Why Acceleration Stays Downward at the Top of a Vertical Throw

A ball is thrown vertically upwards. It slows down, reaches its highest point, and then falls back towards the ground. At the exact highest point, its velocity is zero.

So why is its acceleration still approximately \(9.8\ \mathrm{m\,s^{-2}}\) downwards?

Because the ball has stopped moving upwards, not stopped experiencing gravity. Its weight still acts downwards, producing a downward net force. The ball’s velocity is momentarily zero, but that velocity is still changing.

This distinction is the key to understanding vertical throws. Zero velocity does not mean zero acceleration.

Imagine freezing the ball at the very top of its flight. What will its velocity be a moment later? Downwards. Gravity must therefore be accelerating it downwards as it passes through the turning point.

01The highest point is a turning point, not a stopping point

There are two different quantities to keep separate:

• Velocity describes how quickly an object’s position changes, including its direction.
• Acceleration describes how quickly its velocity changes.

During a vertical throw, the ball behaves as follows:

Here, upwards is defined as positive, so downwards is negative.

Notice that the acceleration never changes direction.

The ball’s velocity changes from positive to zero and then negative. Gravity causes this continuous change.

What happens immediately around the highest point?

Consider an idealised ball with negligible air resistance. Suppose we examine its motion in intervals of \(0.10\) seconds around the highest point.

The following values are illustrative, using \(g=9.8\ \mathrm{m\,s^{-2}}\).

The velocity changes by \(-0.98\ \mathrm{m\,s^{-1}}\) over each \(0.10\)-second interval.

Therefore:

\[ a=\frac{\Delta v}{\Delta t} \]

\[ a=\frac{-0.98}{0.10} \]

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

The negative sign tells us that acceleration points downwards.

The important observation is that the ball’s velocity passes through zero while its acceleration remains negative.

02Gravity does not disappear at the top

Let’s look at the forces acting on the ball.

While the ball is still in your hand, your hand applies an upward force. Once the ball leaves your hand, that contact force disappears.

Assuming air resistance is negligible, the only significant force remaining is the ball’s weight.

Weight is the gravitational force acting on an object:

\[ W=mg \]

where:

• \(W\) is weight in newtons (\(\mathrm{N}\)),
• \(m\) is mass in kilograms (\(\mathrm{kg}\)),
• \(g\) is the magnitude of gravitational acceleration, approximately \(9.8\ \mathrm{m\,s^{-2}}\) near Earth’s surface.

Newton’s second law states:

\[ F_{\mathrm{net}}=ma \]

Taking upwards as positive, the net vertical force is:

\[ -mg=ma \]

Cancel the mass:

\[ a=-g \]

Therefore:

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

This remains true at the highest point.

The ball does not need to be moving downwards for gravity to accelerate it downwards. The force is already acting while the ball travels upwards.

A useful comparison: a ball resting on a table

Here is a situation where an object really does have zero velocity and zero acceleration.

Place a ball on a horizontal table. It remains stationary because two forces balance:

• Weight acts downwards.
• The table exerts an upward normal force.

The net force is zero, so the acceleration is zero.

Compare the two situations:

The difference is the net force.

A ball on a table has a supporting force balancing its weight. A ball at the top of a vertical throw has no supporting surface, so gravity remains unbalanced.

This is why simply knowing that an object’s velocity is zero tells you nothing definite about its acceleration. You must consider the forces acting on it.

For a broader treatment of this distinction, see Can an Object Have Zero Velocity and Nonzero Acceleration?.

What about air resistance?

The constant-acceleration model assumes air resistance is negligible. This is the usual assumption in basic HSC vertical-motion calculations.

In reality, air resistance can change the acceleration during the flight.

For a ball moving through still air:

• While rising, air resistance acts downwards, in the same direction as weight.
• At the exact highest point, the ball’s speed relative to the air is zero, so air resistance is zero.
• While falling, air resistance acts upwards, opposing the motion.

Consequently, the acceleration is not generally constant throughout a real flight.

However, at the exact highest point, gravity is still the only significant force in this situation. The instantaneous acceleration is therefore still approximately \(9.8\ \mathrm{m\,s^{-2}}\) downwards.

The distinction is important: the constant-acceleration equations are an idealised model, but the downward acceleration at the turning point is not an artificial mathematical trick.

03How could an experiment test this?

A photograph of a ball at its highest point cannot tell us its acceleration. It only tells us where the ball is at one instant.

To investigate acceleration, we need to measure how velocity changes over time.

One way to do this is with a vertical launcher and two photogates.

