:

\[ \boxed{\Delta x_{\mathrm{relative}}=u_{\mathrm{rel}}t} \]

This gives the object’s landing position relative to the vehicle’s original launch point.

There are three useful possibilities:

• If the object is simply released, \(u_{\mathrm{rel}}=0\), so it lands alongside the same horizontal position on the vehicle.
• If launched forwards, it lands ahead of the launch point on the vehicle.
• If launched backwards, it lands behind the launch point on the vehicle.

Case 2: The vehicle accelerates after release

Now suppose the vehicle accelerates forwards while the object is airborne.

Does the object automatically accelerate forwards too?

No.

Once released, the object is no longer being pushed by the vehicle. Under our assumptions, its horizontal acceleration is zero.

The vehicle continues accelerating, but the object maintains its original ground-frame horizontal velocity.

If the vehicle has constant horizontal acceleration \(a_v\), its displacement is:

\[ \Delta x_{\mathrm{vehicle}} =Vt+\frac12a_vt^2 \]

The object’s displacement remains:

\[ \Delta x_{\mathrm{object}} =(V+u_{\mathrm{rel}})t \]

Subtracting gives:

\[ \boxed{ \Delta x_{\mathrm{relative}} =u_{\mathrm{rel}}t-\frac12a_vt^2 } \]

Here, positive \(a_v\) means the vehicle accelerates forwards.

This produces an interesting result:

• If the vehicle accelerates forwards, a released object falls behind it.
• If the vehicle brakes, a released object moves ahead of it.
• If the vehicle maintains constant velocity, a simply released object stays horizontally aligned with it.

This matters whenever a question mentions that the vehicle changes speed after the object is released.

05A reliable method for HSC questions

Before attempting a calculation, identify what the question actually wants.

A useful reminder: the horizontal and vertical equations use the same time \(t\).

Don’t calculate two different flight times. The object follows one trajectory, even though we analyse its horizontal and vertical components separately.

06Worked examples

07Common mistakes to avoid

08Questions and solutions

The following questions build from a straightforward release to situations where you must combine reference frames, projectile motion, and vehicle acceleration.

Question 1

A small parcel is released without being pushed from an open tray on a ute travelling east at \(15.0\text{ m s}^{-1}\).

The tray is \(1.25\text{ m}\) above a level road.

Ignoring air resistance, calculate:

a) The time taken for the parcel to reach the road.

b) The horizontal displacement of the parcel relative to the point directly below its release position.

Use \(g=9.8\text{ m s}^{-2}\).

Solution 1

The parcel is airborne for \(0.505\text{ s}\) and lands \(7.58\text{ m}\) east of the point directly below its release position.

Step 1: Find the initial horizontal velocity.

The parcel is released without a push, so its horizontal velocity relative to the ground initially equals the ute’s velocity.

\[ v_{0x}=15.0\text{ m s}^{-1} \]

Step 2: Calculate the time of flight.

The initial vertical velocity is zero.

Using:

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

Substitute the vertical displacement:

\[ -1.25=-\frac12(9.8)t^2 \]

Rearrange:

\[ t^2=\frac{2(1.25)}{9.8} \]

\[ t=\sqrt{\frac{2.50}{9.8}} \]

\[ \boxed{t=0.505\text{ s}} \]

Step 3: Calculate the horizontal displacement.

Using:

\[ \Delta x=v_{0x}t \]

\[ \Delta x=15.0(0.505) \]

\[ \boxed{\Delta x=7.58\text{ m east}} \]

The parcel moves forwards because it retains the ute’s initial horizontal velocity. Gravity only changes its vertical motion in this model.

Question 2

An electric maintenance cart travels east at \(18.0\text{ m s}^{-1}\).

A test ball is launched horizontally east at \(4.00\text{ m s}^{-1}\) relative to the cart. The launcher is \(2.00\text{ m}\) above a level track.

