How to Tell if a Projectile Lands Before or After a Target

A ball rolls off a table at \(4.0\ \mathrm{m\,s^{-1}}\). The floor is \(1.25\ \mathrm{m}\) below the launch point, and a target is marked \(1.80\ \mathrm{m}\) horizontally from the table’s edge. Will the ball land before the target, directly on it, or beyond it?

Predict what happens before calculating. The ball is falling as soon as it leaves the table, so it might seem likely to fall short.

In fact, the ball lands approximately \(2.02\ \mathrm{m}\) from the edge, or \(0.22\ \mathrm{m}\) beyond the target.

The method is to calculate how long the ball takes to reach the landing height using its vertical motion, then use that time to find its horizontal distance. Compare that distance with the target’s position.

The important distinction is that passing horizontally above a target is not the same as landing on it.

01Why horizontal and vertical motion must be calculated separately

Imagine two identical balls starting at the same height. One is simply dropped, while the other is launched horizontally.

Which one reaches the floor first?

Ignoring air resistance, they land at the same time.

Both balls have the same initial vertical velocity: zero. Both experience the same downward acceleration due to gravity. The second ball’s horizontal speed does not delay its fall.

That’s the central idea behind projectile motion.

The ball moves horizontally at a constant velocity while gravity changes its vertical velocity. These two motions happen simultaneously, but we calculate them separately.

Choose a coordinate system

For a ball launched towards the right, use:

• Positive \(x\): horizontally to the right.
• Positive \(y\): vertically upwards.
• \(x=0\): directly below the launch point.
• \(y=0\): the level of the floor.
• \(t=0\): the instant the ball leaves the launcher.

The horizontal equation is

\[ x=u_x t \]

where \(x\) is horizontal displacement in metres, \(u_x\) is the initial horizontal velocity in metres per second, and \(t\) is elapsed time in seconds.

The vertical equation is

\[ y=h+u_y t-\frac12 gt^2 \]

where \(y\) is height above the floor in metres, \(h\) is the initial height in metres, \(u_y\) is the initial vertical velocity in metres per second, and \(g\) is the magnitude of gravitational acceleration, approximately \(9.8\ \mathrm{m\,s^{-2}}\).

The negative sign appears because gravity accelerates downwards, opposite our positive \(y\)-direction.

For a horizontal launch, \(u_y=0\), so the vertical equation becomes

\[ y=h-\frac12 gt^2 \]

Notice that \(u_x\) does not appear in the vertical equation. This is why increasing the horizontal speed makes the ball travel further without increasing the time it stays airborne, provided the launch height and initial vertical velocity remain unchanged.

This model assumes negligible air resistance, constant gravitational acceleration, and no collision before landing. An actual ball can also experience drag and spin effects.

02The decision rule: before, on, or after the target

Suppose a target is marked on level ground at horizontal position \(x_T\).

First calculate the landing time, \(t_L\), from the vertical equation. Then calculate the horizontal landing position:

\[ x_L=u_x t_L \]

Here, \(x_L\) is the horizontal position of the landing point, measured from the same origin as \(x_T\).

The decision follows immediately.

This assumes the target is a point or narrow line, the ball travels towards increasing \(x\), and the floor is the first surface it reaches.

For a target with a finite width, compare the landing position with both edges of the target region instead.

Which vertical equation should you use?

One major source of mistakes is treating every projectile question as if the ball lands at its launch height.

That shortcut only works for a particular arrangement.

The second formula assumes \(u_y>0\), and its flight time is the non-zero solution for returning to the original height.

The third case is not simply another name for a level-ground problem. The height difference changes the flight time, sometimes enough to reverse the answer about whether the ball lands before or after the target.

04How an experiment could test the prediction

The calculation makes a prediction. To test whether the model describes a real ball accurately, we need to measure where it first contacts the floor.

A useful investigation would use a horizontal ball launcher clamped to the end of a bench.

The ball travels through the launcher, leaves its open end horizontally, moves through the air, and strikes a sheet of recording paper on the floor. A plumb line hanging from the launch point identifies \(x=0\), so horizontal distances can be measured without relying on the position of a table leg.

A side-on video recording or two closely spaced photogates near the launch point could measure the horizontal launch speed.

What should the apparatus show?

For the \(1.25\ \mathrm{m}\) launch height and \(4.0\ \mathrm{m\,s^{-1}}\) speed, the ideal model predicts a landing position of \(2.02\ \mathrm{m}\).

If repeated landing marks clustered near this position, that would support the calculated trajectory under those conditions.

A single successful landing would not prove that the model is universally correct. A launch-speed error, a small downward launch angle, or air resistance could also affect the measured range.