Setting up the apparatus

Imagine a ball launched vertically upwards from a fixed point.

Attach a small, lightweight opaque marker to the ball. Arrange two photogates across its path:

• The lower photogate is \(1.00\ \mathrm{m}\) above the launch point.
• The upper photogate is \(2.00\ \mathrm{m}\) above the launch point.

A digital timer records when the ball interrupts each light beam.

The ball passes through the lower gate, then the upper gate, before reaching its highest point. It subsequently passes through the upper gate and lower gate again while descending.

The photogates measure the time for the marker to interrupt each beam. Knowing the marker’s length allows us to estimate the ball’s speed at each gate.

Why isn’t one photogate enough?

Suppose we initially use only one photogate.

The ball passes through it while rising and later passes through the same gate while falling.

Under the ideal free-fall model, the speed magnitudes at those two crossings are equal. However, the directions are opposite.

This tells us something about the velocity, but it does not directly measure how that velocity changes between the two crossings.

The revised design uses two gates at different heights. We can now compare the velocities at separate positions and measure the time between them.

Illustrative photogate data

Suppose the ball is launched at \(7.00\ \mathrm{m\,s^{-1}}\), with air resistance neglected.

Its maximum rise is:

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

\[ h_{\max}=\frac{7.00^2}{2(9.8)} \]

\[ h_{\max}=2.50\ \mathrm{m} \]

Both photogates are therefore below the highest point.

The idealised model predicts the following velocities:

These are illustrative theoretical values, not measurements from a performed experiment.

The time between the two gate crossings is approximately \(0.234\ \mathrm{s}\) on each pass.

We can estimate the acceleration during the upward pass:

\[ a=\frac{v_f-v_i}{\Delta t} \]

\[ a=\frac{3.13-5.42}{0.234} \]

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

Now consider the downward pass:

\[ a=\frac{-5.42-(-3.13)}{0.234} \]

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

The acceleration is downwards on both sides of the turning point.

The photogates cannot directly measure acceleration at one mathematical instant. They measure changes over finite time intervals. The force analysis from the previous section tells us what happens at the exact highest point.

A real investigation would also need to account for timing uncertainty, gate alignment, the marker’s size, and air resistance.

The important experimental distinction is between measuring a velocity at one position and measuring how velocity changes between positions.

04Using uniform acceleration equations at the highest point

We can now describe the motion mathematically.

For the following calculations, use these assumptions:

• The ball moves vertically.
• Air resistance is negligible.
• Gravitational acceleration is constant.
• Upwards is the positive direction.
• \(g=9.8\ \mathrm{m\,s^{-2}}\).

This gives:

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

The three uniform-acceleration equations are:

\[ v=u+at \]

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

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

where:

• \(u\) is initial velocity in \(\mathrm{m\,s^{-1}}\),
• \(v\) is final velocity in \(\mathrm{m\,s^{-1}}\),
• \(a\) is constant acceleration in \(\mathrm{m\,s^{-2}}\),
• \(t\) is elapsed time in seconds,
• \(s\) is displacement in metres.

For a vertical throw with upwards positive:

\[ v=u-9.8t \]

At the highest point, the vertical velocity is zero.

Therefore:

\[ 0=u-9.8t \]

Rearranging gives:

\[ \boxed{t_{\mathrm{top}}=\frac{u}{9.8}} \]

This calculates the time taken to reach the highest point.

Notice what we have done. We have set velocity equal to zero, not acceleration.

The acceleration remains:

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

What does the velocity-time graph tell us?

A velocity-time graph makes the distinction particularly clear.

For constant acceleration, the gradient of a velocity-time graph represents acceleration.

The ball’s vertical velocity decreases linearly as time increases. The graph crosses the horizontal axis at the highest point, but its gradient does not become zero.

Remember:

• The vertical coordinate represents velocity.
• The gradient represents acceleration.
• Crossing the horizontal axis means velocity is zero.
• A negative gradient means acceleration is downwards when upwards is positive.

The graph does not flatten at the highest point because the ball continues accelerating.

Understanding quadratic roots and the two times at one height

There is another important feature of vertical throws.

A ball can pass the same height twice: once while travelling upwards and again while travelling downwards.

To understand this, start with the displacement equation:

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

Rearranging:

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

This is a quadratic equation in time.

Solving it gives:

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

The two mathematical roots can represent two different physical events.

The negative root in the final case is not an error. It describes the mathematical extension of the trajectory before the chosen launch time.

Always check what the roots represent. A root of the displacement equation is not automatically a time when velocity is zero.