Ignoring air resistance, calculate:

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

b) The time taken to reach the track.

c) The horizontal displacement relative to the ground.

d) How far ahead of the cart’s original launch position the ball lands, assuming the cart maintains constant velocity.

Solution 2

The ball initially travels east at \(22.0\text{ m s}^{-1}\), remains airborne for \(0.639\text{ s}\), and lands \(14.1\text{ m}\) east of its original ground position. It lands \(2.56\text{ m}\) ahead of the cart’s original launch position relative to the cart.

Step 1: Calculate the ball’s initial horizontal velocity relative to the ground.

Both velocities point east, so they are added:

\[ v_{0x}=V+u_{\mathrm{rel}} \]

\[ v_{0x}=18.0+4.00 \]

\[ \boxed{v_{0x}=22.0\text{ m s}^{-1}} \]

Step 2: Calculate the time of flight.

The vertical displacement is:

\[ \Delta y=-2.00\text{ m} \]

Using:

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

\[ -2.00=-4.90t^2 \]

\[ t=\sqrt{\frac{2.00}{4.90}} \]

\[ \boxed{t=0.639\text{ s}} \]

Step 3: Calculate the ground-frame horizontal displacement.

\[ \Delta x_{\mathrm{ball}}=v_{0x}t \]

\[ \Delta x_{\mathrm{ball}}=22.0(0.639) \]

\[ \boxed{\Delta x_{\mathrm{ball}}=14.1\text{ m east}} \]

Step 4: Calculate the cart’s displacement.

The cart maintains its constant velocity:

\[ \Delta x_{\mathrm{cart}}=18.0(0.639) \]

\[ \Delta x_{\mathrm{cart}}=11.5\text{ m east} \]

The ball’s position relative to the cart is:

\[ \Delta x_{\mathrm{relative}} =14.1-11.5 \]

Using unrounded values:

\[ \Delta x_{\mathrm{relative}} =(22.0-18.0)(0.639) \]

\[ \boxed{\Delta x_{\mathrm{relative}}=2.56\text{ m east}} \]

The ball lands ahead of the cart’s original launch position because it was given additional forward velocity relative to the cart.

Question 3

A test ute travels east at \(24.0\text{ m s}^{-1}\).

A horizontal launcher mounted on the ute fires a ball west at \(30.0\text{ m s}^{-1}\) relative to the ute. The launcher is \(3.20\text{ m}\) above a level road.

Ignoring air resistance, determine:

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

b) Its time of flight.

c) Its horizontal displacement relative to the ground.

d) Its horizontal displacement relative to the ute if the ute maintains constant velocity.

Explain whether the ball initially travels forwards or backwards relative to the ground.

Solution 3

The ball initially travels west at \(6.00\text{ m s}^{-1}\) relative to the ground. It remains airborne for \(0.808\text{ s}\), travels \(4.85\text{ m}\) west of its release position, and lands \(24.2\text{ m}\) behind its original position on the ute.

Step 1: Calculate the initial horizontal velocity relative to the ground.

Take east as positive.

The ute’s velocity is:

\[ V=+24.0\text{ m s}^{-1} \]

The launch velocity relative to the ute is:

\[ u_{\mathrm{rel}}=-30.0\text{ m s}^{-1} \]

Using velocity addition:

\[ v_{0x}=V+u_{\mathrm{rel}} \]

\[ v_{0x}=24.0-30.0 \]

\[ \boxed{v_{0x}=-6.00\text{ m s}^{-1}} \]

The negative sign indicates that the ball initially travels west relative to the ground.

This is different from the earlier truck example because the backwards launch speed is now greater than the vehicle’s forward speed.

Step 2: Calculate the time of flight.

The ball is launched horizontally, so its initial vertical velocity is zero.