The limitation of the first design

One launch height and one launch speed cannot tell us how the landing position depends on height.

A stronger test would change the launch height while keeping the horizontal launch speed constant. The launcher should remain horizontal, use the same ball and release setting, and have its speed checked for each set of trials.

For a horizontal launch,

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

The predicted range therefore increases with the square root of the height, not directly in proportion to height.

Here are illustrative predictions, not experimental measurements, for a fixed horizontal launch speed of \(4.0\ \mathrm{m\,s^{-1}}\).

Increasing the height from \(0.80\ \mathrm{m}\) to \(1.80\ \mathrm{m}\) changes the predicted result from landing before the target to landing after it, even though the launch speed stays unchanged.

The observation to look for is whether measured ranges follow this pattern. Repeated trials and launch-speed measurements help separate model behaviour from experimental variability.

06A fast check without solving the full flight time

Sometimes the question only asks whether the ball lands before or after a target, not the exact landing position.

If the target lies on an unobstructed, horizontal landing surface, calculate the time required to reach its horizontal position:

\[ t_T=\frac{x_T}{u_x} \]

Then find the ball’s predicted height at that time:

\[ y(t_T)=h+u_y t_T-\frac12 gt_T^2 \]

Compare this height with the landing surface height, \(y_T\).

This is especially useful for checking a calculated answer.

But it has an important limitation: the surface must actually exist where the ball would contact it.

If the target sits on a raised platform, a ramp, or a surface that begins partway along the path, you must determine the first physical intersection with that surface. A ball may strike a platform’s front wall instead of landing on top.

07Mistakes that reverse the answer

The most useful exam habit is to ask exactly which vertical height the ball must reach before it has landed.

One final distinction: if the calculated landing distance differs from the target by only a few millimetres, the ideal model might classify it precisely, but a real experiment might not. Launch-speed variation, alignment, ball size, and measurement uncertainty all matter when deciding whether the target is reliably hit.

08Questions and solutions

The questions build from ordinary horizontal launches to experimental inference, non-horizontal landing surfaces, moving targets, and a final problem where the first collision changes the answer.

Unless stated otherwise, assume negligible air resistance, a point-like ball, constant gravitational acceleration \(g=9.8\ \mathrm{m\,s^{-2}}\), and no bouncing. Horizontal positions increase to the right.

Question 1

A ball is launched horizontally from a table \(0.80\ \mathrm{m}\) high at \(5.0\ \mathrm{m\,s^{-1}}\).

A target is marked \(2.25\ \mathrm{m}\) horizontally from the point directly below the launch.

Calculate the landing distance and determine whether the ball lands before or after the target.

Solution 1

The ball lands approximately \(2.02\ \mathrm{m}\) from the table, or \(0.23\ \mathrm{m}\) before the target.

Because the launch is horizontal, the vertical motion starts with zero velocity.

Calculate the flight time:

\[ \begin{aligned} t_L&=\sqrt{\frac{2h}{g}}\\ &=\sqrt{\frac{2(0.80)}{9.8}}\\ &=0.404\ \mathrm{s} \end{aligned} \]

Then calculate horizontal displacement:

\[ \begin{aligned} x_L&=u_x t_L\\ &=5.0(0.404)\\ &=2.02\ \mathrm{m} \end{aligned} \]

The target is at \(2.25\ \mathrm{m}\), so

\[ 2.25-2.02=0.23\ \mathrm{m} \]

The landing point is short of the target.

The critical reasoning is that the vertical drop determines the flight time. The horizontal speed determines how far the ball travels during that time, not how long it stays airborne.

Question 2

A camera records a ball launched horizontally from a bench of unknown height.

At \(t=0.500\ \mathrm{s}\), the ball is horizontally \(2.00\ \mathrm{m}\) from the launch point and its centre is \(0.375\ \mathrm{m}\) above the floor.

The horizontal launch speed is constant at \(4.00\ \mathrm{m\,s^{-1}}\).

A floor target lies \(2.20\ \mathrm{m}\) from the launch origin.

Treat all measurements as exact within this idealised question.

(a) Determine the original launch height.

(b) Predict whether the ball lands before or after the target, and by how much.

(c) Keeping the horizontal launch speed unchanged, calculate the new launch height needed to land directly on the target.

Solution 2

The original launch height is \(1.60\ \mathrm{m}\). The ball lands approximately \(0.086\ \mathrm{m}\) after the target, and reducing the launch height to approximately \(1.482\ \mathrm{m}\) would make it land on the target.

(a) Reconstruct the launch height.

The camera gives the ball’s height at a known time.

