How to Find Two Launch Angles That Hit the Same Target

A ball launcher fires towards a small target 16.0 m away. The ball leaves at \(14.0\ \mathrm{m\,s^{-1}}\), and the target is exactly level with the launch point. You find one angle that works. Could a much steeper launch hit the same target?

Yes. Ignoring air resistance, both \(26.6^\circ\) and \(63.4^\circ\) work. The shallow shot travels horizontally faster but spends less time in the air. The steep shot travels horizontally slower but stays airborne longer.

For a fixed launch speed, the two angles that reach a target at the same height are complementary: they add to \(90^\circ\). But that gives you only the relationship between the angles. To calculate their actual values, you also need the horizontal distance and launch speed.

01The two-angle rule

For a projectile launched and received at the same height, the horizontal range is

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

where:

• \(R\) is the horizontal distance to the target, in metres (\(\mathrm{m}\)).
• \(u\) is the initial launch speed, in metres per second (\(\mathrm{m\,s^{-1}}\)).
• \(\theta\) is the launch angle above the horizontal.
• \(g\) is the magnitude of gravitational acceleration, taken as \(9.8\ \mathrm{m\,s^{-2}}\) unless stated otherwise.

Rearrange to isolate the angle:

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

To find both launch angles, calculate

\[ s=\frac{gR}{u^2} \]

Then use

\[ \theta_{\text{low}} =\frac12\sin^{-1}(s) \]

and

\[ \theta_{\text{high}} =90^\circ-\theta_{\text{low}} \]

Use degree mode on your calculator.

The inverse sine calculation gives the smaller launch angle. The second angle comes from the fact that \(\sin A=\sin(180^\circ-A)\).

There are three possible outcomes:

These conditions assume a positive horizontal distance, an upward launch angle between \(0^\circ\) and \(90^\circ\), constant gravity, negligible air resistance, and equal launch and target heights.

If \(s>1\), your calculator isn’t the problem. The target is beyond the maximum possible same-height range at that speed.

02Why two different angles can reach the same target

Imagine throwing a tennis ball almost horizontally. It moves quickly towards its target but falls before travelling very far.

Now throw it almost vertically. It stays in the air much longer, but hardly moves sideways.

Somewhere between these extremes, there’s a balance between horizontal speed and flight time. That balance explains why two different launch angles can produce the same horizontal distance.

Separate the horizontal and vertical motion

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

At launch, the velocity has two components:

\[ \begin{aligned} u_x&=u\cos\theta\\ u_y&=u\sin\theta \end{aligned} \]

Here, \(u_x\) and \(u_y\) are the horizontal and vertical launch velocities, both measured in \(\mathrm{m\,s^{-1}}\).

Gravity accelerates the ball downwards, so \(a_y=-g\). In the ideal projectile model, horizontal acceleration is zero.

After \(t\) seconds, the horizontal displacement is

\[ x=(u\cos\theta)t \]

The vertical displacement from the launch point is

\[ \Delta y=(u\sin\theta)t-\frac12gt^2 \]

Time \(t\), measured in seconds (\(\mathrm{s}\)), connects the two motions. The same ball is moving horizontally and vertically at the same time.

Use the equal-height condition

If the target is at the same height as the launch point, then its vertical displacement is zero:

\[ 0=(u\sin\theta)T-\frac12gT^2 \]

Here, \(T\) is the time taken to reach the target.

There are two mathematical solutions. \(T=0\) describes the launch instant, not arrival at the distant target.

For the later return to launch height, divide by the non-zero time:

\[ 0=u\sin\theta-\frac12gT \]

Therefore,

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

This is why steeper launches stay airborne longer when they return to the same height.

Now use horizontal motion:

\[ R=(u\cos\theta)T \]

Substitute the flight time:

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

The double-angle identity

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

gives

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

The equation contains both effects: horizontal speed and available flight time.

