How to Measure Projectile Launch Speed Using a Horizontal Launch

A ball rolls off the edge of a laboratory bench and lands 1.3 m away. How fast was it moving when it left the bench?

You can’t work that out from the landing distance alone. A faster ball travels further, but so does a ball launched from a greater height. You need to know how long it was in the air.

For a horizontal launch, there’s a way to find that time without using a stopwatch. Measure the ball’s vertical drop \(h\) and horizontal range \(R\), then calculate its launch speed using:

\[
u=R\sqrt{\frac{g}{2h}}
\]

Here, \(u\) is launch speed in metres per second (\(\mathrm{m\,s^{-1}}\)), \(R\) is horizontal distance in metres (m), \(h\) is vertical drop in metres (m), and \(g\) is gravitational acceleration, taken as \(9.8\ \mathrm{m\,s^{-2}}\).

The important condition is that the ball must leave horizontally. If it starts with an upward or downward velocity component, the calculation needs to change.

Before looking at the equipment, make a prediction. Suppose you raise the launcher but keep its launch speed exactly the same. Will the ball land further away?

Yes. It has further to fall, so gravity takes longer to bring it to the floor. The horizontal speed hasn’t changed, but the ball travels horizontally for longer.

That is both the idea behind the measurement and a trap when interpreting experimental results.

01Why a horizontal launch lets you measure speed

Imagine releasing two identical balls from the same height at exactly the same moment.

You simply drop the first ball. You launch the second horizontally.

Which reaches the floor first?

Under the ideal projectile-motion model, they land simultaneously. The second ball’s horizontal motion doesn’t delay its fall because horizontal and vertical motion are independent.

This gives us a useful mental model: the vertical motion provides the time, and the horizontal motion provides the distance.

The floor effectively acts as a clock. It doesn’t measure time directly, but the known falling distance allows us to calculate it.

Step 1: Calculate the flight time from the vertical drop

Choose the ball’s centre at launch as the origin.

Let \(x\) be positive horizontally in the launch direction, and \(y\) be positive upwards.

For a perfectly horizontal launch:

• Initial vertical velocity: \(u_y=0\ \mathrm{m\,s^{-1}}\).
• Vertical acceleration: \(a_y=-g=-9.8\ \mathrm{m\,s^{-2}}\).
• Vertical displacement to impact: \(\Delta y=-h\).

The vertical displacement equation is:

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

Substituting the conditions:

\[
-h=0-\frac12gt^2
\]

Rearranging:

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

The symbol \(t\) represents flight time in seconds (s).

Notice that horizontal launch speed doesn’t appear in this equation. Under our assumptions, the flight time depends only on the vertical drop and gravitational acceleration.

Step 2: Calculate speed from horizontal distance

With negligible air resistance, there is no horizontal acceleration, so the horizontal velocity stays constant.

The ball travels horizontal distance \(R\) in time \(t\):

\[
R=u_xt
\]

Here, \(u_x\) is the initial horizontal velocity.

Rearranging:

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

Substituting the vertical flight-time equation:

\[
u_x=R\sqrt{\frac{g}{2h}}
\]

For an exactly horizontal launch, the initial vertical velocity is zero, so the launch speed \(u\) equals \(u_x\).

That’s the measurement method. We need only two distances to calculate the initial speed, provided the assumptions hold.

For the broader relationship between launch velocity, flight time, and distance, see Horizontal Range of a Projectile: HSC Physics Guide. Here, we’re using that relationship backwards to find an unknown speed from experimental measurements.

02Setting up the horizontal launch experiment

A workable apparatus needs a repeatable launch, a known starting position, and a reliable way to identify where the ball first lands.

Start with a small ball launcher or a rigid track whose final straight section is horizontal. Clamp it securely near the edge of a bench, with enough clearance for the ball to leave without striking the bench.

The ball travels along the launcher, passes the exit, and moves through the air towards the floor.

Place a sheet of white paper under the expected landing area. Carbon paper placed over it can produce a visible mark when the ball first strikes.

Measure the correct launch height

There is a small geometric detail worth getting right.

