Why Two Launch Angles Can Give the Same Projectile Range

A ball launcher fires two identical balls at the same speed: one at \(25^\circ\) above the horizontal, and the other at \(65^\circ\). The steeper shot rises much higher and stays in the air more than twice as long. Wouldn’t it travel further?

Not if both balls land at their launch height. They have exactly the same horizontal range.

The reason is that \(25^\circ\) and \(65^\circ\) are complementary launch angles: they add to \(90^\circ\). At the same launch speed, the steeper shot gains flight time but loses horizontal speed. The shallower shot has less time but moves sideways faster. Under ideal projectile conditions, these effects balance exactly.

That equality is useful, but it has conditions. A small change to the experiment, such as moving the landing surface below the launcher, can destroy it.

01A launcher experiment that could give the wrong answer

Imagine testing the claim with a spring-powered ball launcher.

The launcher sits on a stand, with its barrel pointing towards an open space. A ball travels along the barrel, leaves the muzzle, follows a curved path through the air, and eventually lands on a horizontal surface.

A tape measure records the horizontal distance from the muzzle’s starting position to the landing point. A side-on camera records the ball’s motion.

Set the barrel to \(25^\circ\), fire a ball, and mark where it lands. Then adjust the barrel to \(65^\circ\), use the same spring compression, and fire again.

Should the two landing marks coincide?

You might expect them to. But this first design has two problems.

First, if the barrel rotates around a pivot near its base, changing the angle also changes the height of the muzzle. The two balls no longer start from the same position.

Second, equal spring compression does not necessarily guarantee equal launch speed. Energy losses inside the launcher can affect the speed at which a ball exits.

If the balls land at different positions, you cannot confidently blame the difference on launch angle alone.

A fairer design

A better experiment controls the quantities that the prediction depends on.

• Fix the muzzle’s position, allowing the barrel to rotate around that point.
• Use two closely spaced light gates near the muzzle to measure the ball’s launch speed.
• Adjust the launcher until both angles produce the same measured speed.
• Place a narrow, movable catch tray so that its catching height matches the muzzle height.
• Measure the horizontal distance to the tray, repeat each launch, and compare the landing positions.

The catch tray matters. A ball falling onto the floor below the launcher is not completing the same motion as a ball returning to its release height.

Suppose the controlled launch speed is \(4.90\ \mathrm{m\,s^{-1}}\). What distances and flight times should this apparatus produce?

We can predict them before taking any measurements.

02Why complementary angles produce equal range

Separate the horizontal and vertical motion

When a ball leaves the launcher, it moves both sideways and upwards.

Gravity pulls downwards. In the ideal projectile model, with air resistance neglected, gravity changes the vertical velocity but leaves the horizontal velocity constant.

Think of two independent measurements:

• The horizontal velocity tells you how quickly the ball moves towards the catch tray.
• The vertical motion tells you how long the ball stays above its release height.

The range comes from combining those two measurements.

Choose rightwards as positive \(x\) and upwards as positive \(y\). Put the origin at the muzzle.

Let:

• \(u\) be the launch speed in \(\mathrm{m\,s^{-1}}\),
• \(\theta\) be the launch angle above the horizontal,
• \(u_x\) be the initial horizontal velocity in \(\mathrm{m\,s^{-1}}\),
• \(u_y\) be the initial vertical velocity in \(\mathrm{m\,s^{-1}}\),
• \(t\) be the elapsed time in seconds, and
• \(g=9.80\ \mathrm{m\,s^{-2}}\) be the magnitude of gravitational acceleration.

Resolving the initial velocity gives

\[ u_x=u\cos\theta \]

and

\[ u_y=u\sin\theta. \]

With constant gravity and negligible air resistance, the horizontal equation is

\[ x=u_xt. \]

The vertical equation is

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

Here, \(x\) and \(y\) are displacements from the muzzle in metres. The negative gravitational term appears because upwards is positive.

These equations describe the same moving ball. We separate them because gravity only acts vertically in this model.

The key connection: horizontal speed multiplied by flight time

For a ball returning to its launch height, the final vertical displacement is zero.

If \(T\) is the total flight time,

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

One solution is \(T=0\), the instant of launch. The later return to the original height occurs at

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

The horizontal range \(R\), measured in metres, is the horizontal displacement during that time:

\[ R=u_xT. \]

Substituting the flight time gives

\[ \boxed{} \]

\[ R=\frac{2u_xu_y}{g}. \]

Look at that last equation carefully.