Why the answers are complementary

Suppose the smaller angle is \(30^\circ\). Then

\[ \sin(2\times30^\circ)=\sin60^\circ \]

For a \(60^\circ\) launch,

\[ \sin(2\times60^\circ)=\sin120^\circ \]

These sine values are equal.

More generally, replacing \(\theta\) with \(90^\circ-\theta\) leaves \(\sin(2\theta)\) unchanged.

The result is equal range, not equal flight time or equal maximum height.

03How could an experiment test the prediction?

Suppose you want to investigate the two-angle result using an adjustable ball launcher.

The simple design would be to set the launcher to two calculated angles and record whether the ball reaches a receiver.

But there’s a problem: the equation assumes the same launch speed at both angles. An adjustable launcher might not preserve that speed perfectly.

A better experiment needs to measure it.

Apparatus and measurement

Position a launcher on a fixed stand at the left of an outdoor test area.

Immediately beyond the barrel, place two light gates a known short distance apart, aligned with the direction of launch. The time taken to move between them gives an estimate of the ball’s launch speed.

Further to the right, place a small receiver at a measured horizontal distance, with its centre level with the muzzle. A camera positioned perpendicular to the flight plane can record the ball’s position at successive times.

Run the investigation in this order:

1. Measure the horizontal separation and confirm that the launch point and receiver centre have equal heights.
2. Set the first calculated angle, measure the launch speed, and record the trajectory and arrival time.
3. Repeat at the second angle, checking that the launch speed has not changed.
4. Compare the recorded arrival positions, maximum heights, and travel times with the predictions.

For the earlier \(14.0\ \mathrm{m\,s^{-1}}\) example, the ideal model predicts two hits at 16.0 m, with arrival times of about 1.28 s and 2.56 s.

If a controlled test produced these two trajectories, the results would support the model.

If one angle missed, that observation alone would not disprove the complementary-angle rule. You would first check the actual launch speeds, the relative heights, air resistance, and measurement uncertainty.

The photogates address the most important flaw in the first design: assuming that changing the barrel angle leaves the launch speed unchanged. They do not eliminate air resistance or other experimental errors.

The broader relationship between angle, height, flight time, and range is explained in Projectile Launch Angle: Height, Flight Time and Range.

04What changes when the target is at a different height?

Now move the target upwards.

The projectile must travel horizontally to the target and arrive at a positive vertical displacement. The flight is no longer symmetric about launch height.

Would the two angles still add to \(90^\circ\)?

Not generally.

The complementary-angle rule relies on the projectile returning to its original height. Once the target is above or below the launch point, you need the separate horizontal and vertical equations again.

Importantly, the condition is about the relative heights of the launch point and target, not whether the launcher happens to be on the ground.

A launcher and target both 1.20 m above the ground are still a same-height problem. A launcher 1.20 m above the ground and a target 2.70 m above the ground are not.

The trajectory equation handles the height difference

Start with

\[ x=(u\cos\theta)t \]

Rearrange to find the time required to reach horizontal position \(x\):

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

Substitute this into the vertical equation:

\[ \Delta y=(u\sin\theta)t-\frac12gt^2 \]

After simplifying,

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

This tells us the projectile’s vertical displacement at any chosen horizontal position.

At the target, set \(x=R\), and let

\[ K=\frac{gR^2}{2u^2} \]

Here, \(K\) has units of metres. It represents the gravitational-drop term for a horizontal launch travelling the same horizontal distance at speed \(u\).

Writing \(q=\tan\theta\), the target equation becomes

\[ Kq^2-Rq+(K+\Delta y)=0 \]

This is a quadratic equation in \(q\).

Solve for \(q\), then calculate each possible angle using

\[ \theta=\tan^{-1}(q) \]

Because we have kept the signed displacement \(\Delta y\), the method works for targets above or below the launch point. It can also produce a negative angle, meaning a launch below the horizontal.

05Which method should you use?

