How to Find a Projectile’s Height at a Horizontal Distance

A football is kicked towards a \(3.00\ \mathrm{m}\) fence, \(12.0\ \mathrm{m}\) away. It leaves the ground at \(16.0\ \mathrm{m\,s^{-1}}\), at \(30.0^\circ\) above the horizontal. Will it be high enough to clear the fence?

Knowing how far the ball will eventually travel doesn’t answer that question. We need its height at one particular horizontal position.

The method is simple: use horizontal motion to find the time taken to reach the distance, then use vertical motion to calculate the height at that time. If the projectile starts above the ground, include its initial height as well.

01Find the time horizontally, then the height vertically

Picture the ball moving through the air. It travels forwards while gravity continually changes how quickly it moves upwards or downwards.

Under the usual ideal projectile model, the horizontal velocity stays constant. The vertical velocity changes because gravity acts downwards.

Both parts of the motion share the same elapsed time.

Choose rightwards as positive \(x\) and upwards as positive \(y\). Let the ground be \(y=0\), and suppose the projectile moves to the right.

We use:

Near Earth’s surface, take \(g=9.80\ \mathrm{m\,s^{-2}}\), unless the question specifies otherwise.

Step 1: Separate the initial velocity

For a launch at angle \(\theta\) above the horizontal:

\[ u_x=u\cos\theta \]

\[ u_y=u\sin\theta \]

The horizontal component determines how quickly the ball reaches the specified distance. The vertical component tells us how its height changes during that journey.

Step 2: Find the time to reach the distance

There is no horizontal acceleration in the ideal model, so:

\[ x=u_xt \]

Rearranging:

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

Or, using the launch speed and angle:

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

This is the important move. We don’t need to calculate the ball’s total flight time or find when it reaches maximum height.

We only need the time to reach the specified horizontal position.

Step 3: Calculate the vertical position

The vertical acceleration is \(-g\), because we chose upwards as positive.

Using the vertical displacement equation:

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

Therefore:

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

This equation contains three physical contributions:

• \(y_0\) is the height where the projectile starts.
• \(u_yt\) is the vertical displacement it would have without gravity.
• \(\frac12gt^2\) is how far gravity shifts it downwards relative to that gravity-free motion.

For a projectile launched from ground level, \(y_0=0\). For a projectile launched from a balcony or platform, \(y_0\) must be included.

A direct equation if you already know the distance

You can also eliminate time by substituting \(t=x/u_x\):

\[ \begin{aligned} y(x)&=y_0+\frac{u_y}{u_x}x\\ &\quad-\frac{gx^2}{2u_x^2} \end{aligned} \]

This is the projectile’s trajectory equation, giving height directly from horizontal position.

It’s useful for comparing heights at several distances or finding where a trajectory crosses a particular level. For a full explanation of its structure, see how to derive the parabolic projectile path equation.

For a single numerical height question, the two-step time method is often easier to follow.

02How could we measure the height at a particular distance?

Suppose we want to test the calculation rather than just trust the equations.

Imagine an adjustable ball launcher clamped to a bench, with its muzzle \(1.00\ \mathrm{m}\) above the floor. It fires a small ball to the right at \(5.00\ \mathrm{m\,s^{-1}}\), at \(30.0^\circ\) above the horizontal.

Mark \(x=0\) on the floor directly below the muzzle. Place a vertical recording screen \(1.00\ \mathrm{m}\) horizontally from the launch point. A ball striking the screen would leave a mark showing its height.

What should the original setup show?

Predict the impact height before measuring it.

The horizontal velocity is:

\[ u_x=5.00\cos30^\circ =4.33\ \mathrm{m\,s^{-1}} \]

The vertical velocity at launch is:

\[ u_y=5.00\sin30^\circ =2.50\ \mathrm{m\,s^{-1}} \]

The travel time to the screen is:

\[ \begin{aligned} t&=\frac{1.00}{5.00\cos30^\circ}\\ &=0.231\ \mathrm{s} \end{aligned} \]

The predicted height above the floor is:

\[ \begin{aligned} y&=1.00+2.50(0.231)\\ &\quad-4.90(0.231)^2\\ &\approx1.316\ \mathrm{m} \end{aligned} \]

So the ball’s centre should pass the screen approximately \(1.32\ \mathrm{m}\) above the floor, under the ideal assumptions.