Range depends on the product of the horizontal and vertical launch velocities. Multiplying \(u_x\) by \(u_y\) gives the same result as multiplying \(u_y\) by \(u_x\).

For complementary angles at the same launch speed, the two components exchange values. That’s the physical reason the ranges match.

What would the experiment establish?

A real experiment should produce landing positions close to these predictions if the model is suitable and the controlled quantities remain constant.

The two landing marks might not coincide perfectly. Air resistance, variations in launch speed, imperfect angle settings, and measurement uncertainty can all produce differences.

Repeating the trials would help determine whether any difference is larger than the expected experimental variation.

Agreement would support the ideal model under the tested conditions. It would not establish that every projectile, including one experiencing substantial air resistance, follows exactly the same rule.

03Why the two angles add to 90°

Substitute the velocity components into the range equation:

\[ R=\frac{2u^2\sin\theta\cos\theta}{g}. \]

Using the trigonometric identity

\[ 2\sin\theta\cos\theta=\sin(2\theta), \]

we obtain the familiar level-ground range formula:

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

The important feature is \(\sin(2\theta)\).

Sine has the same value for supplementary angles. For example,

\[ \sin50^\circ=\sin130^\circ. \]

Those double angles come from launch angles of \(25^\circ\) and \(65^\circ\).

More generally, if one launch angle is \(\theta\), its complementary angle is

\[ 90^\circ-\theta. \]

Doubling those angles produces \(2\theta\) and \(180^\circ-2\theta\), which have equal sine values.

For positive ranges, acute launch angles, and the same fixed launch speed, the two distinct level-ground angles are therefore complementary.

Why 45° is the turning point

At \(45^\circ\), the horizontal and vertical components are equal.

The projectile has enough upward velocity to remain airborne for a substantial time, without sacrificing too much horizontal velocity.

For a fixed launch speed, the range formula becomes largest when

\[ \sin(2\theta)=1. \]

That occurs when \(2\theta=90^\circ\), giving

\[ \theta=45^\circ. \]

The maximum horizontal range is

\[ R_{\max}=\frac{u^2}{g}. \]

For launch angles below \(45^\circ\), increasing the angle increases the range. Above \(45^\circ\), increasing the angle decreases it.

This gives a useful decision rule for the ideal, equal-height case:

The two-angle result is not guaranteed for every target. It applies to horizontal range when launch and landing heights match.

04What changes when the ball lands below its launch height?

Now return to the original \(25^\circ\) and \(65^\circ\) experiment.

Instead of catching the balls at muzzle height, allow them to fall onto the floor \(0.80\ \mathrm{m}\) below the muzzle.

Keep the launch speed at \(4.90\ \mathrm{m\,s^{-1}}\).

Would the two balls still land at the same horizontal distance?

No. The shallower shot now travels further.

The mistake would be to apply the equal-height range formula without checking the landing height.

The vertical equation must change first

Our positive direction is still upwards, but the ball’s final displacement is now

\[ y=-0.80\ \mathrm{m}. \]

More generally, let \(h\) be the positive vertical drop from the launch point to the landing surface.

Then

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

Solving this quadratic and choosing the positive flight time gives

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

The horizontal range is still

\[ R=u_xT. \]

Notice that the formula for horizontal motion hasn’t changed. The vertical motion determines a different flight time.

For our two launches, the ideal predictions are:

The \(25^\circ\) shot lands about \(0.77\ \mathrm{m}\) further away.

Why does the shallow launch benefit more?

Both trajectories return to their original launch height at the same horizontal distance, \(1.88\ \mathrm{m}\).

At that moment, both balls are descending. The shallow shot has a smaller downward velocity, so it takes longer to fall the additional \(0.80\ \mathrm{m}\).

It also has a much greater horizontal velocity.

That combination gives the shallow shot a larger amount of extra horizontal travel.

The steeper shot still stays airborne longer overall, but it does not travel as far.

This is not another version of the complementary-angle rule. It is a different landing-height problem, and it needs the full vertical displacement equation.

A quick way to decide which model applies

If the landing height changes, start with the general method in Horizontal Range of a Projectile: HSC Physics Guide, rather than using the level-ground shortcut.

05Questions and solutions

Assume constant \(g=9.80\ \mathrm{m\,s^{-2}}\), negligible air resistance, and point-like projectiles throughout. Unless stated otherwise, angles are measured above the horizontal, and launch and landing heights are equal.

