Why Does a Kinematics Equation Give Two Times? | HSC Physics

A ball is thrown straight upwards past a window. It crosses the bottom of the window, rises out of view, then crosses exactly the same height again on its way down.

If you use a kinematics equation to calculate when the ball is at the window’s height, you might get two positive values for time. Neither is a mistake. The ball really does occupy the same position twice: once while rising and once while falling.

But what if the window is higher than the ball ever reaches? The equation will have no real solutions because the ball never gets there.

Understanding why this happens means looking beyond the algebra and connecting each mathematical root to the actual motion.

01Why can one position have two different times?

Imagine throwing a ball vertically upwards.

For a moment, forget the equations. Think about what the ball actually does.

• It starts at its launch position.
• It moves upwards, passing through a series of heights.
• It reaches its highest point and momentarily has zero velocity.
• It starts moving downwards, passing through many of those same heights again.

Suppose you want to know when the ball is 5.0 m above its launch position.

There are two possibilities:

1. The ball passes 5.0 m while travelling upwards.
2. The ball passes 5.0 m again while travelling downwards.

The position is identical, but the direction of motion is different.

This is the physical reason a kinematics equation can produce two positive times.

What does the motion look like on a graph?

A displacement-time graph makes the explanation particularly clear.

The graph represents the ball’s vertical displacement as time passes.

Notice that the horizontal line representing a particular displacement intersects the curve twice.

Each intersection has a different time coordinate.

• The first intersection represents the ball reaching that displacement on its way up.
• The second represents the ball reaching the same displacement on its way down.

An important distinction: this is a displacement-time graph, not a drawing of the ball’s actual trajectory. The ball moves along a straight vertical path. The graph curves because its displacement changes non-uniformly with time.

02Connecting the two roots to the kinematics equation

Let’s introduce the mathematics.

Assume:

• Upwards is positive.
• The ball is launched from the origin.
• Air resistance is negligible.
• Gravitational acceleration is uniform, with magnitude \(g=9.8\,\mathrm{m\,s^{-2}}\).

The uniform-acceleration displacement equation is:

\[ s=ut+\frac12at^2 \]

where:

• \(s\) is vertical displacement from the launch position, in metres (m).
• \(u\) is initial vertical velocity, in metres per second (\(\mathrm{m\,s^{-1}}\)).
• \(a\) is vertical acceleration, in metres per second squared (\(\mathrm{m\,s^{-2}}\)).
• \(t\) is elapsed time, in seconds (s).

Because upwards is positive, gravitational acceleration is negative:

\[ a=-9.8\,\mathrm{m\,s^{-2}} \]

Therefore:

\[ \boxed{s=ut-4.9t^2} \]

Notice the \(t^2\) term.

This makes the displacement equation a quadratic equation in time. When you rearrange it to solve for \(t\), you can obtain two roots.

For an upward launch with \(u>0\), rearranging gives:

\[ 4.9t^2-ut+s=0 \]

Using the quadratic formula:

\[ \boxed{ t=\frac{u\pm\sqrt{u^2-2gs}}{g} } \]

The plus and minus signs produce the two possible times.

But why do these roots represent the upward and downward parts of the flight?

The connection to the highest point

The highest point is particularly important because it separates the two parts of the motion.

At this instant, the ball’s vertical velocity is zero.

Using the uniform-acceleration velocity equation:

\[ v=u+at \]

and substituting \(a=-g\):

\[ 0=u-gt \]

Therefore, the time at the highest point is:

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

The two roots of the displacement equation have an interesting relationship:

\[ t_1+t_2=\frac{2u}{g} \]

Since \(t_{\text{top}}=u/g\):

\[ \boxed{ \frac{t_1+t_2}{2}=t_{\text{top}} } \]

This means the two crossing times are equally spaced around the time at which the ball reaches its highest point.

For example, if the ball reaches its highest point at 1.5 s, and crosses a particular height at 0.5 s while rising, it will cross that height again at 2.5 s while falling.