05A reliable way to approach vertical-throw questions

When solving an HSC vertical-motion problem, follow this sequence:

1. Choose a positive direction. Upwards is usually convenient for a ball thrown upwards.
2. Identify the forces. If air resistance is negligible, gravity is the only force after release.
3. Assign the acceleration. With upwards positive, use \(a=-9.8\ \mathrm{m\,s^{-2}}\).
4. Identify the quantity being asked for. At the highest point, set \(v=0\), not \(a=0\).
5. Choose the appropriate equation. Use the velocity equation for time to the apex and the displacement equation for a specified height.
6. Interpret every root. Check whether the time is physically possible and use the velocity sign to identify upward or downward motion.

One final distinction: this article deals with a purely vertical throw, so the ball’s total velocity becomes zero at the highest point. For an angled projectile, only the vertical component of velocity becomes zero. Its horizontal velocity can remain nonzero.

For that situation, see Projectile Motion: Horizontal and Vertical Motion Explained.

06Questions and solutions

The following questions move from straightforward applications towards more demanding combinations of forces, velocity, acceleration, and quadratic roots.

Question 1

A ball is thrown vertically upwards from level ground at \(13.0\ \mathrm{m\,s^{-1}}\).

Neglect air resistance and take upwards as positive.

At the highest point:

a) What is the ball’s vertical velocity?

b) What is its acceleration?

c) Calculate its velocity \(0.40\) seconds after reaching the highest point.

Solution 1

The ball has zero velocity at the highest point, but its acceleration is \(-9.8\ \mathrm{m\,s^{-2}}\). After \(0.40\) seconds, its velocity is \(-3.92\ \mathrm{m\,s^{-1}}\).

Part a: Velocity at the highest point

At the highest point, the ball momentarily stops moving upwards before descending.

Therefore:

\[ \boxed{v=0\ \mathrm{m\,s^{-1}}} \]

Part b: Acceleration at the highest point

Gravity continues acting downwards throughout the flight.

With upwards positive:

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

Part c: Velocity \(0.40\) seconds later

Use:

\[ v=u+at \]

Take the highest point as the starting instant:

\[ u=0,\qquad a=-9.8 \]

Substitute:

\[ v=0+(-9.8)(0.40) \]

\[ \boxed{v=-3.92\ \mathrm{m\,s^{-1}}} \]

The negative sign indicates downward motion.

The ball does not remain stationary because gravity continues changing its velocity after it reaches the highest point.

Question 2

A ball is thrown vertically upwards from level ground and returns to its launch height \(3.20\) seconds later.

Neglect air resistance and take upwards as positive.

a) Calculate the initial velocity of the ball.

b) A student claims that the two roots of the displacement equation mean the ball has zero velocity both at launch and when it returns to the ground. Explain why this is incorrect.

c) Calculate the actual time at which the ball’s velocity becomes zero.

Solution 2

The initial velocity is \(+15.68\ \mathrm{m\,s^{-1}}\), and the ball reaches zero velocity after \(1.60\) seconds. The two roots of the displacement equation represent the launch and return times, not two zero-velocity events.

Part a: Initial velocity

At the launch height, the displacement is zero.

Using:

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

With \(a=-9.8\):

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

Factorise:

\[ 0=t(u-4.9t) \]

The two roots are:

\[ t=0 \]

and:

\[ t=\frac{u}{4.9} \]

The second root represents the return to the launch height.

We are told this occurs after \(3.20\) seconds:

\[ 3.20=\frac{u}{4.9} \]

Therefore:

\[ u=4.9(3.20) \]

\[ \boxed{u=+15.68\ \mathrm{m\,s^{-1}}} \]

Part b: Why the student’s interpretation is wrong

The displacement equation describes the ball’s position relative to its starting point.

A displacement of zero means the ball is at its original height. It does not mean the ball is stationary.

At launch, the ball is moving upwards.

At \(3.20\) seconds, it is moving downwards.

The student has confused a zero of the displacement equation with a zero of the velocity equation.

Part c: Actual time at zero velocity

Use:

\[ v=u+at \]

At the highest point:

\[ 0=15.68-9.8t \]

Rearrange:

\[ t=\frac{15.68}{9.8} \]

\[ \boxed{t=1.60\ \mathrm{s}} \]

The ball reaches its highest point halfway through its flight.

Its acceleration remains \(-9.8\ \mathrm{m\,s^{-2}}\) at this instant.

Question 3

Two balls move vertically near a building.