Before calculating any angle, identify the target’s vertical position relative to the launcher.

Once you’ve found an angle, a separate question may remain: does its trajectory actually clear an intervening wall, net, or roof?

That’s when the method in How to Find a Projectile’s Height at a Horizontal Distance becomes particularly useful.

06Questions and solutions

Assume negligible air resistance, constant gravitational acceleration \(g=9.8\ \mathrm{m\,s^{-2}}\), and point-like projectiles unless stated otherwise. All angles are measured from the positive horizontal direction.

Question 1

A ball is launched at \(20.0\ \mathrm{m\,s^{-1}}\) towards a target 30.0 m away at exactly the same height.

Calculate both possible upward launch angles. Explain why the result contains two angles rather than one.

Solution 1

The two launch angles are \(23.7^\circ\) and \(66.3^\circ\) above the horizontal.

Use the same-height range equation:

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

Substitute:

\[ \begin{aligned} \sin(2\theta) &=\frac{9.8(30.0)}{20.0^2}\\ &=0.735 \end{aligned} \]

The smaller angle is

\[ \begin{aligned} \theta_1&=\frac12\sin^{-1}(0.735)\\ &=23.65^\circ \end{aligned} \]

The larger angle is

\[ \begin{aligned} \theta_2&=90^\circ-23.65^\circ\\ &=66.35^\circ \end{aligned} \]

Both angles work because the greater horizontal speed of the shallow shot compensates for its shorter flight time, while the steep shot has less horizontal speed but more time in the air.

A \(45^\circ\) launch would not hit this target at its own height. It would have a greater range:

\[ \begin{aligned} R_{\max}&=\frac{u^2}{g}\\ &=\frac{20.0^2}{9.8}\\ &=40.8\ \mathrm{m} \end{aligned} \]

It would therefore pass above the target’s position and land beyond it if unobstructed.

Question 2

A launcher fires a ball at \(21.0\ \mathrm{m\,s^{-1}}\) towards a target 36.0 m away at the same height.

A position sensor 9.00 m horizontally from the launcher records the ball’s centre at 3.375 m above launch height.

After that shot, a vertical barrier is placed 27.0 m from the launcher. It extends from below the flight path up to 4.00 m above launch height.

Determine:

• Which of the two target-hitting angles produced the recorded shot.
• Whether that shot would clear the new barrier.
• Whether switching to the other target-hitting angle would clear the barrier and still reach the target.
• What height the sensor would then record.

Solution 2

The recorded shot used the lower angle, \(26.6^\circ\). It would strike the new barrier. Switching to \(63.4^\circ\) would clear the barrier and still reach the target, but the sensor would then record a much greater height of \(13.5\ \mathrm{m}\).

First, calculate the two target-hitting angles:

\[ \begin{aligned} \sin(2\theta) &=\frac{9.8(36.0)}{21.0^2}\\ &=0.800 \end{aligned} \]

Therefore,

\[ \theta_1=26.565^\circ \]

and

\[ \theta_2=63.435^\circ \]

Their tangent values are

\[ \tan\theta_1=0.500 \]

and

\[ \tan\theta_2=2.00 \]

We need the height of each trajectory at specific horizontal positions.

Use

\[ y(x)=xq-\frac{gx^2}{2u^2}(1+q^2) \]

where \(q=\tan\theta\), and \(y(x)\) is height relative to the launch point.

At the sensor, \(x=9.00\ \mathrm{m}\):

\[ \frac{gx^2}{2u^2} =\frac{9.8(9.00)^2}{2(21.0)^2} =0.900\ \mathrm{m} \]

For the low-angle shot,

\[ \begin{aligned} y_1(9)&=9(0.5)-0.9(1+0.5^2)\\ &=3.375\ \mathrm{m} \end{aligned} \]

For the high-angle shot,

\[ \begin{aligned} y_2(9)&=9(2)-0.9(1+2^2)\\ &=13.5\ \mathrm{m} \end{aligned} \]

The observed sensor reading therefore identifies the low-angle trajectory.