This symmetry comes from uniform acceleration. It is not a coincidence.

At the highest point, velocity is zero, but acceleration is still directed downwards. Gravity has not switched off. You can explore this distinction further in Can an Object Have Zero Velocity and Nonzero Acceleration?.

03When should you expect two roots?

Not every displacement produces two positive times.

The answer depends on where the target position is relative to the launch point and the highest point.

For an object launched upwards from \(s=0\), consider these possibilities.

Here, \(s_{\max}\) is the maximum vertical displacement above the launch point.

Under uniform gravitational acceleration:

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

The important point is that two mathematical roots do not automatically mean two physically valid future times.

You must check the actual motion.

A negative root is not the same as a negative velocity

This is a common source of confusion.

Suppose you obtain:

\[ t=-0.50\,\mathrm{s} \]

This means the mathematical model places that event 0.50 seconds before the chosen time origin.

It does not mean the object is travelling backwards in time.

Compare this with:

\[ v=-5.0\,\mathrm{m\,s^{-1}} \]

If upwards is positive, this means the object is travelling downwards at \(5.0\,\mathrm{m\,s^{-1}}\).

The signs describe different things:

• Negative time: an event before the chosen time origin.
• Negative velocity: motion opposite to the positive direction.

Whether a negative-time root represents a real event depends on whether the object was actually following the same uniform-acceleration motion before \(t=0\).

We will return to this distinction in a more challenging example.

04Testing the idea: Can a light gate detect two crossing times?

We can investigate the two-root prediction using a vertical ball launcher and optical sensors.

The question is simple: can we detect the same ball passing the same height twice?

Setting up the experiment

Imagine a small ball launched vertically upwards from an electronic launcher.

A vertical metre ruler is mounted beside its path, with the launch position marked as zero.

Two optical gates are positioned horizontally across the ball’s path:

• Gate A is 1.50 m above the launch point.
• Gate B is 1.60 m above the launch point.

A timer starts when the launcher releases the ball. Each optical gate records the instant the ball interrupts its light beam.

A soft catch is positioned above the ball’s predicted maximum height.

What should the experiment show?

Suppose the launcher gives the ball an initial upward velocity of:

\[ u=8.0\,\mathrm{m\,s^{-1}} \]

Assume negligible air resistance.

Start with gate A, positioned 1.50 m above the launch point.

Substituting into the displacement equation:

\[ 1.50=8.0t-4.9t^2 \]

Rearranging and applying the quadratic formula:

\[ t= \frac{8.0\pm\sqrt{8.0^2-2(9.8)(1.50)}}{9.8} \]

Therefore:

\[ \boxed{t=0.216\,\mathrm{s}\quad\text{or}\quad 1.417\,\mathrm{s}} \]

The model predicts two interruptions at gate A.

The first occurs as the ball rises through 1.50 m. The second occurs as it falls through 1.50 m.

We can repeat the calculation for gate B.

The following table shows the predicted readings.

Illustrative idealised predictions, not measurements from an actual experiment.

Notice the order of the interruptions.

During the upward journey:

1. The ball passes gate A.
2. It passes gate B.

During the downward journey:

1. It passes gate B.
2. It passes gate A.

This is exactly what the uniform-acceleration model predicts.

Why revise the original design?

A single optical gate can record two interruptions, but it does not independently identify the direction of motion at each interruption.

Adding a second gate solves this measurement problem. The order in which the gates trigger identifies whether the ball is travelling upwards or downwards.

The predicted readings support the idea that one position can correspond to two different times.

In an actual investigation, the measured times might differ slightly from the predictions. Air resistance, variations in launch speed, gate width, and timing uncertainty could all contribute.

The experiment tests the idealised model; it does not guarantee that every real flight follows it perfectly.

05How to interpret the roots in an HSC question

When you solve a quadratic kinematics equation, don’t immediately write down both answers and move on.

Use this decision process.

Step 1: Identify the time origin.

Ask when \(t=0\) occurs.

Is it when the ball is released, when it passes a marker, or when another event happens?