The ball’s vertical motion describes its centre of mass. For a uniform solid sphere, the centre of mass is at its geometric centre.

The relevant height is therefore the vertical distance travelled by the ball’s centre, from the instant it leaves the launcher until its first contact with the landing surface.

Suppose the ball’s centre is initially \(0.810\ \mathrm{m}\) above the paper, and the ball has a radius of \(0.010\ \mathrm{m}\).

At first contact, its centre is still one radius above the paper. The correct vertical drop is:

\[
\begin{aligned}
h&=0.810-0.010\\
&=0.800\ \mathrm{m}
\end{aligned}
\]

The correction is small, but ignoring it introduces a systematic error.

For a larger ball or a lower launch height, this correction can become more important.

Establish the horizontal origin

A ruler placed along the floor won’t help if you start measuring from the wrong point.

Use a plumb line to identify the point on the floor directly beneath the centre of the ball at the instant of release. Mark this position as \(x=0\).

Measure the horizontal distance from that mark to the ball’s first impact mark, along the direction of launch.

Don’t measure from the bench leg, the end of the bench, or the nearest edge of the launcher unless that point happens to align with the actual launch position.

If the ball drifts sideways, record its horizontal displacement in the intended launch direction rather than automatically treating a diagonal floor measurement as the forward range.

A large sideways drift is also a reason to inspect the launcher alignment.

Collect a useful set of measurements

Keep the launch setting, exit position, landing surface, and launcher alignment fixed.

Launch the ball several times, identifying the first impact mark for each trial. A bounce or roll after the initial collision must not be included in the measured range.

Record each range separately before calculating the mean. Repeated launches reveal variation that a single landing mark hides.

The first version of this experiment might use only one launch and one impact mark. That provides a speed estimate, but it cannot show how repeatable the launch is. Repeating trials is the first improvement.

03Turning the landing marks into a speed

04Improving the experiment: change the launch height

One height and several launches give us a repeatability check. They don’t tell us whether the speed equation itself describes the apparatus well.

A useful next test is to change the vertical drop while trying to keep the launch speed and direction unchanged.

What should happen if the horizontal launch model is correct?

From our original relationship:

\[
R=u\sqrt{\frac{2h}{g}}
\]

If \(u\) and \(g\) remain constant, then:

\[
R\propto\sqrt h
\]

This means the horizontal range increases with the square root of the vertical drop.

If the height becomes four times as large, the flight time doubles, and so does the horizontal range.

It does not become four times as large.

05When the horizontal-launch model gives the wrong speed

The calculation is simple because it makes strong assumptions.

It assumes the ball leaves horizontally, gravitational acceleration is approximately constant, air resistance is negligible, the landing surface is correctly located, and the ball moves without striking anything before first contact.

The most important experimental weakness is often the assumption about launch direction.

A slightly upward launch can look horizontal

Suppose the launcher points a few degrees upwards.

The ball now has a small positive initial vertical velocity. For a fixed vertical drop, gravity needs longer to bring it to the floor than it would for a horizontal launch.

If the horizontal component of launch velocity remains the same, the ball travels further.

A student might measure the larger range and conclude that the ball was launched faster.

That conclusion feels reasonable because greater launch speed can produce greater range. But the extra distance could instead come from additional flight time.

For a non-horizontal launch, the vertical displacement equation becomes:

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

The initial vertical velocity \(u_y\) is no longer zero, so the earlier flight-time shortcut does not apply.

A downward-pointing launcher has the opposite effect on flight time: the ball reaches the floor sooner than a horizontally launched ball with the same vertical drop.

Repeating the same slightly tilted launch many times won’t necessarily reveal this mistake. The marks may be tightly clustered because the error is systematic rather than random.

A practical decision process for experimental results

A side-on video can help check whether the ball starts with a vertical velocity component. Position the camera perpendicular to the plane of motion, with a ruler in the same plane to establish scale.

This provides an independent measurement rather than relying entirely on the assumption that the guide is level.

What the measurement can and cannot establish

A horizontal launch experiment measures speed indirectly. You observe distances, calculate an ideal flight time, and infer launch speed.