Question 1

A ball is launched at \(12.0\ \mathrm{m\,s^{-1}}\) at \(35^\circ\) above horizontal ground.

Find the other acute launch angle that produces the same horizontal range. Calculate the common range and determine which launch stays airborne longer.

Solution 1

The other angle is \(55^\circ\), both launches travel approximately \(13.8\ \mathrm{m}\), and the \(55^\circ\) launch stays airborne longer.

For a fixed speed and equal launch and landing heights, the two distinct angles giving the same range are complementary.

Therefore,

\[ \theta_2=90^\circ-35^\circ=55^\circ. \]

Calculate the range using either angle:

\[ \begin{aligned} R&=\frac{u^2\sin(2\theta)}{g}\\ &=\frac{12.0^2\sin70^\circ}{9.80}\\ &=13.8\ \mathrm{m}. \end{aligned} \]

To compare flight times, use

\[ T=\frac{2u\sin\theta}{g}. \]

For the \(35^\circ\) shot,

\[ \begin{aligned} T_{35}&=\frac{2(12.0)\sin35^\circ}{9.80}\\ &=1.40\ \mathrm{s}. \end{aligned} \]

For the \(55^\circ\) shot,

\[ \begin{aligned} T_{55}&=\frac{2(12.0)\sin55^\circ}{9.80}\\ &=2.01\ \mathrm{s}. \end{aligned} \]

The tempting mistake is to assume that a longer flight must produce a longer range. The higher launch has less horizontal velocity, which exactly compensates for the extra time.

Question 2

A laboratory launcher is adjusted to produce the same initial horizontal velocity of \(8.00\ \mathrm{m\,s^{-1}}\) at two angles: \(30^\circ\) and \(60^\circ\).

Both balls land at their launch height.

A student argues that, because the angles are complementary, the balls must travel the same horizontal distance.

Determine the total launch speed, flight time, and range for each angle. Assess the student’s reasoning.

Solution 2

The student is incorrect. The \(30^\circ\) shot travels approximately \(7.54\ \mathrm{m}\), while the \(60^\circ\) shot travels approximately \(22.6\ \mathrm{m}\), three times as far.

The complementary-angle rule requires equal total launch speeds, not merely equal horizontal components.

Start with

\[ u_x=u\cos\theta. \]

Rearranging,

\[ u=\frac{u_x}{\cos\theta}. \]

At \(30^\circ\),

\[ \begin{aligned} u_{30}&=\frac{8.00}{\cos30^\circ}\\ &=9.24\ \mathrm{m\,s^{-1}}. \end{aligned} \]

At \(60^\circ\),

\[ \begin{aligned} u_{60}&=\frac{8.00}{\cos60^\circ}\\ &=16.0\ \mathrm{m\,s^{-1}}. \end{aligned} \]

The total speeds are different.

Next, find the vertical components. Since \(u_y=u_x\tan\theta\),

\[ \begin{aligned} u_{y,30}&=8.00\tan30^\circ\\ &=4.62\ \mathrm{m\,s^{-1}},\\ u_{y,60}&=8.00\tan60^\circ\\ &=13.9\ \mathrm{m\,s^{-1}}. \end{aligned} \]

The flight times are

\[ \begin{aligned} T_{30}&=\frac{2(4.6188)}{9.80}\\ &=0.943\ \mathrm{s},\\ T_{60}&=\frac{2(13.8564)}{9.80}\\ &=2.83\ \mathrm{s}. \end{aligned} \]

Finally, calculate the horizontal ranges:

\[ \begin{aligned} R_{30}&=8.00(0.9426)\\ &=7.54\ \mathrm{m},\\ R_{60}&=8.00(2.8278)\\ &=22.6\ \mathrm{m}. \end{aligned} \]

The higher launch stays airborne three times as long while moving horizontally at exactly the same speed.

The critical distinction is what the experiment keeps constant. Complementary angles only guarantee equal range when the launch speeds are equal and the other ideal-model conditions are satisfied.

Question 3

Two balls are launched from the same ledge at \(10.0\ \mathrm{m\,s^{-1}}\), one at \(20^\circ\) and the other at \(70^\circ\).

The flat ground below is \(2.00\ \mathrm{m}\) lower than the launch point.

A student calculates the equal-height range for both shots and obtains approximately \(6.56\ \mathrm{m}\). They then claim both balls will land at the same position on the lower ground because the extra \(2.00\ \mathrm{m}\) of falling affects both equally.