Step 2: Solve the quadratic equation.

Calculate both mathematical roots, keeping their signs.

Step 3: Check the physical interval.

If the question describes motion after release, a negative elapsed time is outside that interval.

If the object was already moving before your chosen time origin, a negative root might describe a real earlier event.

Step 4: Identify the direction of motion.

Use:

\[ v=u+at \]

A positive velocity means motion in the positive direction. A negative velocity means motion in the opposite direction.

Step 5: Check whether anything interrupted the motion.

A collision, a catch, or an impact can change the velocity. If that happens, the original uniform-acceleration equation cannot describe the entire motion across the event.

For more practice with assigning signs consistently, see How to Choose Positive Direction in HSC Vertical Motion.

06Worked examples

07Questions and solutions

Use \(g=9.8\,\mathrm{m\,s^{-2}}\) throughout. Unless otherwise stated, ignore air resistance and assume uniform gravitational acceleration.

Question 1

A ball is thrown vertically upwards from ground level with an initial velocity of \(14.0\,\mathrm{m\,s^{-1}}\).

Calculate the two times at which the ball is 8.0 m above its launch position.

Solution 1

The ball reaches 8.0 m at \(0.790\,\mathrm{s}\) while rising and \(2.07\,\mathrm{s}\) while falling.

Step 1: Identify the known quantities.

Take upwards as positive.

\[ \begin{aligned} u&=+14.0\,\mathrm{m\,s^{-1}}\\ s&=+8.0\,\mathrm{m}\\ a&=-9.8\,\mathrm{m\,s^{-2}} \end{aligned} \]

Step 2: Substitute into the displacement equation.

\[ 8.0=14.0t-4.9t^2 \]

Rearranging:

\[ 4.9t^2-14.0t+8.0=0 \]

Step 3: Apply the quadratic formula.

\[ t=\frac{14.0\pm\sqrt{14.0^2-4(4.9)(8.0)}}{9.8} \]

\[ t=\frac{14.0\pm\sqrt{39.2}}{9.8} \]

Therefore:

\[ \boxed{t_1=0.790\,\mathrm{s}} \]

\[ \boxed{t_2=2.07\,\mathrm{s}} \]

Step 4: Interpret the answers.

The first root represents the upward crossing, and the second represents the downward crossing.

Both are positive and occur after launch.

The ball’s maximum height is approximately 10.0 m, so reaching 8.0 m twice is physically possible.

Question 2

A ball is launched vertically upwards from the roof of a building, 15.0 m above the ground, with an initial velocity of \(12.0\,\mathrm{m\,s^{-1}}\).

A horizontal sensor is positioned 20.0 m above the ground.

Calculate:

a) The two times at which the ball passes the sensor.

b) The ball’s velocity at each crossing.

Take upwards as positive and ignore air resistance.

Solution 2

The ball passes the sensor at \(0.532\,\mathrm{s}\) while rising and \(1.92\,\mathrm{s}\) while falling. Its velocities at these times are approximately \(+6.78\,\mathrm{m\,s^{-1}}\) and \(-6.78\,\mathrm{m\,s^{-1}}\), respectively.

Step 1: Find the displacement from the roof.

The sensor is at 20.0 m, but the ball starts at 15.0 m.

Therefore:

\[ s=20.0-15.0 \]

\[ s=+5.0\,\mathrm{m} \]

The initial velocity and acceleration are:

\[ \begin{aligned} u&=+12.0\,\mathrm{m\,s^{-1}}\\ a&=-9.8\,\mathrm{m\,s^{-2}} \end{aligned} \]

Step 2: Apply the displacement equation.

\[ 5.0=12.0t-4.9t^2 \]

Rearranging:

\[ 4.9t^2-12.0t+5.0=0 \]

Step 3: Solve for time.

\[ t=\frac{12.0\pm\sqrt{12.0^2-4(4.9)(5.0)}}{9.8} \]

\[ t=\frac{12.0\pm\sqrt{46}}{9.8} \]

Therefore:

\[ \boxed{t_1=0.532\,\mathrm{s}} \]

\[ \boxed{t_2=1.92\,\mathrm{s}} \]

Both roots are positive, so both describe future crossings.

