How to Find the Minimum Launch Speed to Clear a Wall

A ball might rise higher than a wall and still hit it. The problem is where it reaches that height. If the ball is already descending when it arrives at the wall, its maximum height tells you very little about whether it will get over.

To find the minimum launch speed needed to clear a wall, calculate the projectile’s height at the wall’s horizontal position, set that height equal to the wall’s height, and solve for the initial speed. The result is the speed at which the projectile just grazes the top. A slightly greater speed provides clearance, assuming the launch angle stays fixed.

Here’s something worth predicting before we calculate anything. If the launcher points too low, can you always make the ball clear the wall by increasing its speed?

No. A faster ball spends less time falling under gravity, but it still cannot travel above the straight line in which the launcher points. That gives us an important limit before we even start solving equations.

01Find the height at the wall, not the maximum height

Picture a ball launcher on the left and a vertical wall some distance to the right. The ball leaves the launcher at an angle above the horizontal.

We’ll use these symbols:

Take rightwards as positive horizontally and upwards as positive vertically. Ground level is \(y=0\).

Use \(g=9.80\ \mathrm{m\,s^{-2}}\) unless the question gives another value.

We’ll assume negligible air resistance, constant gravitational acceleration, and a point-like projectile. Initially, the wall has negligible thickness.

Step 1: Separate horizontal and vertical motion

The ball has two components of initial velocity:

\[ u_x=u\cos\theta \]

\[ u_y=u\sin\theta \]

Here, \(u_x\) is the horizontal velocity and \(u_y\) is the initial vertical velocity, both measured in \(\mathrm{m\,s^{-1}}\).

Horizontal velocity stays constant in the ideal model. Vertical velocity changes because gravity acts downwards.

Both components share the same elapsed time. That’s the connection we need.

Step 2: Find the time taken to reach the wall

Horizontally, the projectile travels a distance \(d\) at constant speed \(u\cos\theta\).

Using distance = speed × time:

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

Rearrange:

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

This is the time at which the ball reaches the wall’s horizontal position.

We don’t need the total flight time. The ball might hit the ground much later.

Step 3: Calculate the vertical height at that time

The vertical displacement equation is:

\[ y-y_0=u_yt-\frac12gt^2 \]

Substitute \(u_y=u\sin\theta\):

\[ y=y_0+(u\sin\theta)t-\frac12gt^2 \]

Now substitute the time needed to reach the wall:

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

After simplifying:

\[ \boxed{\begin{aligned} y(d)&=y_0+d\tan\theta\\ &\quad-\frac{gd^2}{2u^2\cos^2\theta} \end{aligned}} \]

This gives the ball’s vertical position at the wall.

The expression has three parts:

• \(y_0\): where the ball starts.
• \(d\tan\theta\): how much height the straight launch direction would gain over the horizontal distance \(d\).
• \(\frac{gd^2}{2u^2\cos^2\theta}\): the amount gravity pulls the ball below that straight line.

The first two terms describe where the ball would be if gravity did not act. The last term corrects that prediction.

For a known launch speed, substitute the values and compare \(y(d)\) with \(h\). If \(y(d)>h\), the ideal projectile clears a thin wall.

For an unknown minimum speed, we do the reverse.

Step 4: Set the ball’s height equal to the wall’s height

At the theoretical minimum speed, the projectile just reaches the wall top:

\[ y(d)=h \]

Therefore:

\[ h=y_0+d\tan\theta- \frac{gd^2}{2u^2\cos^2\theta} \]

Rearrange:

\[ y_0+d\tan\theta-h =\frac{gd^2}{2u^2\cos^2\theta} \]

To keep the final expression readable, let:

\[ A=y_0+d\tan\theta-h \]

Here, \(A\) is the vertical gap between the straight launch-direction line and the wall top, measured in metres.

Solving for the threshold speed gives:

\[ \boxed{ u_{\text{min}}= \sqrt{\frac{gd^2}{2A\cos^2\theta}} } \]

This formula applies when \(A>0\), \(d>0\), and the projectile is launched towards the wall with \(0^\circ\leq\theta<90^\circ\).

The quantity \(u_{\text{min}}\) is the theoretical grazing threshold. For strict clearance, use \(u>u_{\text{min}}\).

02Before calculating: can the launch direction reach above the wall?