Ball A is thrown vertically upwards from ground level at \(19.6\ \mathrm{m\,s^{-1}}\).

At the same instant, ball B is thrown vertically downwards at \(4.9\ \mathrm{m\,s^{-1}}\) from a platform \(29.4\ \mathrm{m}\) above the ground.

Neglect air resistance.

a) Determine whether the balls meet.

b) If they meet, calculate the time after launch and their height above the ground.

c) Calculate the velocity of each ball at the instant they meet.

Solution 3

The balls meet \(1.20\) seconds after launch, at a height of approximately \(16.5\ \mathrm{m}\) above the ground. Ball A is moving upwards at \(7.84\ \mathrm{m\,s^{-1}}\), while ball B is moving downwards at \(16.66\ \mathrm{m\,s^{-1}}\).

Part a: Determine whether the balls meet

Take upwards as positive and measure displacement from ground level.

Both balls experience the same acceleration:

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

For ball A:

\[ y_A=19.6t-4.9t^2 \]

For ball B, the initial height is \(29.4\ \mathrm{m}\), and its initial velocity is negative:

\[ y_B=29.4-4.9t-4.9t^2 \]

The balls meet when their positions are equal.

Therefore:

\[ 19.6t-4.9t^2 = 29.4-4.9t-4.9t^2 \]

Notice that the acceleration terms cancel.

Rearrange:

\[ 19.6t+4.9t=29.4 \]

\[ 24.5t=29.4 \]

\[ \boxed{t=1.20\ \mathrm{s}} \]

Both balls are still in the air at this time, so the meeting is physically possible.

Part b: Find the meeting height

Substitute \(t=1.20\) into ball A’s displacement equation:

\[ y_A=19.6(1.20)-4.9(1.20)^2 \]

\[ y_A=23.52-7.056 \]

\[ \boxed{y=16.46\ \mathrm{m}} \]

The balls meet approximately \(16.5\ \mathrm{m}\) above the ground.

Part c: Find their velocities

For ball A:

\[ v_A=u_A+at \]

\[ v_A=19.6-9.8(1.20) \]

\[ \boxed{v_A=+7.84\ \mathrm{m\,s^{-1}}} \]

Ball A is moving upwards.

For ball B:

\[ v_B=-4.9-9.8(1.20) \]

\[ \boxed{v_B=-16.66\ \mathrm{m\,s^{-1}}} \]

Ball B is moving downwards.

An interesting feature of this problem is that both balls have identical acceleration. Their relative acceleration is zero, so their relative velocity remains constant.

The tempting approach is to treat their motions as completely separate and solve two complicated quadratic equations. Comparing their positions directly reveals that the identical acceleration terms cancel.

Question 4

A ball is thrown vertically upwards through still air.

Consider its motion:

a) While it is rising.

b) At the exact highest point.

c) While it is descending, before reaching terminal velocity.

For each stage, describe the direction of the forces and acceleration. Explain whether the acceleration remains constant throughout the entire flight.

Solution 4

The acceleration points downwards at all three stages, but its magnitude changes because of air resistance. At the exact highest point, the acceleration is approximately \(9.8\ \mathrm{m\,s^{-2}}\) downwards.

Part a: While rising

The ball is moving upwards.

Weight acts downwards, and air resistance also acts downwards because it opposes the upward motion.

The net force is therefore downwards and greater than the ball’s weight alone.

Consequently, acceleration is downwards, with a magnitude greater than \(9.8\ \mathrm{m\,s^{-2}}\).

Part b: At the highest point

The ball’s instantaneous velocity is zero.

Because the ball is momentarily stationary relative to the still air, air resistance is zero at this instant.

Weight is therefore the only significant force.

The acceleration is:

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

This is the key moment. Zero velocity does not mean zero acceleration.

Part c: While descending

The ball is now moving downwards.

Weight acts downwards, but air resistance acts upwards.

Before terminal velocity is reached, weight is greater than air resistance, so the net force and acceleration remain downwards.

However, the magnitude of acceleration is less than \(9.8\ \mathrm{m\,s^{-2}}\).

Does acceleration remain constant?

No.

Air resistance changes as the ball’s velocity changes. Therefore, the net force and acceleration also change.

The standard HSC uniform-acceleration model neglects air resistance, allowing us to use a constant downward acceleration of \(9.8\ \mathrm{m\,s^{-2}}\).

The real-world situation is more complicated, but the acceleration at the exact turning point is still downward.

Question 5

This final problem combines quadratic roots, symmetry in vertical motion, and the effect of identical gravitational accelerations.