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

\[ -3.20=-4.90t^2 \]

\[ t=\sqrt{\frac{3.20}{4.90}} \]

\[ \boxed{t=0.808\text{ s}} \]

Step 3: Calculate the horizontal displacement relative to the ground.

\[ \Delta x_{\mathrm{ball}}=v_{0x}t \]

\[ \Delta x_{\mathrm{ball}}=(-6.00)(0.808) \]

\[ \boxed{\Delta x_{\mathrm{ball}}=-4.85\text{ m}} \]

The ball lands \(4.85\text{ m}\) west of its original ground position.

Step 4: Calculate the ute’s displacement.

The ute continues travelling east:

\[ \Delta x_{\mathrm{ute}}=24.0(0.808) \]

\[ \Delta x_{\mathrm{ute}}=19.4\text{ m east} \]

Therefore, the ball’s position relative to the ute is:

\[ \Delta x_{\mathrm{relative}} =-4.85-19.4 \]

\[ \boxed{\Delta x_{\mathrm{relative}}=-24.2\text{ m}} \]

The ball lands \(24.2\text{ m}\) behind its original position relative to the ute.

The important point is that the ball moves backwards relative to both the ground and the ute in this particular case. That happens because its backwards launch speed exceeds the ute’s initial forwards speed.

Question 4

A drone flies horizontally east at \(16.0\text{ m s}^{-1}\), \(4.90\text{ m}\) above an open field.

It releases a lightweight marker without giving it any additional horizontal push.

Immediately after release, the drone accelerates east at \(1.20\text{ m s}^{-2}\) and maintains this acceleration until the marker reaches the field.

Ignoring air resistance, calculate:

a) The time taken for the marker to reach the field.

b) The marker’s horizontal displacement relative to the ground.

c) The drone’s horizontal displacement during this time.

d) The marker’s horizontal position relative to the drone at impact.

Solution 4

The marker travels \(16.0\text{ m}\) east relative to the ground and lands \(0.600\text{ m}\) behind the drone’s horizontal position at impact.

Step 1: Calculate the time of flight.

The marker’s initial vertical velocity is zero.

Using:

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

\[ -4.90=-4.90t^2 \]

Therefore:

\[ t^2=1 \]

\[ \boxed{t=1.00\text{ s}} \]

Step 2: Calculate the marker’s horizontal displacement.

At release, the marker has the same horizontal velocity as the drone:

\[ v_{0x}=16.0\text{ m s}^{-1} \]

The marker experiences no horizontal acceleration.

Therefore:

\[ \Delta x_{\mathrm{marker}}=v_{0x}t \]

\[ \Delta x_{\mathrm{marker}}=16.0(1.00) \]

\[ \boxed{\Delta x_{\mathrm{marker}}=16.0\text{ m east}} \]

Step 3: Calculate the drone’s displacement.

The drone accelerates east at \(1.20\text{ m s}^{-2}\).

Use:

\[ \Delta x_{\mathrm{drone}} =Vt+\frac12a_vt^2 \]

Substitute:

\[ \Delta x_{\mathrm{drone}} =16.0(1.00)+\frac12(1.20)(1.00)^2 \]

\[ \Delta x_{\mathrm{drone}}=16.0+0.600 \]

\[ \boxed{\Delta x_{\mathrm{drone}}=16.6\text{ m east}} \]

Step 4: Find the marker’s position relative to the drone.

Subtract the drone’s displacement from the marker’s displacement:

\[ \Delta x_{\mathrm{relative}} =16.0-16.6 \]

\[ \boxed{\Delta x_{\mathrm{relative}}=-0.600\text{ m}} \]

The marker lands \(0.600\text{ m}\) behind the drone’s position at impact.

Why?

The marker keeps travelling at its original horizontal velocity of \(16.0\text{ m s}^{-1}\). The drone accelerates and reaches a higher speed.

The drone therefore moves further forwards than the marker.

The trap is assuming that the marker continues accelerating just because the drone does. It doesn’t.

Question 5

An open rail test carriage travels east at \(20.0\text{ m s}^{-1}\).