Notice that its vertical displacement from launch is only \(0.316\ \mathrm{m}\). That is not its total height above the floor.

Why one impact mark isn’t enough

A mark near \(1.32\ \mathrm{m}\) would be consistent with the calculation. But a single mark can’t establish whether the ball was rising or falling, or whether horizontal velocity remained constant throughout its flight.

The screen also stops the ball. Moving it to another distance requires another launch, possibly with a slightly different initial speed.

A better design uses a calibrated background grid and a stationary camera positioned directly side-on. Video frames from one flight can provide horizontal and vertical positions at known times without interrupting the trajectory.

Here are illustrative predictions for the same \(5.00\ \mathrm{m\,s^{-1}}\), \(30.0^\circ\) launch. These are calculated model values, not experimental measurements.

The predicted horizontal spacing is almost \(0.433\ \mathrm{m}\) for every \(0.10\ \mathrm{s}\). Meanwhile, the height rises and eventually begins to fall.

The ball reaches its maximum height between \(0.20\) and \(0.30\ \mathrm{s}\). That is why its heights at those two times are so similar despite its substantially different horizontal positions.

Real camera measurements could test these predictions. Differences might reveal inconsistent launch speed, air resistance, spin, or measurement errors such as parallax.

The key experimental idea is that a measured horizontal position identifies an elapsed time. That same time connects the horizontal and vertical observations.

03Worked examples: calculating height at a fixed distance

04Three mistakes that change the answer

1. Confusing height with vertical displacement

Compare two projectiles with identical initial velocities. One launches from ground level. The other launches from a platform \(5.00\ \mathrm{m}\) higher.

At the same horizontal distance, their times, vertical displacements from launch, and vertical velocities are identical in the ideal model.

But their heights above the ground differ by \(5.00\ \mathrm{m}\).

The useful distinction is:

\[ \Delta y=y-y_0 \]

Vertical displacement \(\Delta y\) is measured from the launch point. Height \(y\) is measured from the chosen vertical origin.

Never silently replace one with the other.

2. Using maximum-height or range formulas for the wrong question

At a specified horizontal position, the projectile might be rising, at its highest point, or falling.

You cannot assume its vertical velocity is zero unless the position is actually the apex.

Similarly, the familiar same-height range formula tells you where the ball returns to its launch level, not its height at an intermediate horizontal position.

Choose the equation from the information you need, not simply because it looks familiar.

3. Extending the trajectory past an impact

Suppose a calculation predicts \(y=-2.0\ \mathrm{m}\) at \(x=30.0\ \mathrm{m}\), with flat ground at \(y=0\).

Does that mean the ball is two metres underground?

No. The mathematical parabola continues below the ground, but the physical ball must have hit the ground earlier.

A calculated height at a given distance is physically meaningful only if the projectile reaches that position before hitting the ground or another object.

This gives us a reliable decision process:

These methods assume constant downward gravitational acceleration, negligible air resistance, and non-zero horizontal velocity towards the target. A vertical launch needs separate treatment because \(u_x=0\).

05Questions and solutions

The questions increase in difficulty. Questions 2-5 use \(g=10.0\ \mathrm{m\,s^{-2}}\) for simpler arithmetic. Treat projectiles as point particles, and neglect air resistance unless stated otherwise.

Question 1

A ball rolls horizontally off a balcony \(5.00\ \mathrm{m}\) above level ground at \(10.0\ \mathrm{m\,s^{-1}}\).

Calculate:

(a) Its height above the ground at a horizontal distance of \(6.00\ \mathrm{m}\).

(b) Its vertical velocity at that distance.

Use \(g=9.80\ \mathrm{m\,s^{-2}}\).

Solution 1

(a) The ball is \(3.24\ \mathrm{m}\) above the ground at \(x=6.00\ \mathrm{m}\). (b) Its vertical velocity is \(-5.88\ \mathrm{m\,s^{-1}}\), so it is falling.

A horizontal launch has zero initial vertical velocity, but gravity begins changing that velocity immediately.