The quantity \(A\) provides a useful decision rule.

Why is \(A=0\) not enough?

Even at a very high launch speed, gravity causes some downward displacement during the non-zero travel time. The ball approaches the straight launch-direction line as speed increases, but never reaches that line at a wall a positive distance away.

For example, suppose a ball leaves ground level at \(20^\circ\), towards a wall \(10.0\ \mathrm{m}\) away and \(4.00\ \mathrm{m}\) high.

The straight launch direction reaches:

\[ \begin{aligned} y_{\text{line}} &=10.0\tan20^\circ\\ &=3.64\ \mathrm{m} \end{aligned} \]

Even without gravity, the ball would be below the top.

Increasing speed cannot fix this launch angle. The launcher must point higher, or the geometry must change.

This is the first calculation you should make when a question asks whether a certain wall can be cleared at a fixed angle. There’s no point solving for a speed if the launch direction itself is too low.

03How could you test the prediction experimentally?

A range measurement alone won’t settle whether a projectile clears a particular wall.

Imagine building a test with an adjustable ball launcher, a pair of light gates, a marked horizontal distance, and a vertical measurement screen.

The original test: measure where the ball lands

Suppose we release a ball from the launcher and measure its horizontal range using the landing mark on the ground.

Would a landing point beyond the wall’s position prove that the ball could clear the wall?

No. Two launches can have the same range but follow very different arcs. One may pass high above an obstacle while another passes below its top.

The landing mark gives us a final horizontal position. It does not tell us the ball’s height at every earlier position.

The revised test: measure height where the wall would be

Keep the launcher fixed at a known angle and height.

Place two light gates close to the muzzle to estimate launch speed. Each gate detects when the ball interrupts its light beam, and a timer measures the interval between those detections.

For an illustrative estimate, suppose the gates are \(0.200\ \mathrm{m}\) apart and the time between interruptions is \(0.0160\ \mathrm{s}\).

Then:

\[ \begin{aligned} u&\approx\frac{0.200}{0.0160}\\ &\approx12.5\ \mathrm{m\,s^{-1}} \end{aligned} \]

This treats the ball’s speed as approximately constant over the short gate spacing and assumes that spacing represents the distance travelled between the two detections.

Next, mark the wall’s horizontal position on the ground. Temporarily use a vertical measurement screen or calibrated side-on video rather than a solid wall.

The ball travels from the launcher, through the gates, and then across the marked vertical plane. Record its centre height at that crossing.

Now compare the measured height with the model prediction.

If the launch angle remains fixed, the model predicts that increasing launch speed raises the crossing height, but with progressively smaller gains as the path approaches the straight launch-direction line.

If actual measurements disagreed consistently, you would investigate the assumptions: air resistance, changing launch angle, gate timing, or measurement error.

This describes a proposed test, not observations from a performed experiment. The distinction matters because calculated trajectories are predictions until measurements test them.

06Three checks that prevent most wall-clearance mistakes

1. Don’t use maximum height as the clearance test

A ball can reach a maximum height greater than \(h\) but be lower than \(h\) when it reaches \(x=d\).

The relevant height is always \(y(d)\), not the highest point anywhere on the trajectory.

If a question gives you the launch speed and asks only whether an obstacle is cleared, use the forward calculation described in How to Find a Projectile’s Height at a Horizontal Distance.

2. Don’t use the level-ground range formula automatically

The familiar range equation is:

\[ R=\frac{u^2\sin2\theta}{g} \]

where \(R\) is the horizontal distance between launch and landing, measured in metres.

This form assumes the projectile lands at the same height from which it was launched, with no earlier collision.

Even when those assumptions hold, a range greater than the wall distance does not prove clearance. The ball might pass below the wall top on its way to a landing point further away.

3. Check whether the obstacle has thickness

A thin wall has one relevant horizontal position. A wide, flat-topped obstacle occupies an entire interval of horizontal positions.

The ball must stay above the top throughout that interval.

For a parabolic trajectory under constant gravity, the height curve is concave downwards. Over a closed horizontal interval, its lowest height occurs at one of the two ends.

That means checking both edges is sufficient for an ideal point projectile travelling over a horizontal, flat-topped obstacle. Checking only the near edge is not.