A small horizontal launch platform is mounted inside the carriage, \(1.25\text{ m}\) above its flat deck.

At the instant two identical balls leave the platform, the carriage begins braking with constant acceleration.

• Ball A is simply released without a horizontal push.
• Ball B is launched horizontally west at an unknown speed relative to the carriage.

Ignore air resistance. Assume both balls have zero horizontal acceleration after release, and the carriage maintains constant braking acceleration throughout their flight.

Both balls land on the carriage deck.

The following positions are measured relative to the carriage’s original launch mark at the instant each ball lands:

• Ball A lands \(0.255\text{ m}\) east of the mark.
• Ball B lands \(2.78\text{ m}\) west of the mark.

The launch mark is \(3.00\text{ m}\) from the rear edge of the deck and \(1.00\text{ m}\) from the front edge.

Determine:

a) The magnitude of the carriage’s braking acceleration.

b) The initial horizontal launch speed of ball B relative to the carriage.

c) Whether ball B initially travels east or west relative to the ground.

d) Whether both balls can land on the carriage deck, given its dimensions.

Explain why a ball thrown backwards relative to the carriage can still travel forwards relative to the ground.

Solution 5

The carriage brakes at \(2.00\text{ m s}^{-2}\). Ball B is launched west at approximately \(6.01\text{ m s}^{-1}\) relative to the carriage, but initially travels east at approximately \(14.0\text{ m s}^{-1}\) relative to the ground. Both balls land within the deck.

The key is to use ball A to determine the carriage’s braking acceleration before analysing ball B.

Step 1: Calculate the common time of flight.

Both balls leave the same height with zero initial vertical velocity.

Therefore, they have the same time of flight.

Using:

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

\[ -1.25=-4.90t^2 \]

\[ t=\sqrt{\frac{1.25}{4.90}} \]

\[ \boxed{t=0.505\text{ s}} \]

This common time is important because the carriage’s displacement and both balls’ displacements must be compared at the same instant.

Step 2: Determine the carriage’s braking acceleration using ball A.

Take east as positive.

Let \(b\) represent the magnitude of the carriage’s braking acceleration.

The carriage’s acceleration is therefore:

\[ a_{\mathrm{carriage}}=-b \]

At the instant of release, ball A has the same horizontal velocity as the carriage:

\[ v_{0x,A}=20.0\text{ m s}^{-1} \]

After release, ball A has zero horizontal acceleration.

Its ground-frame displacement is:

\[ \Delta x_A=20.0t \]

The carriage, however, is braking:

\[ \Delta x_{\mathrm{carriage}} =20.0t-\frac12bt^2 \]

The relative displacement of ball A is the difference:

\[ \Delta x_{A/\mathrm{carriage}} =\Delta x_A-\Delta x_{\mathrm{carriage}} \]

Substituting:

\[ \Delta x_{A/\mathrm{carriage}} =20.0t-\left(20.0t-\frac12bt^2\right) \]

Simplifying:

\[ \Delta x_{A/\mathrm{carriage}} =\frac12bt^2 \]

We know ball A lands \(0.255\text{ m}\) east of the mark.

Therefore:

\[ 0.255=\frac12b(0.505)^2 \]

Rearrange:

\[ b=\frac{2(0.255)}{(0.505)^2} \]

\[ \boxed{b=2.00\text{ m s}^{-2}} \]

The carriage’s acceleration is therefore:

\[ \boxed{a_{\mathrm{carriage}}=-2.00\text{ m s}^{-2}} \]

The negative sign indicates braking while travelling east.

Why did ball A land ahead of its original position on the carriage?

The ball maintained its original horizontal velocity, but the carriage slowed down. The carriage therefore travelled a shorter horizontal distance than the ball.

Step 3: Determine ball B’s launch speed relative to the carriage.

Let \(w\) represent the magnitude of ball B’s initial westward velocity relative to the carriage.