First, find the travel time:

\[ \begin{aligned} t&=\frac{x}{u_x}\\ &=\frac{6.00}{10.0}\\ &=0.600\ \mathrm{s} \end{aligned} \]

Because \(u_y=0\):

\[ \begin{aligned} y&=y_0-\frac12gt^2\\ &=5.00-4.90(0.600)^2\\ &=5.00-1.764\\ &=3.236\ \mathrm{m}\\ &\approx3.24\ \mathrm{m} \end{aligned} \]

For the vertical velocity:

\[ \begin{aligned} v_y&=u_y-gt\\ &=0-9.80(0.600)\\ &=-5.88\ \mathrm{m\,s^{-1}} \end{aligned} \]

The tempting mistake is to treat a horizontal launch as motion at constant height. The launch velocity is horizontal, but the acceleration is not.

Question 2

A small ball is launched from ground level at \(20.0\ \mathrm{m\,s^{-1}}\), at \(45.0^\circ\) above the horizontal.

A horizontal overhead beam extends from \(x=10.0\ \mathrm{m}\) to \(x=30.0\ \mathrm{m}\). Its underside is \(9.00\ \mathrm{m}\) above the ground.

A student calculates that the ball’s unobstructed trajectory is below \(9.00\ \mathrm{m}\) at both ends of the beam and concludes that it will pass safely underneath.

Is the student correct? Find the first horizontal position where the ball would hit the beam. Assume the beam is solid, the ball is a point particle, and \(g=10.0\ \mathrm{m\,s^{-2}}\).

Solution 2

The student is incorrect. The ball first hits the beam at approximately \(x=13.7\ \mathrm{m}\).

Checking the two endpoints does not establish what happens between them. The ball can rise above both endpoint heights before falling again.

Resolve the launch velocity:

\[ \begin{aligned} u_x&=20.0\cos45^\circ\\ &=10\sqrt2\ \mathrm{m\,s^{-1}}\\ u_y&=20.0\sin45^\circ\\ &=10\sqrt2\ \mathrm{m\,s^{-1}} \end{aligned} \]

Substitute these into the trajectory equation. With \(x\) and \(y\) in metres:

\[ y(x)=x-\frac{x^2}{40} \]

At the beginning of the beam:

\[ y(10)=10-\frac{100}{40} =7.50\ \mathrm{m} \]

At the far end:

\[ y(30)=30-\frac{900}{40} =7.50\ \mathrm{m} \]

Both endpoint calculations are correct, but they don’t settle the question.

Find where the ideal trajectory reaches beam height:

\[ 9=x-\frac{x^2}{40} \]

Rearrange:

\[ x^2-40x+360=0 \]

Using the quadratic formula:

\[ \begin{aligned} x&=\frac{40\pm\sqrt{160}}{2}\\ &=20\pm2\sqrt{10} \end{aligned} \]

The two positions are:

\[ x\approx13.7\ \mathrm{m} \]

and

\[ x\approx26.3\ \mathrm{m} \]

The first is where the ball rises into the beam. The second would be where the unobstructed mathematical trajectory drops back through the same height.

Only the first collision occurs in the stated physical situation.

The crucial interaction is between the beam’s entire horizontal extent and the curved trajectory. Being below the beam at both ends is not enough.

Question 3

A ball is launched to the right and upwards from a point \(1.00\ \mathrm{m}\) above level ground. Its initial speed and angle are unknown.

Two non-contact measuring gates record the following illustrative positions for the same uninterrupted flight:

• At \(x=4.00\ \mathrm{m}\), the ball’s centre is \(3.40\ \mathrm{m}\) above the ground.
• At \(x=12.0\ \mathrm{m}\), the ball’s centre is again \(3.40\ \mathrm{m}\) above the ground.

Assume ideal projectile motion with \(g=10.0\ \mathrm{m\,s^{-2}}\).

Determine:

(a) The horizontal position of the highest point.

(b) The maximum height above the ground.

(c) The initial horizontal and vertical velocity components.

(d) Whether the ball is rising or falling at each measuring gate.