For a horizontal launch,

\[ y=h-\frac12 gt^2 \]

Therefore,

\[ h=y+\frac12 gt^2 \]

Substitute:

\[ \begin{aligned} h&=0.375+4.9(0.500)^2\\ &=0.375+1.225\\ &=1.600\ \mathrm{m} \end{aligned} \]

The ball began \(1.60\ \mathrm{m}\) above the floor.

(b) Determine the landing position.

Calculate the total flight time:

\[ \begin{aligned} t_L&=\sqrt{\frac{2(1.600)}{9.8}}\\ &=0.5714\ \mathrm{s} \end{aligned} \]

Then:

\[ \begin{aligned} x_L&=4.00(0.5714)\\ &=2.286\ \mathrm{m} \end{aligned} \]

Compared with the target:

\[ 2.286-2.200=0.086\ \mathrm{m} \]

The ball lands approximately \(8.6\ \mathrm{cm}\) after the target.

(c) Adjust the launch height.

For a direct hit, horizontal travel must equal \(2.20\ \mathrm{m}\).

The required time is

\[ t=\frac{2.20}{4.00}=0.550\ \mathrm{s} \]

The ball must fall the entire launch height in this time:

\[ \begin{aligned} h_{\text{new}}&=\frac12 gt^2\\ &=4.9(0.550)^2\\ &=1.48225\ \mathrm{m} \end{aligned} \]

So the required new height is approximately \(1.482\ \mathrm{m}\), a reduction of

\[ 1.600-1.48225=0.11775\ \mathrm{m} \]

or about \(11.8\ \mathrm{cm}\).

A tempting mistake is to assume the ball must already have landed because it has travelled beyond \(2.00\ \mathrm{m}\). But at the recorded position, it is still \(0.375\ \mathrm{m}\) above the floor.

The camera reading describes an intermediate position, not the landing point. It also allows us to reconstruct the earlier launch height and design a change that produces the required landing time.

Question 3

A ball is launched horizontally at \(6.00\ \mathrm{m\,s^{-1}}\) from a point \(2.00\ \mathrm{m}\) above level ground.

The ground remains flat until \(x=3.00\ \mathrm{m}\). From that position onwards, it rises as a straight ramp with equation

\[ y=0.50(x-3.00) \]

where \(x\) and \(y\) are measured in metres from the launch origin and the original ground level.

A target is marked on the ramp at \(x=3.70\ \mathrm{m}\).

(a) Determine the ball’s first landing position and whether it is before or after the target.

(b) A student ignores the ramp and calculates where the ball would land on the original flat ground. Show why this predicts a different outcome.

Solution 3

The ball first lands on the ramp at approximately \(x=3.56\ \mathrm{m}\), which is \(0.14\ \mathrm{m}\) before the target. Ignoring the ramp would incorrectly predict landing after the target.

(a) Find the first contact with the actual ground.

The ball’s horizontal position is

\[ x=6.00t \]

Therefore,

\[ t=\frac{x}{6.00} \]

Its vertical position is

\[ y=2.00-4.90t^2 \]

Substituting for time gives the ball’s height as a function of horizontal position:

\[ y_b=2.00-0.1361x^2 \]

The ramp height, for \(x\geq3.00\), is

\[ y_r=0.50(x-3.00) \]

At contact, \(y_b=y_r\):

\[ 2.00-0.1361x^2=0.50(x-3.00) \]

Rearrange:

\[ 0.1361x^2+0.50x-3.50=0 \]

Solving the quadratic gives the positive root

\[ x\approx3.56\ \mathrm{m} \]

The negative root is outside the physical region of the ramp.

We must also check that the ball reaches the start of the ramp without hitting the flat ground.

At \(x=3.00\ \mathrm{m}\),

\[ \begin{aligned} t&=\frac{3.00}{6.00}=0.500\ \mathrm{s}\\ y&=2.00-4.90(0.500)^2\\ &=0.775\ \mathrm{m} \end{aligned} \]

The ball is still above the flat ground when it reaches the ramp.

It therefore contacts the rising ramp at approximately \(3.56\ \mathrm{m}\).

The target is at \(3.70\ \mathrm{m}\), so

\[ 3.70-3.56=0.14\ \mathrm{m} \]

The ball lands approximately \(14\ \mathrm{cm}\) before the target.

(b) Compare with the incorrect flat-ground model.

If the ground had remained flat at \(y=0\), the flight time would be

\[ \begin{aligned} t&=\sqrt{\frac{2(2.00)}{9.8}}\\ &=0.639\ \mathrm{s} \end{aligned} \]

The corresponding horizontal range would be

\[ x=6.00(0.639)=3.83\ \mathrm{m} \]

This would suggest landing \(0.13\ \mathrm{m}\) after the target.