A real ball also has a radius, and real walls may have thickness or uneven tops. The ball’s centre needs additional clearance, and contact can occur before its centre reaches the wall’s face. Unless a question specifies the ball’s size, HSC projectile calculations usually treat it as a point.

07Questions and solutions

These questions progress from finding one threshold speed to situations where several conditions must hold at once. Take \(g=9.80\ \mathrm{m\,s^{-2}}\) throughout, assume negligible air resistance and a point-like projectile, and use strict inequalities when a question says the ball must clear or pass beneath something without touching it.

Question 1

A ball is launched from level ground at \(45.0^\circ\) above the horizontal.

A thin vertical wall is \(8.00\ \mathrm{m}\) away and \(1.20\ \mathrm{m}\) high.

Calculate the theoretical minimum launch speed needed to reach the wall top. Explain whether that exact speed provides clearance.

Solution 1

The theoretical minimum speed is \(9.60\ \mathrm{m\,s^{-1}}\). At exactly this speed, the projectile grazes the wall top, so a slightly greater speed is needed for strict clearance.

First calculate:

\[ \begin{aligned} A&=y_0+d\tan\theta-h\\ &=0+8.00\tan45.0^\circ-1.20\\ &=6.80\ \mathrm{m} \end{aligned} \]

Then:

\[ \begin{aligned} u_{\text{min}}^2 &=\frac{gd^2}{2A\cos^2\theta}\\ &=\frac{9.80(8.00)^2} {2(6.80)\cos^245.0^\circ}\\ &=92.24\ \mathrm{m^2\,s^{-2}} \end{aligned} \]

Therefore:

\[ u_{\text{min}}=9.60\ \mathrm{m\,s^{-1}} \]

The tempting mistake is to treat the calculated threshold as a speed that guarantees some clearance.

But the equation was solved using \(y(d)=h\). That describes a grazing trajectory, not a trajectory passing above the top.

For strict clearance in the ideal model, the speed must exceed \(9.60\ \mathrm{m\,s^{-1}}\).

Question 2

A ball is launched from level ground at \(14.0\ \mathrm{m\,s^{-1}}\).

The launcher can be set to either \(30.0^\circ\) or \(60.0^\circ\) above the horizontal.

A thin wall is \(12.0\ \mathrm{m}\) from the launcher and \(2.30\ \mathrm{m}\) high.

1. Show that the two launch angles produce the same horizontal range on level ground.
2. Determine whether each shot clears the wall at \(12.0\ \mathrm{m}\).
3. The wall is moved to \(8.00\ \mathrm{m}\) from the launcher without changing its height. Determine whether each shot now clears it, and explain the change.

Solution 2

Both angles give a horizontal range of \(17.3\ \mathrm{m}\), but they do not give the same height at the wall. At \(12.0\ \mathrm{m}\), the \(30.0^\circ\) shot fails and the \(60.0^\circ\) shot clears. Moving the wall to \(8.00\ \mathrm{m}\) allows both shots to clear it.

Part 1: Calculate the ranges.

For launch and landing at the same height:

\[ R=\frac{u^2\sin2\theta}{g} \]

For \(30.0^\circ\):

\[ \begin{aligned} R&=\frac{(14.0)^2\sin60.0^\circ}{9.80}\\ &=17.3\ \mathrm{m} \end{aligned} \]

For \(60.0^\circ\):

\[ \begin{aligned} R&=\frac{(14.0)^2\sin120.0^\circ}{9.80}\\ &=17.3\ \mathrm{m} \end{aligned} \]

Since \(\sin60^\circ=\sin120^\circ\), the ranges match.

This is why complementary launch angles can give the same level-ground range.

Part 2: Calculate the heights at \(12.0\ \mathrm{m}\).

For ground-level launch:

\[ y(d)=d\tan\theta- \frac{gd^2}{2u^2\cos^2\theta} \]

At \(30.0^\circ\):

\[ \begin{aligned} y(12.0) &=12.0\tan30.0^\circ\\ &\quad-\frac{9.80(12.0)^2} {2(14.0)^2\cos^230.0^\circ}\\ &=2.13\ \mathrm{m} \end{aligned} \]

Since \(2.13<2.30\), this ball hits the wall.