Step 4: Calculate the velocity at each crossing.

Use:

\[ v=u+at \]

For the first crossing:

\[ v_1=12.0-9.8(0.532) \]

\[ \boxed{v_1=+6.78\,\mathrm{m\,s^{-1}}} \]

For the second crossing:

\[ v_2=12.0-9.8(1.92) \]

\[ \boxed{v_2=-6.78\,\mathrm{m\,s^{-1}}} \]

The positive velocity indicates upward motion. The negative velocity indicates downward motion.

The important trap: The 20.0 m given in the question is a height above ground, not the displacement from the launch point. Using \(s=20.0\) m would give the wrong equation.

Question 3

A ball is launched vertically upwards from a position defined as \(y=0\).

Two optical gates are positioned at unknown heights above the launch point.

The following idealised timing data are recorded:

Assume the ball experiences constant downward acceleration of \(9.8\,\mathrm{m\,s^{-2}}\), with no air resistance.

Calculate:

a) The ball’s initial upward velocity.

b) The height of each optical gate above the launch point.

c) The maximum height reached by the ball.

d) Whether the timing data are consistent with one uninterrupted uniform-acceleration flight.

Solution 3

The initial upward velocity is \(14.7\,\mathrm{m\,s^{-1}}\). Gate A is 5.096 m above the launch point, Gate B is 8.624 m above it, and the maximum height is 11.025 m. The timing data are consistent with one uninterrupted uniform-acceleration flight.

This question is more interesting because we are working backwards from the crossing times rather than calculating times from known heights.

Step 1: Identify the relationship between the two roots.

The displacement equation is:

\[ s=ut-4.9t^2 \]

Rearranging:

\[ 4.9t^2-ut+s=0 \]

For a quadratic with roots \(t_1\) and \(t_2\), the sum of the roots is:

\[ t_1+t_2=\frac{u}{4.9} \]

Therefore:

\[ u=4.9(t_1+t_2) \]

Step 2: Calculate the initial velocity using Gate A.

The two recorded times are:

\[ t_1=0.40\,\mathrm{s} \]

\[ t_2=2.60\,\mathrm{s} \]

Substituting:

\[ u=4.9(0.40+2.60) \]

\[ u=4.9(3.00) \]

\[ \boxed{u=14.7\,\mathrm{m\,s^{-1}}} \]

We can check this using Gate B:

\[ u=4.9(0.80+2.20) \]

\[ u=14.7\,\mathrm{m\,s^{-1}} \]

Both gates give the same initial velocity.

Step 3: Calculate the height of Gate A.

The product of the roots of the quadratic is:

\[ t_1t_2=\frac{s}{4.9} \]

Rearranging:

\[ s=4.9t_1t_2 \]

For Gate A:

\[ s_A=4.9(0.40)(2.60) \]

\[ \boxed{s_A=5.096\,\mathrm{m}} \]

Step 4: Calculate the height of Gate B.

For Gate B:

\[ s_B=4.9(0.80)(2.20) \]

\[ \boxed{s_B=8.624\,\mathrm{m}} \]

Both gates are above the launch point and below the ball’s maximum height.

Step 5: Calculate the maximum height.

First, find the time at the highest point:

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

\[ t_{\text{top}}=\frac{14.7}{9.8} \]

\[ t_{\text{top}}=1.50\,\mathrm{s} \]

Now use the displacement equation:

\[ s=ut-4.9t^2 \]

\[ s=14.7(1.50)-4.9(1.50)^2 \]

\[ s=22.05-11.025 \]

\[ \boxed{s_{\max}=11.025\,\mathrm{m}} \]

Step 6: Check the consistency of the timing data.

Both pairs of crossing times have the same midpoint:

\[ \frac{0.40+2.60}{2}=1.50\,\mathrm{s} \]

\[ \frac{0.80+2.20}{2}=1.50\,\mathrm{s} \]

This is the predicted time of the highest point.