Explain why this reasoning fails. Determine which ball lands further away and calculate both ground-level ranges.

Solution 3

The \(20^\circ\) ball lands further away, travelling approximately \(10.1\ \mathrm{m}\) compared with \(7.22\ \mathrm{m}\) for the \(70^\circ\) ball.

The extra vertical drop does not affect their ranges equally. The balls reach their original launch height with different downward velocities and different horizontal velocities.

First, consider the student’s \(6.56\ \mathrm{m}\) result.

\[ \begin{aligned} R_0&=\frac{u^2\sin40^\circ}{g}\\ &=\frac{10.0^2\sin40^\circ}{9.80}\\ &=6.56\ \mathrm{m}. \end{aligned} \]

This is correct for both trajectories at the instant they return to their original launch height.

However, the balls continue falling another \(2.00\ \mathrm{m}\).

For a downward displacement of magnitude \(h=2.00\ \mathrm{m}\),

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

For the \(20^\circ\) shot,

\[ \begin{aligned} u_x&=10.0\cos20^\circ\\ &=9.397\ \mathrm{m\,s^{-1}},\\ u_y&=10.0\sin20^\circ\\ &=3.420\ \mathrm{m\,s^{-1}}. \end{aligned} \]

Its total flight time is

\[ \begin{aligned} T_{20} &=\frac{3.420+\sqrt{3.420^2+39.2}}{9.80}\\ &=1.077\ \mathrm{s}. \end{aligned} \]

Its range is

\[ \begin{aligned} R_{20}&=9.397(1.077)\\ &=10.1\ \mathrm{m}. \end{aligned} \]

For the \(70^\circ\) shot,

\[ \begin{aligned} u_x&=10.0\cos70^\circ\\ &=3.420\ \mathrm{m\,s^{-1}},\\ u_y&=10.0\sin70^\circ\\ &=9.397\ \mathrm{m\,s^{-1}}. \end{aligned} \]

The total flight time is

\[ \begin{aligned} T_{70} &=\frac{9.397+\sqrt{9.397^2+39.2}}{9.80}\\ &=2.111\ \mathrm{s}. \end{aligned} \]

The corresponding range is

\[ \begin{aligned} R_{70}&=3.420(2.111)\\ &=7.22\ \mathrm{m}. \end{aligned} \]

The deeper explanation comes from what happens after each ball crosses its original launch height on the way down.

At that instant, the \(20^\circ\) ball is moving downwards more slowly than the \(70^\circ\) ball. It therefore takes longer to fall the remaining \(2.00\ \mathrm{m}\).

During that additional time, it also moves horizontally much faster.

The \(20^\circ\) ball gains about \(3.56\ \mathrm{m}\) beyond the equal-height range, while the \(70^\circ\) ball gains only about \(0.66\ \mathrm{m}\).

The student’s original range calculation was not mathematically wrong. It answered a different question: where the balls would be when they returned to launch height, not where they would hit the lower ground.

Question 4

Two balls are fired from the same point at \(14.0\ \mathrm{m\,s^{-1}}\), one at \(30^\circ\) and one at \(60^\circ\). Both return to the launch height.

A camera operator wants to record the balls colliding in mid-air. They propose adjusting the launch times until both balls arrive at a position halfway along their common horizontal range at the same instant.

Determine the height of each trajectory at this halfway position.

Could changing the launch times make the balls collide anywhere above the ground?

If both balls must instead arrive at their common landing point simultaneously, determine which ball should be launched first and by how much.

Solution 4

The two balls are \(2.50\ \mathrm{m}\) and \(7.50\ \mathrm{m}\) above launch height at the halfway position. No launch delay can make them collide above ground. To arrive at their shared landing point simultaneously, the \(60^\circ\) ball must be launched approximately \(1.05\ \mathrm{s}\) before the \(30^\circ\) ball.

The key is that sharing a horizontal range does not mean sharing the same path.