At \(60.0^\circ\):

\[ \begin{aligned} y(12.0) &=12.0\tan60.0^\circ\\ &\quad-\frac{9.80(12.0)^2} {2(14.0)^2\cos^260.0^\circ}\\ &=6.38\ \mathrm{m} \end{aligned} \]

This ball clears the wall comfortably.

Part 3: Move the wall to \(8.00\ \mathrm{m}\).

Using the same height equation with \(d=8.00\ \mathrm{m}\):

For \(30.0^\circ\):

\[ y(8.00)=2.49\ \mathrm{m} \]

For \(60.0^\circ\):

\[ y(8.00)=7.46\ \mathrm{m} \]

Both heights exceed \(2.30\ \mathrm{m}\), so both shots now clear the wall.

Why did the lower-angle shot improve when the wall moved closer?

Its maximum height is:

\[ \begin{aligned} H_{\text{max}} &=\frac{(u\sin\theta)^2}{2g}\\ &=\frac{(14.0\sin30.0^\circ)^2} {2(9.80)}\\ &=2.50\ \mathrm{m} \end{aligned} \]

Here, \(H_{\text{max}}\) is the maximum height above the launch level.

The lower trajectory can rise above a \(2.30\ \mathrm{m}\) wall, but only over a limited part of its flight. At \(12.0\ \mathrm{m}\), it has already dropped below that height. At \(8.00\ \mathrm{m}\), it is still sufficiently high.

The trap is assuming that equal horizontal ranges mean equal trajectories. They don’t. The vertical position at the obstacle is the deciding measurement.

Question 3

A ball is launched from level ground at \(14.0\ \mathrm{m\,s^{-1}}\), at \(45.0^\circ\) above the horizontal.

It approaches a solid rectangular obstacle with a flat horizontal top \(4.00\ \mathrm{m}\) above the ground.

The obstacle begins at \(x=10.0\ \mathrm{m}\) and ends at \(x=16.0\ \mathrm{m}\), measured from the launch point.

1. Does the ball pass above the near edge? Does it clear the entire obstacle?
2. Assuming the ball follows its ideal trajectory until first contact, calculate where it first touches the top after passing the near edge.
3. Find the theoretical minimum launch speed needed to pass over the entire obstacle without entering it, keeping the angle at \(45.0^\circ\).

Solution 3

The ball passes above the near edge but strikes the obstacle’s top at approximately \(x=14.47\ \mathrm{m}\). The theoretical grazing threshold for clearing the entire obstacle is \(14.46\ \mathrm{m\,s^{-1}}\).

Part 1: Calculate the height at both edges.

At \(45.0^\circ\), we have:

\[ \tan45^\circ=1 \]

and:

\[ \cos^245^\circ=\frac12 \]

Therefore the trajectory becomes:

\[ y(x)=x-\frac{g}{u^2}x^2 \]

For \(u=14.0\ \mathrm{m\,s^{-1}}\):

\[ \frac{g}{u^2} =\frac{9.80}{(14.0)^2} =0.0500\ \mathrm{m^{-1}} \]

So:

\[ y(x)=x-(0.0500\ \mathrm{m^{-1}})x^2 \]

At the near edge:

\[ \begin{aligned} y(10.0)&=10.0-0.0500(10.0)^2\\ &=5.00\ \mathrm{m} \end{aligned} \]

The ball is \(1.00\ \mathrm{m}\) above the top.

At the far edge:

\[ \begin{aligned} y(16.0)&=16.0-0.0500(16.0)^2\\ &=3.20\ \mathrm{m} \end{aligned} \]

The ideal trajectory is \(0.80\ \mathrm{m}\) below the top at the far edge.

Therefore the ball cannot clear the entire obstacle.

Part 2: Locate the collision.

After entering above the near edge, the ideal trajectory eventually descends to the height \(y=4.00\ \mathrm{m}\).

Set:

\[ 4.00=x-0.0500x^2 \]

Using metres for \(x\), rearrange:

\[ x^2-20x+80=0 \]

Solve the quadratic:

\[ \begin{aligned} x&=\frac{20\pm\sqrt{400-320}}{2}\\ &=10\pm\sqrt{20} \end{aligned} \]

The two positions are:

\[ x=5.53\ \mathrm{m} \]

and:

\[ x=14.47\ \mathrm{m} \]

The first position is where the projectile rises through the \(4.00\ \mathrm{m}\) height, before reaching the obstacle.