But the ramp rises into the ball’s descending path. The ball encounters the ramp earlier than it would encounter the original ground level.

The issue is not an incorrect projectile equation. It is an incorrect landing condition.

Question 4

A ball is launched horizontally from a table \(1.25\ \mathrm{m}\) above the floor at \(5.00\ \mathrm{m\,s^{-1}}\).

At the instant of launch, a small projected light target is located on the floor at \(x=2.70\ \mathrm{m}\). The light target moves horizontally towards the launcher at a constant speed of \(0.60\ \mathrm{m\,s^{-1}}\).

(a) Does the ball land before or after the moving target, measured relative to the target’s position at landing?

(b) Find the target’s required constant speed towards the launcher for the ball to land exactly on it, assuming both begin at the stated positions.

Solution 4

The ball lands approximately \(0.128\ \mathrm{m}\) after the moving target. A target speed of approximately \(0.346\ \mathrm{m\,s^{-1}}\) towards the launcher would produce a direct hit.

(a) Compare both positions at the same instant.

The ball’s flight time depends on its vertical motion:

\[ \begin{aligned} t_L&=\sqrt{\frac{2(1.25)}{9.8}}\\ &=0.5051\ \mathrm{s} \end{aligned} \]

Its landing position is

\[ \begin{aligned} x_b&=5.00(0.5051)\\ &=2.525\ \mathrm{m} \end{aligned} \]

The moving target starts at \(2.70\ \mathrm{m}\) and travels left.

Its position at the instant of landing is

\[ x_T=2.70-0.60t_L \]

Therefore,

\[ \begin{aligned} x_T&=2.70-0.60(0.5051)\\ &=2.397\ \mathrm{m} \end{aligned} \]

The difference is

\[ \begin{aligned} x_b-x_T&=2.525-2.397\\ &=0.128\ \mathrm{m} \end{aligned} \]

The ball lands approximately \(12.8\ \mathrm{cm}\) after the moving target.

Notice the surprising reversal. If the target had remained stationary at \(2.70\ \mathrm{m}\), the ball would have landed before it.

The moving target travels back past the eventual landing position before the ball reaches the ground.

(b) Find the speed required for a direct hit.

Let \(v_T\) be the positive magnitude of the target’s speed towards the launcher.

For a direct hit,

\[ x_b=2.70-v_Tt_L \]

Rearrange:

\[ v_T=\frac{2.70-x_b}{t_L} \]

Substitute:

\[ \begin{aligned} v_T&=\frac{2.70-2.525}{0.5051}\\ &\approx0.346\ \mathrm{m\,s^{-1}} \end{aligned} \]

The target must move towards the launcher at approximately \(0.346\ \mathrm{m\,s^{-1}}\).

The governing constraint is simultaneous position. Comparing the ball’s final position with the target’s initial position is not enough when the target itself is moving.

Question 5

A ball is launched from a height of \(2.00\ \mathrm{m}\) above a horizontal floor.

Its initial velocity components are

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

and

\[ u_y=+6.00\ \mathrm{m\,s^{-1}} \]

For this question, use \(g=10.0\ \mathrm{m\,s^{-2}}\).

A rigid raised deck has a horizontal top at \(y=3.00\ \mathrm{m}\). Its top extends from \(x=a\) to \(x=8.00\ \mathrm{m}\), and it has a vertical front wall at \(x=a\) extending from the floor to the top.

A narrow target is painted on top of the deck, exactly \(1.00\ \mathrm{m}\) horizontally beyond its front edge. Its position is therefore \(x=a+1.00\).

The deck’s front edge can be moved. Its position satisfies

\[ 0.50<a<6.50 \]

with \(a\) measured in metres.

(a) Determine the values of \(a\) for which the ball first lands on the horizontal top of the deck rather than striking its front wall.

(b) For those top-landing arrangements, determine when the ball lands before, directly on, or after the painted target.

(c) Explain what happens at \(a=1.00\ \mathrm{m}\) and \(a=5.00\ \mathrm{m}\), and why using a height equation alone is not sufficient.

Assume the ball behaves as a point particle and that its trajectory ends at its first contact with the deck or floor. You do not need to analyse any motion after a collision.

Solution 5

The ball lands on top of the deck only when \(1.00<a<5.00\ \mathrm{m}\), excluding the corner-contact cases at the endpoints. Within that interval, it lands after the target for \(1.00<a<4.00\), directly on the target at \(a=4.00\), and before the target for \(4.00<a<5.00\). Outside the top-landing interval, it hits the deck’s front wall first.

