How to Find the Time Between Two Passes of the Same Height

A ball is thrown vertically upwards and passes a horizontal marker twice: once on the way up and once on the way down. If its initial vertical velocity is \(u\), the marker is \(h\) metres above the launch point, and gravitational acceleration has magnitude \(g\), the time between the two passes is:

\[
\boxed{\Delta t=\frac{2\sqrt{u^2-2gh}}{g}}
\]

This equation gives the time interval directly. You don’t need to calculate both crossing times separately.

It assumes that the ball reaches the marker twice during one uninterrupted flight, air resistance is negligible, and gravitational acceleration is constant. The key is recognising that both crossings are symmetric about the highest point of the motion.

01The experiment: timing two passes at the same height

Imagine throwing a ball vertically upwards through a fixed horizontal line. The ball crosses that line while rising, reaches its highest point, then crosses the same line while falling.

How could you measure the time between those two crossings?

Setting up the measurement

A suitable investigation would use:

• A small ball thrown vertically upwards.
• A vertical metre ruler positioned behind the ball’s path.
• A horizontal reference line marked at a known height above the launch point.
• A high-speed camera positioned side-on to record the motion.

The ruler and the ball’s path should lie in the same vertical plane. The camera should face this plane directly to minimise parallax error.

The camera records the ball throughout its flight. By tracking the centre of the ball, we can identify the instant it crosses the reference line on the way up and the instant it crosses again on the way down.

The difference between these timestamps is the required time interval.

What should the experiment show?

Before recording, consider two possible predictions.

One prediction is that the ball spends more time travelling through the region above the marker than below it. Another is that the two crossings occur at equal time intervals on either side of the highest point.

The second prediction is correct under the uniform-acceleration model.

Gravity acts downwards throughout the flight. As the ball rises, its upward velocity decreases. At the highest point, its vertical velocity reaches zero momentarily. It then gains downward velocity as it falls.

For a given height, the ball has the same speed on the way up and down, but its velocity has the opposite sign.

This means the time taken to travel from the marker to the highest point equals the time taken to return from the highest point to that marker.

Illustrative results

Suppose the ball is launched upwards at \(8.00\,\mathrm{m\,s^{-1}}\), and the reference line is \(1.20\,\mathrm{m}\) above the launch point.

Take gravitational acceleration as \(g=9.80\,\mathrm{m\,s^{-2}}\).

The idealised model predicts:

The interval between crossings is therefore:

\[
\Delta t=1.466-0.167
\]

\[
\Delta t\approx1.299\,\mathrm{s}
\]

These are illustrative theoretical values, not measurements from an actual experiment.

Notice that the average of the two crossing times is approximately \(0.816\,\mathrm{s}\), which is the time at the highest point.

That relationship is the experimental pattern we would look for.

Why the measurement method matters

A handheld stopwatch is not ideal for this investigation. Human reaction time makes it difficult to identify the precise instant of each crossing.

A low-frame-rate video can also introduce uncertainty because the ball may cross the reference line between two recorded frames.

A better approach is to use a high-frame-rate recording, a fixed camera, and frame-by-frame tracking of the ball’s centre. Repeating the trial also helps identify inconsistent readings.

In a real experiment, air resistance and measurement uncertainty mean the two crossing times might not be perfectly symmetric. The ideal mathematical model assumes negligible air resistance.

02Why the two crossing times are symmetric

We can now derive the equation that gives the time between the two passes.

Step 1: Choose a positive direction

Take upwards as positive.

For an object moving vertically under gravity:

• Initial vertical velocity is positive.
• Gravitational acceleration is negative.
• Upward displacement is positive.
• Downward velocity is negative.

Therefore:

\[
a=-g
\]

where \(g=9.8\,\mathrm{m\,s^{-2}}\) is the positive magnitude of gravitational acceleration.

If you need a fuller explanation of choosing signs in vertical-motion problems, see How to Choose Positive Direction in HSC Vertical Motion.