The second position is where it descends through that height while over the obstacle.

So first contact with the obstacle’s top occurs at:

\[ x=14.47\ \mathrm{m} \]

The projectile’s motion would then change because of the collision; we cannot continue treating it as an undisturbed projectile after that point.

Part 3: Find the speed needed to clear the full width.

At \(45.0^\circ\), the required speed threshold at horizontal position \(d\) is:

\[ u_{\text{min}}^2=\frac{gd^2}{d-h} \]

The far edge is at \(d=16.0\ \mathrm{m}\), so:

\[ \begin{aligned} u_{\text{min}}^2 &=\frac{9.80(16.0)^2}{16.0-4.00}\\ &=209.1\ \mathrm{m^2\,s^{-2}} \end{aligned} \]

Thus:

\[ u_{\text{min}}=14.46\ \mathrm{m\,s^{-1}} \]

At this speed, the predicted height at the near edge is about \(5.31\ \mathrm{m}\), while the height at the far edge is \(4.00\ \mathrm{m}\).

Because the trajectory is concave downwards, it cannot dip below both endpoint heights in between. The far-edge grazing condition therefore controls the threshold for the whole flat-topped obstacle.

A speed greater than \(14.46\ \mathrm{m\,s^{-1}}\) is required for strict clearance.

The tempting shortcut was checking only the near face. That works for a negligibly thin wall, but not for a wide obstacle.

Question 4

A ball is launched from a platform \(1.00\ \mathrm{m}\) above level ground. Its launch speed and angle are unknown.

Two measurements of its centre height are provided:

Treat these as exact illustrative data consistent with the ideal projectile model, not actual experimental readings.

A thin wall \(3.80\ \mathrm{m}\) high stands at \(x=12.0\ \mathrm{m}\).

1. Find the equation of the projectile’s path in the form \(y(x)=y_0+px-qx^2\), where \(p\) and \(q\) are constants.
2. Determine the initial horizontal and vertical velocity components, the launch speed, and the launch angle.
3. Decide whether the ball clears the wall, and explain why extending a straight line through the two measured positions would give a misleading answer.

Solution 4

The ball does not clear the wall. Its predicted height at \(x=12.0\ \mathrm{m}\) is \(3.40\ \mathrm{m}\), which is \(0.40\ \mathrm{m}\) below the top. The inferred launch speed is \(12.68\ \mathrm{m\,s^{-1}}\) at approximately \(38.7^\circ\).

Part 1: Determine the trajectory equation.

The general path equation is:

\[ y(x)=y_0+px-qx^2 \]

where:

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

and:

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

The coefficient \(p\) is dimensionless, while \(q\) has units of \(\mathrm{m^{-1}}\).

Using numerical distances in metres, with \(y_0=1.00\ \mathrm{m}\), the measurements give:

\[ \begin{aligned} 4p-16q&=2.40\\ 8p-64q&=3.20 \end{aligned} \]

Double the first equation:

\[ 8p-32q=4.80 \]

Subtract the second equation:

\[ 32q=1.60 \]

Therefore:

\[ q=0.0500\ \mathrm{m^{-1}} \]

Substituting back:

\[ \begin{aligned} 4p-16(0.0500)&=2.40\\ p&=0.800 \end{aligned} \]

The trajectory equation is:

\[ \boxed{ y(x)=1.00+0.800x-0.0500x^2 } \]

Here, \(x\) and \(y\) are numerical values in metres.

Part 2: Find the launch velocity.

Since:

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

we have:

\[ u_x=\sqrt{\frac{g}{2q}} \]

Substituting:

\[ \begin{aligned} u_x&=\sqrt{\frac{9.80}{2(0.0500)}}\\ &=9.90\ \mathrm{m\,s^{-1}} \end{aligned} \]

From \(p=u_y/u_x\):

\[ \begin{aligned} u_y&=pu_x\\ &=0.800(9.90)\\ &=7.92\ \mathrm{m\,s^{-1}} \end{aligned} \]

The launch speed is:

\[ \begin{aligned} u&=\sqrt{u_x^2+u_y^2}\\ &=12.68\ \mathrm{m\,s^{-1}} \end{aligned} \]

The angle is:

\[ \begin{aligned} \theta&=\tan^{-1}\left(\frac{u_y}{u_x}\right)\\ &=\tan^{-1}(0.800)\\ &=38.7^\circ \end{aligned} \]

Part 3: Test the wall.