The lower gate is crossed at 0.40 s and 2.60 s, while the upper gate is crossed at 0.80 s and 2.20 s.

The order is physically sensible:

• On the way up, the ball passes A before B.
• On the way down, it passes B before A.

The data are therefore consistent with the same launch speed, the same gravitational acceleration, and one uninterrupted flight.

The deeper idea: The sum of the two roots tells us about the initial velocity, while their product tells us about the displacement of the sensor. Two crossing times contain more information than just when an object reaches a particular position.

Question 4

A small trolley moves along a straight horizontal track.

At \(t=0\), it is at \(x=0\), travelling to the right at \(5.0\,\mathrm{m\,s^{-1}}\). It experiences a constant acceleration of \(-2.0\,\mathrm{m\,s^{-2}}\).

The track extends in both directions, with no obstacles.

Calculate:

a) The two times at which the trolley is at \(x=+6.0\) m.

b) The velocity of the trolley at each time.

c) Explain why the trolley can occupy the same position twice despite experiencing constant acceleration.

Solution 4

The trolley reaches \(x=+6.0\) m at \(2.0\,\mathrm{s}\) and \(3.0\,\mathrm{s}\). It is moving right at the first crossing and left at the second.

Step 1: Identify the known quantities.

Take right as positive.

\[ \begin{aligned} u&=+5.0\,\mathrm{m\,s^{-1}}\\ s&=+6.0\,\mathrm{m}\\ a&=-2.0\,\mathrm{m\,s^{-2}} \end{aligned} \]

Here, \(s\) is the trolley’s displacement from its starting position.

Step 2: Substitute into the displacement equation.

\[ s=ut+\frac12at^2 \]

\[ 6.0=5.0t-1.0t^2 \]

Rearranging:

\[ t^2-5.0t+6.0=0 \]

Step 3: Factorise.

\[ (t-2)(t-3)=0 \]

Therefore:

\[ \boxed{t_1=2.0\,\mathrm{s}} \]

\[ \boxed{t_2=3.0\,\mathrm{s}} \]

Step 4: Calculate the velocity at each crossing.

Use:

\[ v=u+at \]

At \(t=2.0\) s:

\[ v_1=5.0-2.0(2.0) \]

\[ \boxed{v_1=+1.0\,\mathrm{m\,s^{-1}}} \]

At \(t=3.0\) s:

\[ v_2=5.0-2.0(3.0) \]

\[ \boxed{v_2=-1.0\,\mathrm{m\,s^{-1}}} \]

Step 5: Explain why the trolley returns to the same position.

The acceleration points left, opposite to the initial velocity.

Initially, the trolley is travelling right but slowing down.

Eventually, it reaches zero velocity and reverses direction.

We can find when this happens:

\[ 0=5.0-2.0t \]

\[ t=2.5\,\mathrm{s} \]

Its position at that instant is:

\[ s=5.0(2.5)-1.0(2.5)^2 \]

\[ s=12.5-6.25 \]

\[ s=6.25\,\mathrm{m} \]

The trolley travels slightly beyond the 6.0 m marker, turns around, and crosses the marker again.

The tempting mistake is to assume negative acceleration means the trolley must be moving left. It doesn’t. Acceleration describes how velocity changes, not necessarily the direction of current motion.

Question 5

A ball is at \(y=0\) at \(t=0\), moving upwards with a velocity of \(+6.0\,\mathrm{m\,s^{-1}}\).

Its acceleration is constant at \(-9.8\,\mathrm{m\,s^{-2}}\).

A horizontal sensor is positioned at \(y=-1.0\) m.

A student uses the displacement equation and obtains two roots:

\[ t_1=-0.149\,\mathrm{s} \]

\[ t_2=1.37\,\mathrm{s} \]

The student claims that the negative root must always be discarded because time cannot be negative.

Is this claim correct?

Consider two different physical histories.