Step 2: Write the displacement equation

The uniform-acceleration displacement equation is:

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

For vertical motion, substitute \(a=-g\):

\[
s=ut-\frac12gt^2
\]

Here:

• \(s\) is vertical displacement from the launch point, in metres.
• \(u\) is initial vertical velocity, in \(\mathrm{m\,s^{-1}}\).
• \(t\) is elapsed time after launch, in seconds.
• \(g\) is gravitational acceleration magnitude, in \(\mathrm{m\,s^{-2}}\).

Let the marker be at vertical displacement \(h\) above the launch point.

At both crossings, the displacement is the same:

\[
h=ut-\frac12gt^2
\]

Rearranging:

\[
\frac12gt^2-ut+h=0
\]

This is a quadratic equation in time.

Step 3: Find the two roots

Using the quadratic formula:

\[
t=\frac{-b\pm\sqrt{b^2-4ac}}{2a}
\]

we obtain:

\[
t_1=\frac{u-\sqrt{u^2-2gh}}{g}
\]

\[
t_2=\frac{u+\sqrt{u^2-2gh}}{g}
\]

The smaller root represents the earlier crossing, when the object is moving upwards.

The larger root represents the later crossing, when the object is moving downwards.

Both roots are positive when the marker is above or at the launch height and below the maximum height.

Step 4: Subtract the crossing times

The question asks for the time between the two passes, not the time from launch to either pass.

Therefore:

\[
\Delta t=t_2-t_1
\]

Substitute the two roots:

\[
\Delta t=
\frac{u+\sqrt{u^2-2gh}}{g}
–
\frac{u-\sqrt{u^2-2gh}}{g}
\]

The \(u\) terms cancel:

\[
\boxed{\Delta t=\frac{2\sqrt{u^2-2gh}}{g}}
\]

This is the direct formula for the time between two passes of the same height.

An alternative formula using maximum height

Sometimes a question gives the maximum height rather than the initial velocity.

We can use:

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

where \(H\) is the maximum vertical rise above the launch point.

Rearranging:

\[
u^2=2gH
\]

Substitute into the time-interval equation:

\[
\Delta t=\frac{2\sqrt{2gH-2gh}}{g}
\]

Therefore:

\[
\boxed{\Delta t=2\sqrt{\frac{2(H-h)}{g}}}
\]

This version makes the physical meaning especially clear.

The time interval depends on how far the marker is below the maximum height.

The closer the marker is to the highest point, the shorter the interval between crossings.

03Check whether two physical crossings are possible

A quadratic equation can produce two mathematical roots without necessarily giving two future events.

This is one of the most important things to check.

The maximum vertical rise is:

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

For an object launched upwards, the possibilities are:

The important misconception: a negative root is not a second future pass

Consider a ball thrown upwards at \(8.0\,\mathrm{m\,s^{-1}}\). A marker is positioned \(1.00\,\mathrm{m}\) below the launch point.

Substituting \(h=-1.00\,\mathrm{m}\):

\[
t=\frac{8.0\pm\sqrt{8.0^2-2(9.8)(-1.00)}}{9.8}
\]

\[
t=\frac{8.0\pm\sqrt{83.6}}{9.8}
\]

The two roots are approximately:

\[
t_1=-0.116\,\mathrm{s}
\]

\[
t_2=1.75\,\mathrm{s}
\]

Only the second root represents an event after launch.

The negative root describes when the ball would have been at that height if we extended the mathematical model backwards in time.

The ball starts above the marker and initially moves upwards, so it cannot cross the lower marker on its way up.

It passes that height only once, while falling.

Never turn a negative time into a positive time by taking its absolute value. The sign tells you whether the event occurs before or after your chosen starting instant.

04Worked examples

05Questions and solutions

Use \(g=9.8\,\mathrm{m\,s^{-2}}\) throughout unless stated otherwise. Ignore air resistance except where a collision or other interruption is explicitly described.

Question 1

A ball is thrown vertically upwards from ground level with an initial velocity of \(10.0\,\mathrm{m\,s^{-1}}\).

A horizontal marker is positioned \(3.00\,\mathrm{m}\) above the launch point.

Calculate the time between the ball passing the marker while rising and passing it again while falling.

Solution 1

The time between the two crossings is approximately \(1.31\,\mathrm{s}\).

Step 1: Identify the known quantities.