Solution 3

The apex is at \(x=8.00\ \mathrm{m}\), its height is \(4.20\ \mathrm{m}\), and the initial velocity components are \(u_x=10.0\ \mathrm{m\,s^{-1}}\) and \(u_y=8.00\ \mathrm{m\,s^{-1}}\). The ball is rising at the first gate and falling at the second.

The useful clue is the pair of equal heights.

With constant horizontal velocity and constant downward acceleration, a projectile’s vertical position is a quadratic function of horizontal position. Two points at the same height lie equally far horizontally from the axis of symmetry.

Therefore:

\[ x_{\mathrm{apex}} =\frac{4.00+12.0}{2} =8.00\ \mathrm{m} \]

To find the height and launch velocity, use the trajectory equation.

Write:

\[ y=1.00+Ax-Bx^2 \]

where \(A=u_y/u_x\) is dimensionless and \(B=g/(2u_x^2)\) has units of \(\mathrm{m^{-1}}\).

At \(x=4.00\ \mathrm{m}\):

\[ 4A-16B=2.40 \]

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

\[ 12A-144B=2.40 \]

Multiply the first equation by three and subtract the second:

\[ \begin{aligned} 96B&=4.80\\ B&=0.0500\ \mathrm{m^{-1}} \end{aligned} \]

Substitute into the first equation:

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

So the trajectory is:

\[ y=1.00+0.800x-0.0500x^2 \]

At the apex, \(x=8.00\ \mathrm{m}\):

\[ \begin{aligned} y_{\max}&=1.00+0.800(8.00)\\ &\quad-0.0500(8.00)^2\\ &=1.00+6.40-3.20\\ &=4.20\ \mathrm{m} \end{aligned} \]

Now recover the launch components.

Since \(B=g/(2u_x^2)\):

\[ \begin{aligned} u_x&=\sqrt{\frac{g}{2B}}\\ &=\sqrt{\frac{10.0}{0.100}}\\ &=10.0\ \mathrm{m\,s^{-1}} \end{aligned} \]

And because \(A=u_y/u_x\):

\[ \begin{aligned} u_y&=Au_x\\ &=0.800(10.0)\\ &=8.00\ \mathrm{m\,s^{-1}} \end{aligned} \]

Finally, use \(t=x/u_x\) and \(v_y=u_y-gt\).

At the first gate:

\[ \begin{aligned} t_1&=\frac{4.00}{10.0}=0.400\ \mathrm{s}\\ v_{y1}&=8.00-10.0(0.400)\\ &=+4.00\ \mathrm{m\,s^{-1}} \end{aligned} \]

At the second gate:

\[ \begin{aligned} t_2&=\frac{12.0}{10.0}=1.20\ \mathrm{s}\\ v_{y2}&=8.00-10.0(1.20)\\ &=-4.00\ \mathrm{m\,s^{-1}} \end{aligned} \]

The ball passes the same height once while rising and once while falling.

A single height observation could not have determined the complete launch velocity. The second observation, together with the known launch height and gravitational acceleration, supplies enough information to reconstruct the ideal trajectory.

Question 4

A ball is launched from a ledge \(2.00\ \mathrm{m}\) above level ground at \(20.0\ \mathrm{m\,s^{-1}}\), \(45.0^\circ\) above the horizontal.

A building has a flat roof \(8.00\ \mathrm{m}\) above the ground. The roof extends horizontally from \(x=18.0\ \mathrm{m}\) to \(x=36.0\ \mathrm{m}\). Its front wall rises vertically from the ground at \(x=18.0\ \mathrm{m}\).

Assume ideal projectile motion, \(g=10.0\ \mathrm{m\,s^{-2}}\), and no rebound after impact.

Does the ball hit the front wall, land on the roof, or pass beyond the building? Determine the first contact position.

Solution 4

The ball clears the building’s front wall and lands on the roof at approximately \(x=32.6\ \mathrm{m}\).

We must consider the first surface the ball actually reaches. Solving for a particular height without checking where that surface exists can give the wrong collision point.