History A: The ball is held stationary at \(y=0\) until \(t=0\), then thrown upwards at \(6.0\,\mathrm{m\,s^{-1}}\).

History B: The ball is already moving freely under gravity at \(t=0\). The instant \(t=0\) is simply when it passes \(y=0\) while travelling upwards. It has been in uninterrupted free flight from at least \(t=-0.20\) s until \(t=1.40\) s.

For each history, determine which sensor crossings actually occur. Explain why the mathematical roots are identical but their physical interpretations differ.

Solution 5

The two mathematical roots are \(-0.149\) s and \(1.37\) s.

The negative root is not a real sensor crossing in History A, but it is a genuine earlier crossing in History B.

This is because the equation describes the ball’s motion only over an interval in which the assumed uniform acceleration applies.

Step 1: Establish the displacement equation.

Upwards is positive.

\[ \begin{aligned} u&=+6.0\,\mathrm{m\,s^{-1}}\\ s&=-1.0\,\mathrm{m}\\ a&=-9.8\,\mathrm{m\,s^{-2}} \end{aligned} \]

Substitute:

\[ -1.0=6.0t-4.9t^2 \]

Rearranging:

\[ 4.9t^2-6.0t-1.0=0 \]

Step 2: Solve the quadratic.

\[ t= \frac{6.0\pm\sqrt{6.0^2-4(4.9)(-1.0)}}{9.8} \]

\[ t=\frac{6.0\pm\sqrt{55.6}}{9.8} \]

Therefore:

\[ \boxed{t_1=-0.149\,\mathrm{s}} \]

\[ \boxed{t_2=1.37\,\mathrm{s}} \]

The algebra is complete, but we still need to interpret the roots.

Step 3: Consider History A.

In History A, the ball is held stationary at \(y=0\) before \(t=0\).

It is only launched at \(t=0\).

The free-flight equation describes the ball’s motion after this launch. It does not describe the earlier period when the ball was being held.

Therefore, the negative root is outside the physical interval of the model.

The ball crosses the sensor only once:

\[ \boxed{t=1.37\,\mathrm{s}} \]

It passes the sensor while falling.

Step 4: Consider History B.

In History B, the ball is already moving freely under gravity at \(t=0\).

The time origin is simply a reference instant, not the moment of launch.

The ball was therefore also following the same uniform-acceleration motion before \(t=0\).

The negative root is now physically meaningful.

At:

\[ t=-0.149\,\mathrm{s} \]

the ball crosses the sensor while travelling upwards.

It then passes \(y=0\) at \(t=0\), reaches its highest point, and eventually returns to the sensor at:

\[ t=1.37\,\mathrm{s} \]

Both crossings are physically real.

Step 5: Identify the underlying misconception.

The student’s mistake is assuming that negative time is automatically impossible.

Time is measured relative to a chosen origin. An event can occur before that origin.

What matters is whether the object was actually following the assumed motion during that earlier interval.

The same equation produces the same mathematical roots in both histories because the position, velocity, and acceleration at \(t=0\) are identical.

However, the physical history before \(t=0\) is different.

That is why one root is invalid in History A but valid in History B.

This is an important distinction in physics: a mathematical solution is not automatically a description of an actual physical event.

08What to remember when a quadratic gives two times

When a uniform-acceleration displacement equation produces two roots, it is usually describing two possible instants at which an object occupies the same position.

For an object thrown vertically upwards, these are commonly the upward and downward crossings.

The key ideas are:

• The displacement equation is quadratic in time because acceleration is uniform.
• Two distinct roots can represent two crossings of the same position.
• At the earlier crossing, the object may be moving upwards; at the later crossing, it may be moving downwards.
• A repeated root represents the turning point.
• A negative root represents an event before the chosen time origin, if the mathematical model is extended backwards.
• Only roots that belong to the actual physical interval should be accepted.

Once you have identified the valid time or times, use \(v=u+at\) to determine the direction of motion at each instant.

That connection between displacement, velocity, and time is the next step in understanding velocity-time graphs and analysing turning points in HSC kinematics.