Taking upwards as positive:

\[
u=+10.0\,\mathrm{m\,s^{-1}}
\]

\[
h=3.00\,\mathrm{m}
\]

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

Step 2: Apply the time-interval equation.

\[
\Delta t=\frac{2\sqrt{u^2-2gh}}{g}
\]

Substitute:

\[
\Delta t=
\frac{2\sqrt{10.0^2-2(9.8)(3.00)}}{9.8}
\]

\[
=\frac{2\sqrt{100-58.8}}{9.8}
\]

\[
=\frac{2\sqrt{41.2}}{9.8}
\]

\[
=\frac{12.84}{9.8}
\]

Step 3: State the final answer.

\[
\boxed{\Delta t=1.31\,\mathrm{s}}
\]

The ball takes approximately \(1.31\) seconds between the upward and downward crossings.

Both crossings are possible because its maximum rise is:

\[
H=\frac{10.0^2}{19.6}=5.10\,\mathrm{m}
\]

The marker is below the maximum height.

Question 2

A ball is thrown vertically upwards from a platform.

A high-speed camera records the ball crossing a horizontal marker at the following times after launch:

• First crossing: \(0.30\,\mathrm{s}\).
• Second crossing: \(1.90\,\mathrm{s}\).

Assume that the ball remains in uninterrupted free flight.

Calculate:

a) The time taken to reach maximum height.

b) The initial vertical velocity.

c) The height of the marker above the launch point.

Solution 2

The ball reaches maximum height after \(1.10\,\mathrm{s}\), has an initial vertical velocity of \(10.78\,\mathrm{m\,s^{-1}}\), and passes the marker \(2.79\,\mathrm{m}\) above its launch point.

Step 1: Find the time to maximum height.

The two crossing times are symmetric about the highest point.

Therefore, the apex occurs halfway between them:

\[
t_{\mathrm{apex}}=\frac{t_1+t_2}{2}
\]

Substitute:

\[
t_{\mathrm{apex}}=\frac{0.30+1.90}{2}
\]

\[
\boxed{t_{\mathrm{apex}}=1.10\,\mathrm{s}}
\]

Step 2: Calculate the initial vertical velocity.

At maximum height, the vertical velocity is zero.

Use:

\[
v=u+at
\]

With upwards positive, \(a=-9.8\,\mathrm{m\,s^{-2}}\).

Therefore:

\[
0=u-9.8(1.10)
\]

Rearrange:

\[
u=9.8(1.10)
\]

\[
\boxed{u=10.78\,\mathrm{m\,s^{-1}}}
\]

Step 3: Calculate the marker’s height.

Use the displacement equation:

\[
s=ut-\frac12gt^2
\]

We can use either crossing time. Choose the first crossing at \(t=0.30\,\mathrm{s}\).

\[
s=10.78(0.30)-\frac12(9.8)(0.30)^2
\]

\[
s=3.234-0.441
\]

\[
\boxed{s=2.793\,\mathrm{m}}
\]

The marker is approximately \(2.79\,\mathrm{m}\) above the launch point.

The important reasoning here is that the average of the two crossing times identifies the apex time. This allows us to recover the initial velocity without solving a quadratic equation.

Question 3

A ball is thrown vertically upwards from a balcony \(18.0\,\mathrm{m}\) above the ground with an initial velocity of \(15.0\,\mathrm{m\,s^{-1}}\).

A horizontal cable is positioned \(27.0\,\mathrm{m}\) above the ground.

Calculate:

a) The time between the ball passing the cable on its way up and passing it again on its way down.

b) Whether the ball reaches the ground before or after its second crossing of the cable.

Ignore air resistance.

Solution 3

The ball passes the cable twice, separated by approximately \(1.42\,\mathrm{s}\). It reaches the ground after its second crossing.

Step 1: Calculate the vertical displacement to the cable.

Be careful: the cable’s height is measured from the ground, but the ball’s initial position is already \(18.0\,\mathrm{m}\) above the ground.

Therefore, the displacement from the launch point is:

\[
h=27.0-18.0
\]

\[
h=9.00\,\mathrm{m}
\]

Step 2: Apply the time-interval equation.