First calculate the common range:

\[ \begin{aligned} R&=\frac{14.0^2\sin60^\circ}{9.80}\\ &=17.32\ \mathrm{m}. \end{aligned} \]

The halfway position is

\[ x=\frac R2=8.66\ \mathrm{m}. \]

At a horizontal displacement \(x\), the time elapsed along a particular trajectory is

\[ t=\frac{x}{u\cos\theta}. \]

Substituting this into the vertical equation gives

\[ y=x\tan\theta- \frac{gx^2}{2u^2\cos^2\theta}. \]

For \(30^\circ\),

\[ \begin{aligned} y_{30} &=8.66\tan30^\circ\\ &\quad-\frac{9.80(8.66)^2} {2(14.0)^2\cos^230^\circ}\\ &=2.50\ \mathrm{m}. \end{aligned} \]

For \(60^\circ\),

\[ \begin{aligned} y_{60} &=8.66\tan60^\circ\\ &\quad-\frac{9.80(8.66)^2} {2(14.0)^2\cos^260^\circ}\\ &=7.50\ \mathrm{m}. \end{aligned} \]

The two trajectories are separated vertically by \(5.00\ \mathrm{m}\) at the halfway position.

Could they cross somewhere else in the air?

To answer that, compare their heights at the same horizontal coordinate.

Using the trajectory equation with \(x\) and \(y\) measured in metres, subtraction gives

\[ y_{60}-y_{30} =1.1547x-0.06667x^2. \]

For the heights to match, this difference must be zero:

\[ 0=x(1.1547-0.06667x). \]

The solutions are

\[ x=0 \quad\text{or}\quad x=17.32\ \mathrm{m}. \]

These are the launch point and the shared landing point.

At every horizontal position strictly between them, the \(60^\circ\) trajectory lies above the \(30^\circ\) trajectory.

Changing the launch times changes when a ball occupies a position. It does not change its geometric trajectory under the stated assumptions.

Therefore, no launch delay can produce a collision above the launch-height ground.

To reach the landing point simultaneously, compare the flight times.

For \(30^\circ\),

\[ \begin{aligned} T_{30}&=\frac{2(14.0)\sin30^\circ}{9.80}\\ &=1.429\ \mathrm{s}. \end{aligned} \]

For \(60^\circ\),

\[ \begin{aligned} T_{60}&=\frac{2(14.0)\sin60^\circ}{9.80}\\ &=2.474\ \mathrm{s}. \end{aligned} \]

The difference is

\[ \begin{aligned} \Delta T&=2.474-1.429\\ &=1.05\ \mathrm{s}. \end{aligned} \]

Launch the \(60^\circ\) ball first, followed approximately \(1.05\ \mathrm{s}\) later by the \(30^\circ\) ball.

Both will then arrive at the common landing position at the same instant.

Equal range, equal position, and equal arrival time are three different requirements. They should never be treated as interchangeable.

Question 5

A launcher fires a small ball at a fixed speed of \(14.0\ \mathrm{m\,s^{-1}}\).

A small target opening is located \(12.0\ \mathrm{m}\) horizontally from the launcher and \(6.00\ \mathrm{m}\) above the launch point. The ball can travel unobstructed to the opening.

An experimenter finds that two different upward launch angles allow the ball to pass through this opening.

The target can then be moved vertically while its horizontal position remains fixed at \(12.0\ \mathrm{m}\).

Determine both angles that reach the opening at \(6.00\ \mathrm{m}\). Then find the greatest height at which the target can be reached and the launch angle that achieves this limiting height.

At the limiting height, is the ball necessarily at the highest point of its own trajectory when it passes through the opening? Justify your answer using its vertical velocity.

Explain why neither the complementary-angle rule nor the usual \(45^\circ\) maximum-range rule gives the solution directly.

Solution 5

The two launch angles are approximately \(53.1^\circ\) and \(63.4^\circ\). The greatest reachable target height is \(6.40\ \mathrm{m}\), achieved at approximately \(59.0^\circ\). Surprisingly, the ball is already moving downwards when it reaches that limiting position.

The problem is different from finding the greatest horizontal range. We are finding the greatest possible vertical position at one fixed horizontal distance.

First, establish the trajectory equation.

The horizontal equation is

\[ x=u\cos\theta\,t. \]

Therefore,

\[ t=\frac{x}{u\cos\theta}. \]

Substitute this time into the vertical displacement equation:

\[ y=u\sin\theta\,t-\frac12gt^2. \]

After simplifying,

\[ y=x\tan\theta- \frac{gx^2}{2u^2\cos^2\theta}. \]

Using

\[ \frac1{\cos^2\theta}=1+\tan^2\theta, \]

we obtain an equation in terms of the launch angle alone:

\[ y=x\tan\theta- \frac{gx^2}{2u^2}(1+\tan^2\theta). \]

Next, find the two angles that reach the opening.