The key is that reaching the correct height is not enough. The deck must occupy the appropriate horizontal position, and the ball must not have collided with its front wall earlier.

(a) Determine when top landing is possible.

First write the vertical motion:

\[ y=2.00+6.00t-5.00t^2 \]

Since \(x=5.00t\),

\[ t=\frac{x}{5.00} \]

Substitute:

\[ y=2.00+1.20x-0.20x^2 \]

The deck top is at \(y=3.00\ \mathrm{m}\).

Set the ball’s height equal to the deck height:

\[ 3.00=2.00+1.20x-0.20x^2 \]

Rearrange:

\[ x^2-6x+5=0 \]

Factorise:

\[ (x-1)(x-5)=0 \]

Therefore,

\[ x=1.00\ \mathrm{m} \]

or

\[ x=5.00\ \mathrm{m} \]

There are two intersections with the deck’s height.

At \(x=1.00\ \mathrm{m}\), the ball is travelling upwards through \(y=3.00\ \mathrm{m}\).

At \(x=5.00\ \mathrm{m}\), it is travelling downwards through the same height.

Between those positions, the ball is above \(3.00\ \mathrm{m}\). Before the first crossing and after the second, it is below that height.

For the ball to clear the deck’s front wall and then land on top, the front edge must be positioned between the two crossings:

\[ 1.00<a<5.00 \]

In this interval, the ball passes over the front wall and lands on the horizontal deck at \(x=5.00\ \mathrm{m}\).

What if \(a<1.00\ \mathrm{m}\)?

When the ball reaches the front edge, it is still below the deck’s top. It therefore strikes the vertical wall before it can rise above the deck.

What if \(a>5.00\ \mathrm{m}\)?

The ball has already descended below deck height by the time it reaches the front edge. It again hits the vertical wall rather than landing on top.

For the stated range \(a<6.50\ \mathrm{m}\), the ball has not yet reached the floor when it arrives at the front wall, so the wall collision does occur first in these cases.

(b) Compare the top-landing position with the painted target.

Whenever \(1.00<a<5.00\), the ball lands on top at

\[ x_L=5.00\ \mathrm{m} \]

The target moves with the front edge, so its position is

\[ x_T=a+1.00 \]

For the ball to land directly on the target,

\[ 5.00=a+1.00 \]

Therefore,

\[ a=4.00\ \mathrm{m} \]

Now consider the other possible positions within the top-landing interval.

If

\[ 1.00<a<4.00 \]

then \(a+1.00<5.00\), so the target is before the ball’s landing point. The ball lands after the target.

If

\[ 4.00<a<5.00 \]

then \(a+1.00>5.00\), so the target lies beyond the landing point. The ball lands before the target.

If the front edge is outside the top-landing interval, the ball strikes the front wall. Its first contact is horizontally before the painted target, but it does not land on the deck’s top at all.

That is a different physical outcome from landing before the target on the deck.

(c) Examine the boundary cases and the hidden assumption.

At \(a=1.00\ \mathrm{m}\), the ball reaches \(y=3.00\ \mathrm{m}\) exactly at the deck’s upper front corner while rising.

At \(a=5.00\ \mathrm{m}\), it reaches the upper front corner while descending.

In both cases, the first contact is at the corner. Neither is an ordinary landing on the interior of the horizontal deck. The question does not specify how the ball moves after contact, so no further trajectory can be determined from the given information.

The tempting approach is to solve \(y=3.00\), select the descending crossing at \(x=5.00\), and announce that the ball always lands there.

But that ignores the geometry.

If the deck starts too close to the launcher, the ball meets its front wall before rising above deck height.

If the deck starts too far away, the ball has already fallen below deck height and meets the wall on its descent.

This problem reveals a limitation of the standard landing-distance formula.

A mathematical trajectory can intersect a height without intersecting a physical landing surface at that location. It can also encounter another surface first.

The correct landing position is determined by the first physically possible contact, not merely by the first convenient solution of an equation.

08What to decide before your next projectile calculation

When an HSC Physics question asks whether a ball lands before or after a target, identify the actual landing surface before choosing a formula.

For an unobstructed horizontal floor, the method is straightforward: calculate the vertical flight time, calculate the horizontal landing position, and compare it with the target.

If the ball launches and lands at different heights, use the full vertical displacement equation. If the landing surface is sloped, raised, interrupted, or moving, examine its position and the time of first contact rather than relying on a standard range formula.

That leads naturally to a more general projectile-motion skill: finding where a trajectory intersects a physical surface, and deciding whether that intersection can actually occur before anything else gets in the way.