The initial velocity is:

\[
u=+15.0\,\mathrm{m\,s^{-1}}
\]

Therefore:

\[
\Delta t=\frac{2\sqrt{15.0^2-2(9.8)(9.00)}}{9.8}
\]

\[
=\frac{2\sqrt{225-176.4}}{9.8}
\]

\[
=\frac{2\sqrt{48.6}}{9.8}
\]

\[
=\frac{13.94}{9.8}
\]

\[
\boxed{\Delta t=1.42\,\mathrm{s}}
\]

Step 3: Check whether the second crossing happens before the ball reaches the ground.

This is an important physical check.

The ball starts \(18.0\,\mathrm{m}\) above the ground, so its displacement when it reaches the ground is:

\[
s=-18.0\,\mathrm{m}
\]

Use:

\[
s=ut-\frac12gt^2
\]

Substitute:

\[
-18.0=15.0t-4.9t^2
\]

Rearrange:

\[
4.9t^2-15.0t-18.0=0
\]

Using the quadratic formula:

\[
t=\frac{15.0\pm\sqrt{15.0^2+4(4.9)(18.0)}}{9.8}
\]

The positive root gives:

\[
t\approx3.98\,\mathrm{s}
\]

Step 4: Interpret the result.

The second crossing of the cable occurs approximately \(2.24\,\mathrm{s}\) after launch, while the ball reaches the ground after approximately \(3.98\,\mathrm{s}\).

Therefore, both crossings occur before the ball reaches the ground.

The calculated interval of \(1.42\,\mathrm{s}\) is physically valid.

The key lesson is to measure displacement from the launch point, not automatically from ground level.

Question 4

A ball is thrown vertically upwards from the floor with an initial velocity of \(12.0\,\mathrm{m\,s^{-1}}\).

A horizontal marker is positioned \(3.00\,\mathrm{m}\) above the floor. The ceiling is \(5.00\,\mathrm{m}\) above the floor.

The ball rises through the marker, hits the ceiling, and rebounds downwards with the same speed it had immediately before impact.

Assume the collision is instantaneous, and ignore air resistance.

Calculate the time between:

• The ball’s first upward crossing of the marker.
• Its next downward crossing of the marker after rebounding from the ceiling.

A student instead uses the usual time-between-passes equation for an uninterrupted flight. Explain why this gives the wrong result.

Solution 4

The actual interval between the two specified crossings is approximately \(0.500\,\mathrm{s}\). The uninterrupted-flight formula is invalid because the collision changes the ball’s velocity before its second crossing.

Step 1: Calculate the ball’s speed at the marker.

Before the collision, the ball is moving upwards under constant acceleration.

Use:

\[
v^2=u^2+2as
\]

With upwards positive:

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

At the marker:

\[
v^2=12.0^2+2(-9.8)(3.00)
\]

\[
v^2=144-58.8
\]

\[
v=\sqrt{85.2}
\]

\[
v=9.23\,\mathrm{m\,s^{-1}}
\]

The ball is moving upwards at \(9.23\,\mathrm{m\,s^{-1}}\) when it first crosses the marker.

Step 2: Calculate the ball’s speed at the ceiling.

The ceiling is \(5.00\,\mathrm{m}\) above the launch point.

\[
v^2=12.0^2+2(-9.8)(5.00)
\]

\[
v^2=144-98
\]

\[
v=\sqrt{46}
\]

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

Immediately before impact, the ball is moving upwards at \(6.78\,\mathrm{m\,s^{-1}}\).

Immediately after the perfectly elastic collision, it moves downwards at the same speed.

Its velocity therefore changes from:

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

to:

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

Step 3: Calculate the time from the first marker crossing to the ceiling.

Use:

\[
v=u+at
\]

For this part of the motion:

\[
9.23\rightarrow6.78\,\mathrm{m\,s^{-1}}
\]

Therefore:

\[
6.78=9.23-9.8t
\]

Rearrange:

\[
t=\frac{9.23-6.78}{9.8}
\]

\[
t\approx0.250\,\mathrm{s}
\]

Step 4: Calculate the time from the ceiling back to the marker.