The result is only as trustworthy as the assumptions linking those measurements.

In particular, range measurements alone cannot independently establish that the launcher is horizontal, that air resistance is negligible, or that the launch setting has stayed constant.

Those limitations are worth remembering when interpreting the following problems. Some of them involve measurements that look convincing until another physical constraint is considered.

06Questions and solutions

Try each question before reading its solution. The later questions increasingly test whether you can distinguish an actual change in launch speed from a change in the experimental conditions.

Question 1

A small ball leaves a horizontal launcher from a position where its centre is \(0.720\ \mathrm{m}\) above its centre at first impact. Its horizontal range is \(1.08\ \mathrm{m}\).

Assuming negligible air resistance and \(g=9.8\ \mathrm{m\,s^{-2}}\), calculate:

(a) The time of flight.

(b) The initial launch speed.

Solution 1

The time of flight is \(0.383\ \mathrm{s}\), and the initial launch speed is \(2.82\ \mathrm{m\,s^{-1}}\).

For (a), the ball starts with no vertical velocity, so:

\[
\begin{aligned}
t&=\sqrt{\frac{2h}{g}}\\
&=\sqrt{\frac{2(0.720)}{9.8}}\\
&=0.383\ \mathrm{s}
\end{aligned}
\]

For (b), horizontal velocity remains constant:

\[
\begin{aligned}
u&=\frac{R}{t}\\
&=\frac{1.08}{0.383}\\
&=2.82\ \mathrm{m\,s^{-1}}
\end{aligned}
\]

The ball leaves the launcher moving horizontally at approximately \(2.82\ \mathrm{m\,s^{-1}}\).

A common mistake is to use the vertical acceleration to calculate the horizontal velocity change. Gravity acts vertically, so it changes the ball’s downward velocity, not its horizontal velocity in this ideal model.

Question 2

A launcher is used in two arrangements with the same launch setting.

In arrangement A, the vertical drop is \(0.500\ \mathrm{m}\). The ball lands \(1.00\ \mathrm{m}\) to the right of a marked floor origin \(O_A\), directly beneath its launch point.

For arrangement B, the apparatus is raised and accidentally moved horizontally. Its launch point is now \(0.060\ \mathrm{m}\) to the right of its previous position.

The vertical drop in B is \(0.720\ \mathrm{m}\). The new impact mark is \(1.26\ \mathrm{m}\) to the right of the original floor origin \(O_A\).

A student uses ranges of \(1.00\ \mathrm{m}\) and \(1.26\ \mathrm{m}\) to conclude that the ball left arrangement B faster.

Assuming both launches are horizontal and air resistance is negligible:

(a) Determine the correct horizontal range for arrangement B.

(b) Calculate the launch-speed estimate for each arrangement.

(c) Explain why the student’s conclusion is incorrect, and describe how to prevent this error in a real experiment.

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

Solution 2

The correct range for arrangement B is \(1.20\ \mathrm{m}\), and both arrangements give the same launch-speed estimate of \(3.13\ \mathrm{m\,s^{-1}}\). The apparent increase comes from measuring B’s range from the wrong origin.

For (a), the origin must be directly below the ball’s actual launch point in each trial.

The launcher has moved \(0.060\ \mathrm{m}\) towards the landing position.

Therefore:

\[
\begin{aligned}
R_B&=1.26-0.060\\
&=1.20\ \mathrm{m}
\end{aligned}
\]

For (b), calculate the flight time and speed for arrangement A:

\[
\begin{aligned}
t_A&=\sqrt{\frac{2(0.500)}{9.8}}\\
&=0.319\ \mathrm{s}\\
u_A&=\frac{1.00}{0.31944}\\
&=3.13\ \mathrm{m\,s^{-1}}
\end{aligned}
\]

For B:

\[
\begin{aligned}
t_B&=\sqrt{\frac{2(0.720)}{9.8}}\\
&=0.383\ \mathrm{s}\\
u_B&=\frac{1.20}{0.38333}\\
&=3.13\ \mathrm{m\,s^{-1}}
\end{aligned}
\]

For (c), two things have changed: the launch height and the horizontal position of the launcher.