Now check the barrier at \(x=27.0\ \mathrm{m}\).

The gravitational term is

\[ \frac{9.8(27.0)^2}{2(21.0)^2} =8.10\ \mathrm{m} \]

For the low trajectory,

\[ \begin{aligned} y_1(27)&=27(0.5)-8.1(1.25)\\ &=3.375\ \mathrm{m} \end{aligned} \]

That is below the barrier’s 4.00 m top, so the ball would strike it.

For the high trajectory,

\[ \begin{aligned} y_2(27)&=27(2)-8.1(5)\\ &=13.5\ \mathrm{m} \end{aligned} \]

The high trajectory clears the barrier and, without further obstruction, reaches the original target.

The interesting feature is that the two paths have the same height at \(x=9\) and \(x=27\) individually, because these positions are equally spaced on opposite sides of the midpoint of the same-height range.

The tempting approach is to decide using the launch angles alone. But identifying the actual shot, testing its path against an obstacle, and predicting what changes when the angle changes are three different pieces of reasoning.

Question 3

A researcher uses a launcher with an unknown initial speed. Two shots leave the same point and both reach a receiver 14.7 m away at the same height.

The first takes \(1.50\ \mathrm{s}\). The second takes \(2.00\ \mathrm{s}\).

Assuming the two shots have the same initial speed:

• Calculate that common launch speed and both launch angles.
• Show that the two measured flight times satisfy a relationship that depends only on the target distance and gravity.
• On a later trial, the second time is reported as \(2.20\ \mathrm{s}\), while the first time remains \(1.50\ \mathrm{s}\). Explain whether this later pair can describe two distinct equal-speed shots reaching the same-height target under the ideal model.

Treat all recorded times as exact for this question.

Solution 3

The original two shots have the same speed, \(12.25\ \mathrm{m\,s^{-1}}\), despite their different flight times. Their angles are \(36.9^\circ\) and \(53.1^\circ\). The later \(1.50\ \mathrm{s}\) and \(2.20\ \mathrm{s}\) pair is inconsistent with the stated ideal conditions.

Start with horizontal motion:

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

Since the ball returns to its launch height, vertical motion gives

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

For \(T>0\),

\[ u_y=\frac{gT}{2} \]

For the first shot, \(T_1=1.50\ \mathrm{s}\):

\[ \begin{aligned} u_{x1}&=\frac{14.7}{1.50}\\ &=9.80\ \mathrm{m\,s^{-1}}\\ u_{y1}&=\frac{9.8(1.50)}{2}\\ &=7.35\ \mathrm{m\,s^{-1}} \end{aligned} \]

Combine the perpendicular velocity components:

\[ \begin{aligned} u_1&=\sqrt{9.80^2+7.35^2}\\ &=12.25\ \mathrm{m\,s^{-1}} \end{aligned} \]

The angle is

\[ \begin{aligned} \theta_1&=\tan^{-1}\left(\frac{7.35}{9.80}\right)\\ &=36.87^\circ \end{aligned} \]

For the second shot, \(T_2=2.00\ \mathrm{s}\):

\[ \begin{aligned} u_{x2}&=\frac{14.7}{2.00}\\ &=7.35\ \mathrm{m\,s^{-1}}\\ u_{y2}&=\frac{9.8(2.00)}{2}\\ &=9.80\ \mathrm{m\,s^{-1}} \end{aligned} \]

Its speed is

\[ \begin{aligned} u_2&=\sqrt{7.35^2+9.80^2}\\ &=12.25\ \mathrm{m\,s^{-1}} \end{aligned} \]

and its angle is

\[ \begin{aligned} \theta_2&=\tan^{-1}\left(\frac{9.80}{7.35}\right)\\ &=53.13^\circ \end{aligned} \]

The two shots have identical speed magnitudes. Their horizontal and vertical components have simply exchanged values.