The launch components are:

\[ u_x=u_y=10\sqrt2\ \mathrm{m\,s^{-1}} \]

With \(y_0=2.00\ \mathrm{m}\), the trajectory becomes:

\[ y(x)=2.00+x-\frac{x^2}{40} \]

First, check the front wall at \(x=18.0\ \mathrm{m}\):

\[ \begin{aligned} y(18)&=2.00+18.0-\frac{324}{40}\\ &=11.9\ \mathrm{m} \end{aligned} \]

The wall reaches only \(8.00\ \mathrm{m}\), so the ball’s centre clears its top by \(3.90\ \mathrm{m}\).

Next, find where the ball is at roof height:

\[ 8.00=2.00+x-\frac{x^2}{40} \]

Rearrange:

\[ x^2-40x+240=0 \]

Using the quadratic formula:

\[ \begin{aligned} x&=\frac{40\pm\sqrt{640}}{2}\\ &=20\pm4\sqrt{10} \end{aligned} \]

This gives:

\[ x_1\approx7.35\ \mathrm{m} \]

\[ x_2\approx32.65\ \mathrm{m} \]

The smaller solution represents the ball passing through the height \(y=8.00\ \mathrm{m}\) while rising, before the building begins.

There is no roof at \(x=7.35\ \mathrm{m}\), so that position cannot be a roof collision.

The second solution lies between \(18.0\) and \(36.0\ \mathrm{m}\), where the roof actually exists. The ball is descending there, so it lands on the roof at approximately \(32.6\ \mathrm{m}\).

The tempting approach is to accept the first algebraic intersection with \(y=8.00\ \mathrm{m}\).

But a collision needs both the correct height and a physical surface at that horizontal position. Solving the equation is only part of the reasoning.

Question 5

A launcher fires a small ball from ground level at a fixed speed of \(20.0\ \mathrm{m\,s^{-1}}\). Its angle \(\theta\) can be adjusted anywhere strictly between \(0^\circ\) and \(90^\circ\).

The ball travels to the right through a covered lane. A solid horizontal ceiling extends from \(x=0\) to \(x=12.0\ \mathrm{m}\), at a height of \(12.0\ \mathrm{m}\) above the ground.

Beyond the ceiling is open air. A vertical wall stands at \(x=20.0\ \mathrm{m}\), with its top \(14.0\ \mathrm{m}\) above the ground.

Assume ideal projectile motion, \(g=10.0\ \mathrm{m\,s^{-2}}\), and no bouncing. The ball must stay strictly below the ceiling and pass strictly above the wall to succeed.

Determine:

(a) The greatest possible height at \(x=20.0\ \mathrm{m}\) if the ceiling is removed, and the launch angle that produces it.

(b) Whether the ball would be rising or falling at the wall for that launch angle.

(c) Whether any launch angle can clear the wall without first hitting the ceiling. Prove your conclusion rather than testing only one or two angles.

Solution 5

(a) The greatest unobstructed height at the wall is \(15.0\ \mathrm{m}\), achieved at approximately \(63.4^\circ\). (b) The ball would already be falling at the wall. (c) No launch angle can clear the wall without hitting the ceiling first.

The surprise is that a wall can be individually reachable, yet impossible to clear when another obstacle constrains the earlier part of the same trajectory.

Part (a): Find the maximum height at a fixed distance

Use the trajectory equation with \(y_0=0\):

\[ \begin{aligned} y(x)&=x\tan\theta\\ &\quad-\frac{gx^2}{2u^2\cos^2\theta} \end{aligned} \]

Let \(T=\tan\theta\). Using the identity

\[ \frac{1}{\cos^2\theta}=1+T^2 \]

and substituting \(u=20.0\ \mathrm{m\,s^{-1}}\), \(g=10.0\ \mathrm{m\,s^{-2}}\), we get:

\[ y(x)=xT-\frac{x^2}{80}(1+T^2) \]

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

At the wall, \(x=20.0\ \mathrm{m}\):

\[ y(20)=20T-5(1+T^2) \]

Rearrange by completing the square:

\[ \begin{aligned} y(20)&=-5T^2+20T-5\\ &=15-5(T-2)^2 \end{aligned} \]

The square \((T-2)^2\) cannot be negative. Therefore, the greatest possible height occurs when:

\[ T=2 \]

So:

\[ \begin{aligned} \tan\theta&=2\\ \theta&=\tan^{-1}(2)\\ &\approx63.4^\circ \end{aligned} \]

The maximum height at the wall is:

\[ y(20)_{\max}=15.0\ \mathrm{m} \]

Without the ceiling, this is \(1.00\ \mathrm{m}\) above the wall’s top.