The height change explains why B has a longer flight time. The position change explains why the student’s original measurement of B’s range was too large.

Neither change demonstrates increased launch speed.

The key correction is geometric. Raising or moving the launcher requires a new vertical projection of its launch point onto the floor.

A reliable procedure would re-establish the origin using a plumb line after every repositioning.

This question exposes a subtle experimental problem: a measurement can be precise and still be measured from the wrong reference point.

Question 3

A ball leaves a launcher from a vertical drop of \(0.800\ \mathrm{m}\).

A side-on camera shows that its initial velocity components are:

\[
u_x=3.00\ \mathrm{m\,s^{-1}}
\]

\[
u_y=+0.50\ \mathrm{m\,s^{-1}}
\]

Take upwards as positive, neglect air resistance, and use \(g=9.8\ \mathrm{m\,s^{-2}}\).

(a) Calculate the actual time of flight and horizontal range.

(b) Determine the launch speed a student would incorrectly calculate by assuming the launch was horizontal.

(c) Calculate the ball’s actual total launch speed.

(d) Explain why five closely grouped range measurements would not necessarily reveal the incorrect assumption.

Solution 3

The actual flight time is \(0.458\ \mathrm{s}\), and the horizontal range is \(1.37\ \mathrm{m}\). The incorrect horizontal-launch calculation gives \(3.40\ \mathrm{m\,s^{-1}}\), even though the actual total launch speed is only \(3.04\ \mathrm{m\,s^{-1}}\).

For (a), the initial vertical velocity is positive, so we cannot use the horizontal-launch flight-time formula.

Instead:

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

Substituting:

\[
-0.800=0.50t-4.90t^2
\]

Rearranging:

\[
4.90t^2-0.50t-0.800=0
\]

Using the quadratic formula and selecting the positive time:

\[
t=\frac{0.50+\sqrt{15.93}}{9.80}
\]

\[
t=0.458\ \mathrm{s}
\]

Horizontal velocity is constant because air resistance is neglected.

Therefore:

\[
\begin{aligned}
R&=u_xt\\
&=3.00(0.45829)\\
&=1.375\ \mathrm{m}
\end{aligned}
\]

The ball lands approximately \(1.37\ \mathrm{m}\) from the launch origin.

For (b), a student incorrectly assuming a horizontal launch would calculate:

\[
\begin{aligned}
t_{\text{wrong}}
&=\sqrt{\frac{2(0.800)}{9.8}}\\
&=0.404\ \mathrm{s}
\end{aligned}
\]

They would then use the measured range:

\[
\begin{aligned}
u_{\text{wrong}}&=\frac{1.375}{0.404}\\
&=3.40\ \mathrm{m\,s^{-1}}
\end{aligned}
\]

This overestimates the launch speed because it incorrectly attributes the longer range entirely to faster horizontal motion.

For (c), the actual total launch speed is the magnitude of the initial velocity vector:

\[
\begin{aligned}
u&=\sqrt{u_x^2+u_y^2}\\
&=\sqrt{3.00^2+0.50^2}\\
&=3.04\ \mathrm{m\,s^{-1}}
\end{aligned}
\]

For (d), repeated range measurements mainly test the consistency of the launching process.

If every launch has nearly the same small upward velocity component, the impact marks can be closely grouped.

That would indicate good repeatability, not that the launch angle is correct.

The missing check is an independent measurement of the exit direction or vertical velocity. The side-on camera provides precisely that information.

Question 4

A ball leaves a horizontal launcher with an initial horizontal velocity of \(3.00\ \mathrm{m\,s^{-1}}\), verified using light gates immediately after the exit.

The vertical drop to first impact is \(0.800\ \mathrm{m}\).

Two trials are performed:

• Trial A has no fan. Its measured range is \(1.21\ \mathrm{m}\).
• Trial B uses a fan blowing horizontally against the ball’s motion. Its measured range is \(1.09\ \mathrm{m}\).

Video analysis confirms that the flight time in both trials agrees with the ideal vertical free-fall prediction.