Now consider the times.

For complementary angles,

\[ \begin{aligned} T_1&=\frac{2u\sin\theta_1}{g}\\ T_2&=\frac{2u\cos\theta_1}{g} \end{aligned} \]

Multiply:

\[ \begin{aligned} T_1T_2 &=\frac{4u^2}{g^2}\sin\theta_1\cos\theta_1\\ &=\frac{2u^2}{g^2}\sin(2\theta_1)\\ &=\frac{2R}{g} \end{aligned} \]

This gives a useful consistency test:

\[ T_1T_2=\frac{2R}{g} \]

For the measured distance,

\[ \begin{aligned} \frac{2R}{g} &=\frac{2(14.7)}{9.8}\\ &=3.00\ \mathrm{s^2} \end{aligned} \]

The first pair satisfies it:

\[ (1.50)(2.00)=3.00\ \mathrm{s^2} \]

But the later pair gives

\[ (1.50)(2.20)=3.30\ \mathrm{s^2} \]

This cannot describe two distinct equal-speed ideal trajectories reaching the same-height receiver 14.7 m away.

A changed launch speed, a timing error, air resistance, or an incorrect height measurement could explain the discrepancy. The times alone do not establish which cause is responsible.

The tempting assumption is that different flight times automatically mean different launch speeds. They don’t. The first pair demonstrates exactly why: equal total speed can be split into different horizontal and vertical components.

Question 4

A ball is launched from a platform 5.00 m above level ground. A target is on the ground 10.0 m horizontally from the launch point.

The initial speed is \(14.0\ \mathrm{m\,s^{-1}}\).

Find all launch angles between \(-90^\circ\) and \(90^\circ\) that strike the target. Calculate the corresponding flight times.

Explain why restricting the search to upward angles would miss a valid trajectory, and confirm that neither trajectory hits the ground before the target.

Solution 4

There are two valid launch angles: approximately \(-13.3^\circ\) and \(76.7^\circ\). The first is directed below the horizontal, while the second is a steep upward launch. Their flight times are approximately \(0.734\ \mathrm{s}\) and \(3.11\ \mathrm{s}\).

The target is 5.00 m below the launch point, so

\[ \Delta y=-5.00\ \mathrm{m} \]

Use the general trajectory equation rather than the equal-height range shortcut.

First calculate

\[ \begin{aligned} K&=\frac{gR^2}{2u^2}\\ &=\frac{9.8(10.0)^2}{2(14.0)^2}\\ &=2.50\ \mathrm{m} \end{aligned} \]

The quadratic is

\[ Kq^2-Rq+(K+\Delta y)=0 \]

where \(q=\tan\theta\).

Substitution gives

\[ 2.50q^2-10.0q-2.50=0 \]

Divide by 2.50:

\[ q^2-4q-1=0 \]

Solve:

\[ \begin{aligned} q&=\frac{4\pm\sqrt{20}}{2}\\ &=2\pm\sqrt5 \end{aligned} \]

Therefore,

\[ q_1=-0.2361 \]

and

\[ q_2=4.2361 \]

Convert to angles:

\[ \begin{aligned} \theta_1&=\tan^{-1}(-0.2361)\\ &=-13.28^\circ\\[4pt] \theta_2&=\tan^{-1}(4.2361)\\ &=76.72^\circ \end{aligned} \]

The negative angle is physically meaningful. It describes a ball launched downwards towards a target below its starting point.

Now calculate the flight times using horizontal motion:

\[ T=\frac{R}{u\cos\theta} \]

For the downward launch,

\[ \begin{aligned} T_1&=\frac{10.0}{14.0\cos(-13.28^\circ)}\\ &=0.734\ \mathrm{s} \end{aligned} \]

For the steep upward launch,

\[ \begin{aligned} T_2&=\frac{10.0}{14.0\cos76.72^\circ}\\ &=3.11\ \mathrm{s} \end{aligned} \]

Both reach the target with a vertical displacement of \(-5.00\ \mathrm{m}\), as required.