After the collision, the ball starts moving downwards at \(6.78\,\mathrm{m\,s^{-1}}\).

It accelerates downwards at \(9.8\,\mathrm{m\,s^{-2}}\) until its downward speed reaches \(9.23\,\mathrm{m\,s^{-1}}\).

Using the same velocity relationship:

\[
9.23=6.78+9.8t
\]

\[
t=\frac{9.23-6.78}{9.8}
\]

\[
t\approx0.250\,\mathrm{s}
\]

Step 5: Add the two time intervals.

The time between the upward marker crossing and the next downward marker crossing is:

\[
\Delta t=0.250+0.250
\]

\[
\boxed{\Delta t=0.500\,\mathrm{s}}
\]

Why does the usual formula fail?

If the ball had continued moving freely without hitting the ceiling, it would have reached its natural highest point and then returned to the marker.

In that uninterrupted motion, the time interval would have been:

\[
\Delta t=\frac{2\sqrt{85.2}}{9.8}
\]

\[
\Delta t\approx1.88\,\mathrm{s}
\]

But the ball never reaches that natural highest point. The ceiling interrupts the motion and reverses its velocity.

The original quadratic equation no longer describes the entire flight.

This is the crucial distinction: constant gravitational acceleration does not mean the same velocity equation can be applied across a collision without updating the initial velocity.

You must split the motion into separate stages whenever an impact changes the velocity.

Question 5

A ball is thrown vertically upwards. A high-speed camera records the ball passing two horizontal markers.

The markers are positioned at:

• Marker A: \(2.00\,\mathrm{m}\) above the launch point.
• Marker B: \(4.00\,\mathrm{m}\) above the launch point.

The camera’s clock starts before the ball is launched. The launch occurs at camera time \(0.500\,\mathrm{s}\).

The following crossing times are recorded:

All timestamps are rounded to the nearest \(0.001\,\mathrm{s}\).

Assume the ball remains in uninterrupted free flight and gravitational acceleration is constant.

A student makes the following claim:

“Both markers are crossed twice, so the time intervals between their crossings should be equal. The ball is following the same trajectory.”

Determine whether the recorded data are consistent with a single uniform gravitational acceleration and initial vertical velocity.

If they are consistent, calculate:

a) The gravitational acceleration inferred from the measurements.

b) The initial vertical velocity of the ball.

c) The time of maximum height relative to the launch.

Explain why the student’s claim about the two time intervals is incorrect.

Solution 5

The measurements are consistent with a single gravitational acceleration of approximately \(9.80\,\mathrm{m\,s^{-2}}\) and an initial vertical velocity of \(11.76\,\mathrm{m\,s^{-1}}\). The student’s claim is incorrect: the two crossings at each individual height are symmetric about the apex, but different heights produce different time intervals.

This problem requires combining the symmetry of the motion with the relationship between height and crossing time.

Step 1: Calculate the time interval at each marker.

For marker A:

\[
\Delta t_A=2.716-0.684
\]

\[
\Delta t_A=2.032\,\mathrm{s}
\]

For marker B:

\[
\Delta t_B=2.490-0.910
\]

\[
\Delta t_B=1.580\,\mathrm{s}
\]

The intervals are already different.

This does not mean the measurements are inconsistent. In fact, a higher marker should produce a shorter interval because it is closer to the apex.

Step 2: Find the relationship between the two intervals and gravitational acceleration.

Recall the alternative time-interval equation:

\[
\Delta t=2\sqrt{\frac{2(H-h)}{g}}
\]

Squaring both sides:

\[
\Delta t^2=\frac{8(H-h)}{g}
\]

For marker A:

\[
\Delta t_A^2=\frac{8(H-h_A)}{g}
\]

For marker B:

\[
\Delta t_B^2=\frac{8(H-h_B)}{g}
\]

Subtract the second equation from the first:

\[
\Delta t_A^2-\Delta t_B^2
=
\frac{8(h_B-h_A)}{g}
\]

Notice that the maximum height \(H\) cancels.

This is useful because we can determine \(g\) without knowing the maximum height.