Since east is positive:

\[ u_{\mathrm{rel}}=-w \]

Its initial horizontal velocity relative to the ground is:

\[ v_{0x,B}=20.0-w \]

After time \(t\), its ground-frame displacement is:

\[ \Delta x_B=(20.0-w)t \]

The carriage displacement is:

\[ \Delta x_{\mathrm{carriage}} =20.0t-\frac12bt^2 \]

Subtracting gives the displacement of ball B relative to the carriage:

\[ \Delta x_{B/\mathrm{carriage}} =(20.0-w)t- \left(20.0t-\frac12bt^2\right) \]

Simplifying:

\[ \Delta x_{B/\mathrm{carriage}} =-wt+\frac12bt^2 \]

We know ball B lands \(2.78\text{ m}\) west of the mark:

\[ -2.78=-w(0.505)+\frac12(2.00)(0.505)^2 \]

Rearrange:

\[ -2.78=-0.505w+0.255 \]

\[ 0.505w=2.78+0.255 \]

\[ w=\frac{3.035}{0.505} \]

\[ \boxed{w=6.01\text{ m s}^{-1}} \]

Ball B was therefore launched west at approximately \(6.01\text{ m s}^{-1}\) relative to the carriage.

Step 4: Determine ball B’s initial direction relative to the ground.

Now use velocity addition.

The carriage initially travels east at \(20.0\text{ m s}^{-1}\), while ball B is launched west at \(6.01\text{ m s}^{-1}\) relative to it.

Therefore:

\[ v_{0x,B}=20.0-6.01 \]

\[ \boxed{v_{0x,B}=13.99\text{ m s}^{-1}\text{ east}} \]

Ball B initially travels forwards relative to the ground.

This is the surprising part.

Although ball B is launched backwards relative to the carriage, the carriage is initially moving faster forwards than the ball is launched backwards.

The two velocities combine to give a net eastward velocity.

Step 5: Check whether both balls land on the deck.

The launch mark is:

• \(3.00\text{ m}\) from the rear edge.
• \(1.00\text{ m}\) from the front edge.

Ball A lands \(0.255\text{ m}\) east of the mark.

Its remaining distance to the front edge is:

\[ 1.00-0.255=0.745\text{ m} \]

So ball A lands within the deck.

Ball B lands \(2.78\text{ m}\) west of the mark.

Its remaining distance to the rear edge is:

\[ 3.00-2.78=0.220\text{ m} \]

So ball B also lands within the deck.

Both landing positions are physically possible.

There is one final detail worth noticing.

At impact, the carriage’s velocity is:

\[ v_{\mathrm{carriage}} =20.0-2.00(0.505) \]

\[ v_{\mathrm{carriage}}=18.99\text{ m s}^{-1}\text{ east} \]

Ball B still travels east at approximately \(13.99\text{ m s}^{-1}\) relative to the ground.

It is moving more slowly than the carriage, which explains why it moves backwards relative to the carriage and lands near the rear edge.

This question combines three separate ideas:

• Velocity addition at the instant of launch.
• Independent horizontal and vertical projectile motion.
• Relative displacement when the vehicle accelerates.

The important constraint is that the balls stop sharing the carriage’s acceleration as soon as they leave the platform. Their horizontal velocities remain constant, while the carriage continues braking.

09What to take into your next projectile question

When solving a horizontal launch from a moving vehicle, don’t start by calculating the range.

Start by identifying the reference frame.

The most reliable sequence is:

\[ \boxed{v_{0x}=V+u_{\mathrm{rel}}} \]

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

\[ \boxed{\Delta x=v_{0x}t} \]

Then, if the question asks where the object lands relative to the vehicle, calculate the vehicle’s displacement separately and subtract.

Once you’re comfortable with this method, the next useful step is handling projectiles launched at an angle from moving vehicles. You keep the same velocity-addition principle, but now the launch also has an initial vertical velocity component.