To check whether either hits the ground earlier, consider the full path.

For either angle, the height above ground is a downward-opening parabola in horizontal position \(x\). It begins 5.00 m above ground at \(x=0\) and reaches ground at \(x=10.0\ \mathrm{m}\).

A downward-opening parabola lies above the straight line connecting any two points on it. Here, the line from the launch point to the target remains above ground throughout \(0<x<10.0\ \mathrm{m}\).

Therefore, both trajectories remain above ground until they reach the target.

The mistake would be to insist on two positive complementary angles simply because there are two solutions.

Here, the target is lower than the launch point. The signed vertical displacement changes the quadratic, and one valid launch angle is negative.

Question 5

A launcher fires two projectiles at exactly the same speed, \(14.0\ \mathrm{m\,s^{-1}}\), from the same point.

Both are launched towards the right, with different upward launch angles.

Their initial directions satisfy

\[ \tan\theta_1=1 \]

and

\[ \tan\theta_2=\frac73 \]

The two ideal trajectories eventually pass through the same target point \(P\), somewhere to the right of the launcher. The position of \(P\) is unknown.

A vertical wall stands 9.00 m horizontally from the launcher. Its top is 6.00 m above launch height.

Determine:

• The horizontal distance and vertical displacement of \(P\) relative to the launcher.
• Which trajectory, if either, clears the wall.
• Whether the steeper projectile is ascending or descending when it reaches the wall, and whether the shallower projectile is ascending or descending there.
• Whether the steeper projectile can ever pass below the shallower trajectory at any horizontal position strictly between launch and \(P\).

A student argues that the steeper projectile could eventually be below the shallower one because it might already be falling while the shallower projectile is still rising.

Assess that claim. Use the equations of motion to prove your conclusion, rather than relying on a sketch.

Solution 5

The trajectories meet again at a target 12.0 m horizontally from the launcher and 4.80 m above it. Only the steeper trajectory clears the wall. At the wall, the steeper projectile is already descending and the shallower projectile is still ascending, but the steeper trajectory remains above the shallower one everywhere between launch and the target.

The crucial insight is that the two paths cannot cross and then separate again before reaching \(P\).

We can prove this and locate \(P\) using the trajectory equation.

Let

\[ q_1=\tan\theta_1=1 \]

and

\[ q_2=\tan\theta_2=\frac73 \]

Write the vertical displacement at horizontal position \(x\) as

\[ y(x)=xq-Ax^2(1+q^2) \]

where

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

For the given speed,

\[ \begin{aligned} A&=\frac{9.8}{2(14.0)^2}\\ &=0.0250\ \mathrm{m^{-1}} \end{aligned} \]

The two projectile paths are

\[ y_1(x)=xq_1-Ax^2(1+q_1^2) \]

and

\[ y_2(x)=xq_2-Ax^2(1+q_2^2) \]

At an intersection, their heights are equal.

Therefore,

\[ y_2(x)-y_1(x)=0 \]

Subtract the two equations:

\[ 0=x(q_2-q_1)-Ax^2(q_2^2-q_1^2) \]

Factor the difference of two squares:

\[ 0=x(q_2-q_1) [1-Ax(q_1+q_2)] \]

There are two ways this product can be zero.

The first is \(x=0\), where both trajectories start.

Since \(q_1\ne q_2\), the only other possibility is

\[ 1-Ax(q_1+q_2)=0 \]

Hence, the second intersection occurs at

\[ x=\frac{1}{A(q_1+q_2)} \]

Substitute:

\[ \begin{aligned} x&=\frac{1}{0.0250(1+7/3)}\\ &=12.0\ \mathrm{m} \end{aligned} \]

This locates the target horizontally.