Rearrange:

\[
g=\frac{8(h_B-h_A)}
{\Delta t_A^2-\Delta t_B^2}
\]

Step 3: Substitute the measurements.

\[
g=
\frac{8(4.00-2.00)}
{2.032^2-1.580^2}
\]

\[
g=\frac{16.0}{4.129024-2.496400}
\]

\[
g=\frac{16.0}{1.632624}
\]

Therefore:

\[
\boxed{g\approx9.80\,\mathrm{m\,s^{-2}}}
\]

The inferred gravitational acceleration is consistent with the expected value used in the model.

Step 4: Find the time of maximum height relative to launch.

The two crossings of either individual marker are symmetric about the apex.

Using marker A:

\[
t_{\mathrm{apex,clock}}
=
\frac{0.684+2.716}{2}
\]

\[
t_{\mathrm{apex,clock}}=1.700\,\mathrm{s}
\]

However, remember that the camera started before launch.

The launch occurred at camera time \(0.500\,\mathrm{s}\).

Therefore, the time from launch to maximum height is:

\[
t_{\mathrm{apex}}=1.700-0.500
\]

\[
\boxed{t_{\mathrm{apex}}=1.200\,\mathrm{s}}
\]

We could have used marker B instead. Its two crossing times also have a midpoint of \(1.700\,\mathrm{s}\).

That agreement is another consistency check.

Step 5: Calculate the initial vertical velocity.

At maximum height:

\[
v=u-gt
\]

Since \(v=0\):

\[
0=u-gt_{\mathrm{apex}}
\]

Rearrange:

\[
u=gt_{\mathrm{apex}}
\]

Substitute:

\[
u=9.80(1.200)
\]

\[
\boxed{u=11.76\,\mathrm{m\,s^{-1}}}
\]

The ball’s initial vertical velocity is approximately \(11.76\,\mathrm{m\,s^{-1}}\) upwards.

Step 6: Check whether the reconstructed motion agrees with both marker heights.

Use the displacement equation:

\[
h=ut-\frac12gt^2
\]

For marker A, the first crossing occurs \(0.684-0.500=0.184\,\mathrm{s}\) after launch.

\[
h_A=11.76(0.184)-4.9(0.184)^2
\]

\[
h_A\approx1.998\,\mathrm{m}
\]

This agrees with the supplied height of \(2.00\,\mathrm{m}\).

For marker B, the first crossing occurs \(0.910-0.500=0.410\,\mathrm{s}\) after launch.

\[
h_B=11.76(0.410)-4.9(0.410)^2
\]

\[
h_B\approx3.998\,\mathrm{m}
\]

This agrees with the supplied height of \(4.00\,\mathrm{m}\).

The small differences are explained by the timestamps being rounded to the nearest millisecond.

Why is the student’s claim incorrect?

The student’s reasoning confuses two different kinds of symmetry.

For each individual marker, the upward and downward crossings are symmetric about the highest point. Their average time is always the apex time.

However, the distance between the two crossings depends on the marker’s height.

Marker B is higher than marker A, so it is closer to the apex. The ball has less distance to travel between its upward and downward crossings of marker B.

Consequently:

\[
\Delta t_A=2.032\,\mathrm{s}
\]

\[
\Delta t_B=1.580\,\mathrm{s}
\]

The intervals should not be equal.

The deeper result is that both pairs of crossing times share the same midpoint, even though their separations are different.

06What to remember when solving these problems

When a question asks for the time between two passes of the same height, start by identifying the initial vertical velocity and the marker’s displacement relative to the launch point.

For an uninterrupted upward launch under uniform gravitational acceleration, use:

\[
\boxed{\Delta t=\frac{2\sqrt{u^2-2gh}}{g}}
\]

Or, if maximum height is given:

\[
\boxed{\Delta t=2\sqrt{\frac{2(H-h)}{g}}}
\]

Before accepting the answer, check three things:

• The marker is reachable on both the upward and downward parts of the flight.
• Both roots represent physical events after launch.
• No collision or other interruption changes the motion before the second crossing.

The next useful decision is whether the motion remains one continuous uniform-acceleration stage. If it does, the two-root method works directly. If an impact or another event changes the velocity, divide the motion into separate stages and calculate each interval using its own initial conditions.