Assume the fan produces an approximately constant horizontal acceleration throughout trial B, with negligible vertical force. Use \(g=9.8\ \mathrm{m\,s^{-2}}\).

(a) Calculate the approximate horizontal acceleration in trial B.

(b) Determine the horizontal velocity immediately before impact in trial B.

(c) Calculate the apparent launch speed that the standard horizontal-launch range formula would give for trial B. Explain why this is not the actual launch speed.

(d) Predict the range if the fan were reversed, producing the same magnitude of horizontal acceleration in the opposite direction, without changing the vertical motion.

Solution 4

The fan produces an approximate horizontal acceleration of \(-1.50\ \mathrm{m\,s^{-2}}\). In trial B, the ball reaches the floor with a horizontal velocity of about \(2.40\ \mathrm{m\,s^{-1}}\), while the usual range formula incorrectly suggests a launch speed of \(2.70\ \mathrm{m\,s^{-1}}\). Reversing the fan would increase the predicted range to approximately \(1.33\ \mathrm{m}\).

For (a), first calculate the flight time from vertical motion:

\[
\begin{aligned}
t&=\sqrt{\frac{2(0.800)}{9.8}}\\
&=0.40406\ \mathrm{s}
\end{aligned}
\]

The fan causes horizontal acceleration, so the appropriate horizontal displacement equation is:

\[
R=u_xt+\frac12a_xt^2
\]

Rearranging:

\[
a_x=\frac{2(R-u_xt)}{t^2}
\]

Substituting trial B’s range:

\[
a_x=
\frac{2[1.09-3.00(0.40406)]}
{(0.40406)^2}
\]

\[
a_x=-1.50\ \mathrm{m\,s^{-2}}
\]

The negative sign means the acceleration acts opposite to the positive horizontal launch direction.

For (b), calculate the final horizontal velocity:

\[
v_x=u_x+a_xt
\]

\[
\begin{aligned}
v_x&=3.00-1.497(0.40406)\\
&=2.40\ \mathrm{m\,s^{-1}}
\end{aligned}
\]

The ball is still moving forwards at impact, but more slowly than it was at launch.

For (c), the standard horizontal-launch formula would give:

\[
\begin{aligned}
u_{\text{apparent}}&=\frac{1.09}{0.40406}\\
&=2.70\ \mathrm{m\,s^{-1}}
\end{aligned}
\]

This is smaller than the independently measured initial horizontal velocity of \(3.00\ \mathrm{m\,s^{-1}}\).

Why?

The usual calculation divides horizontal displacement by flight time. That equals the average horizontal velocity over the flight.

When horizontal acceleration is zero, average horizontal velocity equals initial horizontal velocity.

When the fan slows the ball, they are no longer equal.

The light gates measure its actual initial velocity, while the range calculation gives its average velocity during the flight.

For (d), reversing the fan changes the horizontal acceleration to approximately \(+1.497\ \mathrm{m\,s^{-2}}\).

Using the unchanged flight time:

\[
R=u_xt+\frac12a_xt^2
\]

\[
\begin{aligned}
R&=3.00(0.40406)\\
&\quad+\frac12(1.497)(0.40406)^2\\
&=1.33\ \mathrm{m}
\end{aligned}
\]

The ball now travels further because the fan increases its horizontal velocity during flight.

The more important lesson is experimental: a measured range can change even when the launch speed is unchanged.

Here, the independent light-gate measurement prevents us from incorrectly attributing that change to the launcher.

Question 5

A precision projectile launcher is intended to launch a small ball horizontally, but its exit direction cannot be measured directly.

Instead, a movable catching platform is used. Only one platform position is present during each trial, so neither arrangement obstructs the other.

The launcher is mechanically locked, guaranteeing the same initial velocity vector in both trials.

Two exact measurements are obtained:

Both drops are measured between the ball’s centre at launch and its centre at first contact. Both ranges are measured from directly beneath the same launch point.

Assume the ball moves in a vertical plane, horizontal velocity is initially positive, air resistance is negligible, and gravitational acceleration is constant at \(10.0\ \mathrm{m\,s^{-2}}\).

A student claims that both launches were horizontal and calculates a separate speed from each range.