Substitute \(x=12.0\ \mathrm{m}\):

\[ \begin{aligned} y(12.0) &=1.00+0.800(12.0)\\ &\quad-0.0500(12.0)^2\\ &=3.40\ \mathrm{m} \end{aligned} \]

The wall is \(3.80\ \mathrm{m}\) high, so the projectile is \(0.40\ \mathrm{m}\) too low.

Why not just continue the line between the two measured positions?

Between \(x=4.00\ \mathrm{m}\) and \(x=8.00\ \mathrm{m}\), the height rises by \(0.80\ \mathrm{m}\). A straight-line extrapolation would predict another \(0.80\ \mathrm{m}\) rise over the next \(4.00\ \mathrm{m}\), giving \(y=5.00\ \mathrm{m}\).

That would incorrectly predict clearance.

The missing constraint is gravity. The ball’s vertical velocity is continually decreasing, so successive equal horizontal intervals do not produce equal vertical changes.

In fact, the inferred trajectory has its maximum height at \(x=8.00\ \mathrm{m}\). The ball is already descending towards the wall.

Question 5

A launcher stands on a flat field. A ball is launched at an adjustable speed and angle and must land at a marked point on the same level ground.

There are two obstacles:

• A negligibly thin vertical wall at \(x=6.00\ \mathrm{m}\), with its top \(5.00\ \mathrm{m}\) above ground. The ball must pass strictly above it.
• A narrow overhead beam at \(x=12.0\ \mathrm{m}\), with its underside \(5.50\ \mathrm{m}\) above ground. The ball must pass strictly below it while airborne. Assume the beam’s supports do not obstruct the flight.

Initially, the landing marker is exactly \(20.0\ \mathrm{m}\) from the launcher.

1. Is there any combination of launch speed and angle that clears the wall, passes below the beam, and lands exactly at \(20.0\ \mathrm{m}\)? Justify your conclusion.
2. The landing marker is moved to \(18.0\ \mathrm{m}\), with both obstacles unchanged. Determine the full interval of permissible launch angles.
3. Find the limiting lower launch speed for the new arrangement. Does any strictly clearing trajectory actually achieve that limiting speed? Explain.

Solution 5

No launch can satisfy all three conditions with the landing marker at \(20.0\ \mathrm{m}\). Moving the marker to \(18.0\ \mathrm{m}\) makes the problem possible, but only for launch angles strictly between approximately \(51.34^\circ\) and \(53.97^\circ\). The lower speed limit is \(13.45\ \mathrm{m\,s^{-1}}\), and that exact limit does not strictly clear the wall.

The surprising constraint is that fixing the landing range also constrains the heights at both obstacles. You cannot independently adjust the height at one obstacle without affecting the other.

Part 1: Express the trajectory using its required range.

For launch and landing at the same height:

\[ R=\frac{u^2\sin2\theta}{g} \]

We also know:

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

Using the range relationship:

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

Therefore:

\[ \boxed{ y(x)=\frac{x(R-x)}{R}\tan\theta } \]

This equation describes the path of a projectile that lands exactly at horizontal distance \(R\).

It also reveals something important. Once \(R\) is fixed, the heights at different horizontal positions depend on the same value of \(\tan\theta\).

We cannot choose them separately.

Test the original landing distance: \(R=20.0\ \mathrm{m}\).

At the wall, \(x=6.00\ \mathrm{m}\):

\[ \begin{aligned} y(6.00) &=\frac{6.00(20.0-6.00)}{20.0} \tan\theta\\ &=4.20\tan\theta \end{aligned} \]

The ball must pass above \(5.00\ \mathrm{m}\):

\[ 4.20\tan\theta>5.00 \]

Thus:

\[ \tan\theta>1.1905 \]

Now consider the overhead beam at \(x=12.0\ \mathrm{m}\):

\[ \begin{aligned} y(12.0) &=\frac{12.0(20.0-12.0)}{20.0} \tan\theta\\ &=4.80\tan\theta \end{aligned} \]

To pass below the beam:

\[ 4.80\tan\theta<5.50 \]

Therefore:

\[ \tan\theta<1.1458 \]

These two requirements contradict each other:

\[ \tan\theta>1.1905 \]

and:

\[ \tan\theta<1.1458 \]

No angle can satisfy both.