Notice the competing effects. Increasing launch angle initially makes the ball rise more steeply, but it also reduces horizontal velocity. At a large enough angle, reaching the wall takes so long that gravity removes more height than the steeper launch adds.

That is why aiming closer and closer to vertically does not make the projectile higher at every fixed horizontal distance.

Part (b): Is the optimum shot still rising?

At \(\tan\theta=2\):

\[ \begin{aligned} u_x&=\frac{20}{\sqrt5}\\ u_y&=\frac{40}{\sqrt5} \end{aligned} \]

Both components are measured in \(\mathrm{m\,s^{-1}}\).

The time to the wall is:

\[ \begin{aligned} t&=\frac{20}{20/\sqrt5}\\ &=\sqrt5\ \mathrm{s} \end{aligned} \]

The vertical velocity there is:

\[ \begin{aligned} v_y&=u_y-gt\\ &=\frac{40}{\sqrt5}-10\sqrt5\\ &=-4.47\ \mathrm{m\,s^{-1}} \end{aligned} \]

So the ball is descending at the wall.

This exposes a subtle misconception: maximising the height at a particular horizontal distance does not mean arranging for the projectile’s highest point to occur at that distance.

In fact, this trajectory reaches its apex earlier, at \(x=16.0\ \mathrm{m}\). It then descends to the still-impressive height of \(15.0\ \mathrm{m}\) at the wall.

Part (c): Can a different angle avoid the ceiling?

We must test all possible wall-clearing launch angles, not just the optimal one.

To pass strictly above the wall:

\[ y(20)>14 \]

Using our expression:

\[ 15-5(T-2)^2>14 \]

Therefore:

\[ (T-2)^2<0.20 \]

This restricts \(T\) to:

\[ 2-\sqrt{0.20}<T<2+\sqrt{0.20} \]

Approximately:

\[ 1.553<T<2.447 \]

Those are the only launch-angle values that could clear the wall in unobstructed flight.

Now inspect the height at the end of the ceiling, \(x=12.0\ \mathrm{m}\):

\[ y(12)=12T-1.8(1+T^2) \]

The wall-clearance requirement can also be written as:

\[ 1+T^2<4T-2.8 \]

Substituting this bound into the expression for \(y(12)\):

\[ \begin{aligned} y(12)&=12T-1.8(1+T^2)\\ &>12T-1.8(4T-2.8)\\ &=4.8T+5.04 \end{aligned} \]

But every wall-clearing angle has:

\[ T>2-\sqrt{0.20} \]

Hence:

\[ \begin{aligned} y(12)&>4.8(2-\sqrt{0.20})\\ &\quad+5.04\\ &>12.49\ \mathrm{m} \end{aligned} \]

This is the decisive constraint.

Every unobstructed trajectory that would clear the \(14.0\ \mathrm{m}\) wall must already be more than \(12.49\ \mathrm{m}\) high at \(x=12.0\ \mathrm{m}\).

But the ceiling is only \(12.0\ \mathrm{m}\) high and extends all the way to that position.

Because the ball starts below the ceiling, any such trajectory must cross the ceiling’s height before reaching its end. It therefore hits the ceiling first.

The strongest tempting answer is to choose the \(63.4^\circ\) launch because it gives the greatest height at the wall.

That solves the wall problem in isolation. It does not solve the complete problem.

The ceiling removes every angle capable of clearing the wall. Even though the far wall is reachable in the unconstrained projectile model, the entire obstacle arrangement is not.

06What this method lets you decide next

The main habit is to treat horizontal position and vertical position as two measurements connected by the same time.

When a question gives a horizontal distance, find when the projectile gets there. Then use vertical motion to calculate its height, check whether it is rising or falling if needed, and make sure the position occurs before any collision.

The next useful distinction is between height at a particular distance and the distance at which a projectile actually lands. When the question changes from an obstacle’s height to a landing position, the vertical motion must instead determine the available flight time.

That is the starting point for calculating the horizontal range of a projectile, particularly when launch and landing heights differ.