(a) Show whether the two measurements are consistent with identical, perfectly horizontal launches.

(b) Without assuming a horizontal launch, determine the actual initial horizontal and vertical velocity components using the two measurements.

(c) Calculate the actual total launch speed.

(d) Imagine replacing the lower catching platform with a very narrow catcher positioned at exactly the original launch height, without obstructing the launch. How far horizontally from the launch point would the ball cross that height while descending?

(e) Explain why the result in (d) is a particularly strong test of whether the original launch was horizontal.

Solution 5

The launch was not horizontal. Its initial velocity components were \(u_x=3.00\ \mathrm{m\,s^{-1}}\) and \(u_y=+0.500\ \mathrm{m\,s^{-1}}\), giving a total launch speed of \(3.04\ \mathrm{m\,s^{-1}}\). The ball crosses its original launch height again \(0.300\ \mathrm{m}\) horizontally from the exit.

The surprising part is that the small upward velocity component is almost invisible from the launch geometry, but it changes what the range measurements mean.

Part (a): Test the horizontal-launch model.

If both launches were horizontal, the inferred speed from each measurement would be:

\[
u=R\sqrt{\frac{g}{2h}}
\]

For trial 1:

\[
\begin{aligned}
u_1&=1.200\sqrt{\frac{10.0}{2(0.600)}}\\
&=3.46\ \mathrm{m\,s^{-1}}
\end{aligned}
\]

For trial 2:

\[
\begin{aligned}
u_2&=1.800\sqrt{\frac{10.0}{2(1.500)}}\\
&=3.29\ \mathrm{m\,s^{-1}}
\end{aligned}
\]

The calculated speeds are different.

Because the data are specified as exact and the initial velocity vector is guaranteed to be identical, both horizontal-launch estimates cannot be correct.

At least one assumption in the horizontal-launch calculation must fail.

Under the stated model, the remaining possibility is a nonzero initial vertical velocity.

Part (b): Reconstruct both velocity components.

We now have two unknown initial velocity components, \(u_x\) and \(u_y\).

We also have two independent range-and-height measurements.

The vertical displacement equation is:

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

Horizontal velocity is constant, so:

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

Substitute this expression for time into the vertical equation:

\[
-h=\frac{u_yR}{u_x}
-\frac{gR^2}{2u_x^2}
\]

Rearranging:

\[
h=\frac{gR^2}{2u_x^2}
-\frac{u_yR}{u_x}
\]

Dividing by \(R\):

\[
\frac{h}{R}=
\frac{gR}{2u_x^2}
-\frac{u_y}{u_x}
\]

This equation matters because the two trials share the same \(u_x\) and \(u_y\).

To simplify the simultaneous equations, define:

\[
A=\frac{g}{2u_x^2}
\]

and:

\[
B=\frac{u_y}{u_x}
\]

Here, \(A\) has units of \(\mathrm{m^{-1}}\), while \(B\) is dimensionless.

Our equation becomes:

\[
\frac{h}{R}=AR-B
\]

For trial 1:

\[
\frac{0.600}{1.200}=1.200A-B
\]

\[
0.500=1.200A-B
\]

For trial 2:

\[
\frac{1.500}{1.800}=1.800A-B
\]

\[
0.83333=1.800A-B
\]

Subtract the first equation from the second:

\[
\begin{aligned}
0.33333&=0.600A\\
A&=0.55556\ \mathrm{m^{-1}}
\end{aligned}
\]

Now recover the horizontal velocity:

\[
A=\frac{g}{2u_x^2}
\]

Therefore:

\[
\begin{aligned}
u_x&=\sqrt{\frac{g}{2A}}\\
&=\sqrt{\frac{10.0}{2(0.55556)}}\\
&=3.00\ \mathrm{m\,s^{-1}}
\end{aligned}
\]

Next, substitute \(A\) into the first trial equation:

\[
0.500=1.200(0.55556)-B
\]

\[
B=0.16667
\]

Since \(B=u_y/u_x\):

\[
\begin{aligned}
u_y&=Bu_x\\
&=0.16667(3.00)\\
&=+0.500\ \mathrm{m\,s^{-1}}
\end{aligned}
\]

The positive vertical component shows that the launcher points slightly upwards.