The supplied values are

\[ \begin{aligned} u&=14.0\ \mathrm{m\,s^{-1}},\\ x&=12.0\ \mathrm{m},\\ y&=6.00\ \mathrm{m}. \end{aligned} \]

The coefficient of the final term is

\[ \begin{aligned} \frac{gx^2}{2u^2} &=\frac{9.80(12.0)^2}{2(14.0)^2}\\ &=3.60\ \mathrm{m}. \end{aligned} \]

Let \(p=\tan\theta\), a dimensionless quantity. Substitution gives

\[ 6.00=12.0p-3.60(1+p^2). \]

Rearranging and simplifying,

\[ \begin{aligned} 3.60p^2-12.0p+9.60&=0,\\ 3p^2-10p+8&=0,\\ (3p-4)(p-2)&=0. \end{aligned} \]

Therefore,

\[ p=\frac43 \quad\text{or}\quad p=2. \]

Since \(p=\tan\theta\),

\[ \begin{aligned} \theta_1&=\tan^{-1}\left(\frac43\right)\\ &=53.1^\circ,\\ \theta_2&=\tan^{-1}(2)\\ &=63.4^\circ. \end{aligned} \]

Both angles reach the same elevated opening.

They do not add to \(90^\circ\), because the target is not at the launch height.

Now find the highest reachable position at \(x=12.0\ \mathrm{m}\).

Keep the horizontal displacement fixed but allow the target height \(y\) to vary.

We already have

\[ y=12.0p-3.60(1+p^2). \]

Rearrange this expression by completing the square:

\[ y=6.40-3.60\left(p-\frac53\right)^2. \]

The squared expression can never be negative.

Therefore, the greatest possible value of \(y\) occurs when

\[ p=\frac53. \]

At this value,

\[ y_{\max}=6.40\ \mathrm{m}. \]

The corresponding launch angle is

\[ \begin{aligned} \theta&=\tan^{-1}\left(\frac53\right)\\ &=59.0^\circ. \end{aligned} \]

At \(6.40\ \mathrm{m}\), the two possible launch-angle solutions merge into one. Any higher target at that same horizontal distance is unreachable at the given launch speed.

This is a boundary between two possible trajectories, one possible trajectory, and no possible trajectory.

Finally, determine whether the ball is at its apex.

At the limiting angle,

\[ \theta\approx59.0^\circ. \]

The horizontal velocity is

\[ \begin{aligned} u_x&=14.0\cos59.0^\circ\\ &\approx7.20\ \mathrm{m\,s^{-1}}. \end{aligned} \]

The time to reach \(x=12.0\ \mathrm{m}\) is

\[ \begin{aligned} t&=\frac{12.0}{u_x}\\ &\approx1.67\ \mathrm{s}. \end{aligned} \]

The vertical velocity at that instant is

\[ v_y=u\sin\theta-gt. \]

Using the unrounded angle and time,

\[ \begin{aligned} v_y&=14.0\sin59.0^\circ\\ &\quad-9.80(1.666)\\ &\approx-4.32\ \mathrm{m\,s^{-1}}. \end{aligned} \]

The negative sign means the ball is descending.

So the ball is not at its own maximum height when it passes through the highest reachable target position.

This is the conceptual trap.

The greatest height a particular projectile reaches anywhere along its path is not the same as the greatest height any projectile can reach at a specified horizontal distance.

An even steeper launch might achieve a greater maximum height overall, but it could lose too much horizontal velocity to remain sufficiently high by the time it reaches the target’s horizontal position.

Similarly, \(45^\circ\) maximises horizontal range for equal launch and landing heights. It does not maximise vertical height at \(x=12.0\ \mathrm{m}\).

In fact, a \(45^\circ\) launch here reaches only

\[ \begin{aligned} y&=12.0\tan45^\circ\\ &\quad-3.60(1+\tan^245^\circ)\\ &=4.80\ \mathrm{m} \end{aligned} \]

at the target’s horizontal position.

The \(59.0^\circ\) launch reaches \(6.40\ \mathrm{m}\) there, even though it has a shorter level-ground range than the \(45^\circ\) launch.

The governing principle is always the same: horizontal motion determines when the projectile reaches a particular horizontal position, while vertical motion determines its height at that instant.

Complementary angles are one special result of combining those motions when the launch and landing heights are equal. For an elevated target, a moving target, or a different landing height, the separate motion equations tell you which trajectories are actually possible.