Ball A is thrown vertically upwards from ground level.

A motion tracker records that ball A passes a horizontal balcony level:

• At \(t=1.00\ \mathrm{s}\), while travelling upwards.
• At \(t=3.00\ \mathrm{s}\), while travelling downwards.

Exactly at \(t=2.00\ \mathrm{s}\), ball B is released from rest from the balcony.

Both balls are identical, air resistance is negligible, and gravitational acceleration is constant.

Answer the following:

a) Determine the initial velocity of ball A.

b) Determine the position of ball A relative to the balcony when ball B is released.

c) Calculate the acceleration of both balls at this instant.

d) Determine whether the balls will ever meet after ball B is released. Justify your answer mathematically.

Solution 5

The balls never meet after ball B is released. They remain separated by \(4.90\ \mathrm{m}\) because, at the release instant, both have zero velocity and identical downward acceleration.

Part a: Determine the initial velocity of ball A

Take upwards as positive.

For ball A:

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

At \(t=1.00\ \mathrm{s}\):

\[ s=u-4.9 \]

At \(t=3.00\ \mathrm{s}\):

\[ s=3u-4.9(9) \]

Because ball A passes the same balcony height at both times, the displacements are equal.

Therefore:

\[ u-4.9=3u-44.1 \]

Rearrange:

\[ 39.2=2u \]

\[ \boxed{u=19.6\ \mathrm{m\,s^{-1}}} \]

Ball A was launched upwards at \(19.6\ \mathrm{m\,s^{-1}}\).

Part b: Find ball A’s position when ball B is released

The time at which ball A reaches its highest point is found using:

\[ v=u+at \]

At the highest point, \(v=0\):

\[ 0=19.6-9.8t \]

\[ \boxed{t=2.00\ \mathrm{s}} \]

This is exactly when ball B is released.

Therefore, ball A is at its highest point when ball B begins falling.

We now need to find how far above the balcony ball A is.

At \(t=1.00\ \mathrm{s}\), ball A is passing the balcony while rising.

Its velocity at that instant is:

\[ v=19.6-9.8(1.00) \]

\[ v=9.8\ \mathrm{m\,s^{-1}} \]

It takes another \(1.00\) second to reach its highest point.

Using the displacement equation for this interval:

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

\[ s=9.8(1.00)-4.9(1.00)^2 \]

\[ \boxed{s=4.90\ \mathrm{m}} \]

Ball A is therefore \(4.90\ \mathrm{m}\) above the balcony when ball B is released.

Part c: Determine the acceleration of both balls

At \(t=2.00\ \mathrm{s}\), ball A is at its highest point.

Therefore:

\[ v_A=0 \]

Ball B is released from rest, so:

\[ v_B=0 \]

Both balls experience the same gravitational acceleration:

\[ \boxed{a_A=a_B=-9.8\ \mathrm{m\,s^{-2}}} \]

Both are momentarily stationary, but both are accelerating downwards.

Part d: Will the balls ever meet?

No. Their separation remains constant at \(4.90\ \mathrm{m}\).

This is the important part of the problem.

A tempting prediction is that ball B will catch ball A because ball A is at its highest point and appears to be stopped.

However, ball A is not actually stopped in the sense of having zero acceleration. It is already accelerating downwards.

At the instant ball B is released:

• Ball A is \(4.90\ \mathrm{m}\) above the balcony.
• Ball B is at balcony level.
• Both have zero velocity.
• Both have the same downward acceleration.

Let \(\tau\) represent the time elapsed after ball B is released.

Taking the balcony as the zero-height position, ball A’s position is:

\[ y_A=4.90-4.9\tau^2 \]

Ball B’s position is:

\[ y_B=-4.9\tau^2 \]

Calculate their separation:

\[ y_A-y_B = (4.90-4.9\tau^2)-(-4.9\tau^2) \]

The time-dependent terms cancel:

\[ \boxed{y_A-y_B=4.90\ \mathrm{m}} \]

The separation does not change.

The surprising result is that ball A and ball B fall together without ever meeting, even though one was thrown upwards and the other was simply released from rest.

The reason is that, from the instant ball B is released, both objects have identical velocities and identical accelerations. Their relative position therefore cannot change.

This is a powerful consequence of the fact that acceleration at the highest point never becomes zero.

The next useful step is applying the same reasoning to angled projectiles, where the vertical velocity becomes zero at the highest point but horizontal motion can continue. That requires separating horizontal and vertical components while keeping the gravitational acceleration directed downwards.