We can verify the result by calculating the flight times from horizontal motion:

\[
\begin{aligned}
t_1&=\frac{1.200}{3.00}=0.400\ \mathrm{s}\\
t_2&=\frac{1.800}{3.00}=0.600\ \mathrm{s}
\end{aligned}
\]

For the first trial, the vertical displacement is:

\[
\begin{aligned}
\Delta y_1&=0.500(0.400)\\
&\quad-5.00(0.400)^2\\
&=-0.600\ \mathrm{m}
\end{aligned}
\]

For the second trial:

\[
\begin{aligned}
\Delta y_2&=0.500(0.600)\\
&\quad-5.00(0.600)^2\\
&=-1.500\ \mathrm{m}
\end{aligned}
\]

Both measurements are reproduced exactly.

The two ranges therefore reveal the non-horizontal initial velocity under the stated assumptions.

Part (c): Find the total launch speed.

The initial velocity components are perpendicular, so use Pythagoras:

\[
\begin{aligned}
u&=\sqrt{u_x^2+u_y^2}\\
&=\sqrt{3.00^2+0.500^2}\\
&=3.04\ \mathrm{m\,s^{-1}}
\end{aligned}
\]

The true launch speed is \(3.04\ \mathrm{m\,s^{-1}}\).

Notice that neither of the two earlier horizontal-launch estimates was correct.

Part (d): Find where the ball returns to launch height.

The ball initially rises because \(u_y\) is positive.

To find when it returns to its original height, set vertical displacement to zero:

\[
0=u_yt-\frac12gt^2
\]

Factorising:

\[
0=t\left(u_y-\frac12gt\right)
\]

There are two solutions.

The first is \(t=0\), corresponding to the instant of launch.

The second is:

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

Substituting:

\[
\begin{aligned}
t&=\frac{2(0.500)}{10.0}\\
&=0.100\ \mathrm{s}
\end{aligned}
\]

The horizontal distance is:

\[
\begin{aligned}
R&=u_xt\\
&=3.00(0.100)\\
&=0.300\ \mathrm{m}
\end{aligned}
\]

The ball returns to launch height after travelling \(0.300\ \mathrm{m}\) horizontally.

Part (e): Explain the conceptual difference.

A truly horizontal projectile has zero initial vertical velocity.

As soon as positive time passes, gravity moves it below the launch height.

It cannot leave horizontally, rise above its starting height, and later return to that same height under gravity alone.

Our reconstructed projectile can return because it initially moves upwards.

In fact, its maximum rise is only:

\[
\begin{aligned}
\Delta y_{\max}
&=\frac{u_y^2}{2g}\\
&=\frac{0.500^2}{2(10.0)}\\
&=0.0125\ \mathrm{m}
\end{aligned}
\]

That’s just 1.25 cm.

The launcher could look almost perfectly level, and the ball’s early trajectory might be difficult to distinguish from a horizontal launch by eye.

Yet that tiny upward motion is enough to invalidate the basic range-to-speed shortcut.

The lesson is not that the shortcut is unreliable. It’s that a good experiment tests the physical conditions that make the shortcut valid.

07What to measure next

A horizontal-launch experiment gives a useful estimate of initial speed from a ruler, an impact mark, and a known vertical drop.

But its strongest result comes from combining that calculation with checks on the apparatus.

Repeated landing marks test consistency. Changing the launch height tests whether the range follows the predicted relationship with falling distance. Checking the launch direction tests the assumption that initial vertical velocity is zero.

If these checks agree, the indirect speed measurement becomes much more convincing. If they disagree, don’t immediately adjust the speed calculation to make the numbers match. Work out which physical assumption or measurement has failed.

The next useful step is to measure the ball’s motion independently using calibrated side-on video. Comparing its horizontal position at successive times can test whether horizontal velocity stays constant, while its early vertical displacement can reveal a small upward or downward launch component.

That turns a simple launch-speed calculation into an experiment that can distinguish a correct model from a convincing-looking result.