Consequently, no launch speed can satisfy all the original conditions either.

A faster launch at the same angle would change the landing range. If we adjust the angle to keep the landing range at \(20.0\ \mathrm{m}\), we are still trapped by the incompatible height requirements.

Part 2: Move the landing marker to \(R=18.0\ \mathrm{m}\).

Recalculate the heights at the two obstacles.

At \(x=6.00\ \mathrm{m}\):

\[ \begin{aligned} y(6.00) &=\frac{6.00(18.0-6.00)}{18.0} \tan\theta\\ &=4.00\tan\theta \end{aligned} \]

At \(x=12.0\ \mathrm{m}\):

\[ \begin{aligned} y(12.0) &=\frac{12.0(18.0-12.0)}{18.0} \tan\theta\\ &=4.00\tan\theta \end{aligned} \]

The heights are now equal.

This isn’t a coincidence. The projectile reaches its maximum height halfway through a level-ground flight, at \(R/2=9.00\ \mathrm{m}\).

The wall at \(x=6.00\ \mathrm{m}\) and the beam at \(x=12.0\ \mathrm{m}\) are equally far from that midpoint. A parabolic trajectory is symmetric about its highest point.

The obstacle requirements are now:

\[ 4.00\tan\theta>5.00 \]

and:

\[ 4.00\tan\theta<5.50 \]

Combining them:

\[ 5.00<4.00\tan\theta<5.50 \]

So:

\[ 1.250<\tan\theta<1.375 \]

Taking inverse tangents:

\[ \boxed{ 51.34^\circ<\theta<53.97^\circ } \]

Every angle in this open interval gives a geometrically acceptable trajectory, provided the speed is chosen to make the ball land at \(18.0\ \mathrm{m}\).

Part 3: Find the limiting lower speed.

We need the ball to land at \(R=18.0\ \mathrm{m}\):

\[ R=\frac{u^2\sin2\theta}{g} \]

Therefore:

\[ u=\sqrt{\frac{gR}{\sin2\theta}} \]

For a fixed range, the required speed is smallest when \(\sin2\theta\) is largest.

The permitted angles are all greater than \(45^\circ\). Over this interval, \(\sin2\theta\) decreases as \(\theta\) increases.

So the lowest possible speed is approached at the smallest permitted angle:

\[ \tan\theta\to1.250 \]

Using the identity:

\[ \sin2\theta= \frac{2\tan\theta}{1+\tan^2\theta} \]

we get:

\[ \begin{aligned} \sin2\theta &\to\frac{2(1.250)}{1+(1.250)^2}\\ &=0.97561 \end{aligned} \]

Now calculate the limiting speed:

\[ \begin{aligned} u_{\text{limit}} &=\sqrt{\frac{9.80(18.0)}{0.97561}}\\ &=13.45\ \mathrm{m\,s^{-1}} \end{aligned} \]

But at this precise limiting angle:

\[ y(6.00)=4.00(1.250)=5.00\ \mathrm{m} \]

The ball touches the wall top.

That means the limiting trajectory does not strictly clear the wall. There is no smallest speed that achieves strict clearance: the required speed can be made arbitrarily close to \(13.45\ \mathrm{m\,s^{-1}}\) from above, while remaining within the permitted angle interval.

If grazing the wall were allowed, \(13.45\ \mathrm{m\,s^{-1}}\) would be an attained theoretical minimum. With strict clearance, it is only the lower bound.

The first instinct might be to keep increasing speed until the ball clears the wall. But the exact landing requirement prevents that from being an independent adjustment.

At a fixed landing range, the heights at the two obstacles are linked by the shape of the parabola. In the original arrangement, those linked heights make success impossible. Moving the landing position changes the geometry enough for a solution to exist.

That’s the wider lesson behind wall-clearance problems: first identify what must happen at each horizontal position, then determine which quantities are actually free to change. A single wall usually gives a minimum-speed threshold. A wide obstacle, a second height restriction, or a fixed landing point can turn the same two motion equations into a much more demanding problem.