Now substitute \(x=12.0\ \mathrm{m}\) into the shallower path, for which \(q_1=1\):

\[ \begin{aligned} y_P&=12.0(1)-0.0250(12.0)^2(2)\\ &=12.0-7.20\\ &=4.80\ \mathrm{m} \end{aligned} \]

So \(P\) is 4.80 m above the launch point.

Notice that the angles are not complementary: \(\theta_1=45^\circ\), while \(\theta_2\approx66.8^\circ\).

This makes sense because the common target is above launch height. The equal-height rule does not apply.

Now inspect the wall at \(x=9.00\ \mathrm{m}\).

For the shallower path:

\[ \begin{aligned} y_1(9) &=9(1)-0.0250(9)^2(2)\\ &=4.95\ \mathrm{m} \end{aligned} \]

For the steeper path:

\[ \begin{aligned} y_2(9) &=9\left(\frac73\right) -0.0250(9)^2\left(1+\frac{49}{9}\right)\\ &=7.95\ \mathrm{m} \end{aligned} \]

The wall’s top is 6.00 m above launch level.

Therefore:

• The shallower trajectory reaches only 4.95 m and strikes the wall.
• The steeper trajectory reaches 7.95 m and clears it by 1.95 m.

Next, determine whether the projectiles are rising or falling at the wall.

At the highest point of each trajectory, vertical velocity is zero.

The time to the highest point is

\[ t_{\text{top}}=\frac{u\sin\theta}{g} \]

Multiplying by horizontal velocity gives the horizontal position of the apex:

\[ x_{\text{top}} =\frac{u^2}{g} \frac{q}{1+q^2} \]

For the shallower trajectory, \(q_1=1\):

\[ \begin{aligned} x_{\text{top},1} &=\frac{14.0^2}{9.8}\frac{1}{2}\\ &=10.0\ \mathrm{m} \end{aligned} \]

For the steeper trajectory, \(q_2=7/3\):

\[ \begin{aligned} x_{\text{top},2} &=\frac{14.0^2}{9.8} \frac{7/3}{1+49/9}\\ &=7.24\ \mathrm{m} \end{aligned} \]

The wall is at \(x=9.00\ \mathrm{m}\).

So the shallower projectile has not yet reached its apex, while the steeper projectile has already passed its apex.

The student’s proposed explanation now seems plausible: one ball is rising while the other is falling. Could their height ordering have reversed?

No.

Return to the difference between their trajectories:

\[ y_2(x)-y_1(x) =x(q_2-q_1)[1-Ax(q_1+q_2)] \]

We already established that their second intersection is at \(X=12.0\ \mathrm{m}\), so

\[ A(q_1+q_2)=\frac1X \]

Substitute this relationship:

\[ y_2(x)-y_1(x) =(q_2-q_1)x\left(1-\frac{x}{X}\right) \]

Between the launch point and target,

\[ 0<x<X \]

All three factors on the right are positive:

• \(q_2-q_1>0\), because the steeper launch has the larger tangent.
• \(x>0\).
• \(1-x/X>0\), because the projectile has not reached the common target.

Therefore,

\[ y_2(x)-y_1(x)>0 \]

at every horizontal position strictly between launch and target.

The steeper trajectory is always higher over that interval.

The student’s mistake is confusing the direction of vertical motion with vertical position.

A projectile can be descending and still be higher than another projectile that is ascending. What matters is how much height each has already gained or lost, not simply the sign of its current vertical velocity.

In fact, this reasoning gives a stronger result: for two distinct ideal trajectories launched from the same point at the same speed that later meet at a common target to the right, the higher-angle trajectory stays above the lower-angle trajectory until that target.

The familiar complementary-angle rule is one special case, when that common target happens to be at launch height.

That is the next useful decision in a real target problem: finding two angles tells you which paths can reach the target, but comparing the full trajectories tells you which one can actually be used.