How to Calculate Projectile Speed at the Top of Its Path

A football reaches the highest point of its flight. For an instant, it stops rising before it starts falling. Does that mean its speed is zero?

No. At the top of a projectile’s path, its vertical velocity is zero, but its horizontal velocity is still present. If air resistance is negligible, the horizontal velocity stays constant throughout the flight.

So the speed at the top is:

\[
v_{\text{top}}=|u\cos\theta|
\]

Here, \(u\) is the initial launch speed in metres per second (\(\mathrm{m\,s^{-1}}\)), and \(\theta\) is the launch angle measured from the horizontal.

The absolute value makes the result positive because speed has no direction. For a projectile launched upwards and to the right, you can simply use \(v_{\text{top}}=u\cos\theta\).

01Why the projectile doesn’t stop at the top

Imagine watching a ball fly across a field. Two things happen simultaneously:

• It travels horizontally across the field.
• It rises, slows vertically, and then falls.

Gravity changes the second motion. Under the standard projectile model, it doesn’t change the first.

To describe these separately, choose right as the positive \(x\)-direction and up as the positive \(y\)-direction.

Let:

• \(u_x\) be the initial horizontal velocity, in \(\mathrm{m\,s^{-1}}\).
• \(u_y\) be the initial vertical velocity, in \(\mathrm{m\,s^{-1}}\).
• \(v_x\) and \(v_y\) be the corresponding velocity components at a later time.
• \(t\) be the time since launch, in seconds.
• \(g=9.8\ \mathrm{m\,s^{-2}}\) be the approximate magnitude of gravitational acceleration near Earth’s surface.

With negligible air resistance, the only force acting on the ball after release is its weight, directed downwards.

Horizontally, acceleration is zero:

\[
a_x=0
\]

Therefore:

\[
v_x=u_x
\]

Vertically, acceleration is \(-g\):

\[
a_y=-g
\]

Therefore:

\[
v_y=u_y-gt
\]

These are separate equations because gravity changes the vertical component without changing the horizontal component.

At the highest point, the ball changes from moving upwards to moving downwards. Its vertical velocity must pass through zero:

\[
v_y=0
\]

But its horizontal velocity remains unchanged.

The total speed at any instant is the magnitude of the two perpendicular velocity components:

\[
v=\sqrt{v_x^2+v_y^2}
\]

At the top, \(v_y=0\), so:

\[
\begin{aligned}
v_{\text{top}}&=\sqrt{v_x^2+0^2}\\
&=|v_x|\\
&=|u_x|
\end{aligned}
\]

If the initial velocity is given as a speed \(u\) and angle \(\theta\) above the horizontal, then:

\[
u_x=u\cos\theta
\]

This gives the formula from the opening:

\[
\boxed{v_{\text{top}}=|u\cos\theta|}
\]

The important point isn’t the cosine itself. It’s that you’re finding the horizontal component of the launch velocity and keeping it unchanged until the apex.

If resolving the launch velocity is unfamiliar, the separate guide to projectile initial velocity components covers that starting step.

02How could we measure the speed at the top?

The calculation makes a prediction: even while the projectile’s vertical velocity approaches zero, its horizontal velocity should remain constant.

How could we test that instead of just trusting the equation?

The problem with one photograph

Suppose a camera takes a photograph at the instant a ball appears to reach its highest point.

You can measure the ball’s position, but not its instantaneous velocity from that single image. A ball moving sideways at \(12\ \mathrm{m\,s^{-1}}\) could look exactly like one moving sideways at \(6\ \mathrm{m\,s^{-1}}\).

We need positions recorded at known time intervals.

A better design: Side-on video tracking

Place a ball launcher on the left of a large measurement grid. Position a fixed camera perpendicular to the plane of the ball’s flight, with the grid behind the trajectory.

The ball travels from left to right across the grid. The camera records its position in successive frames.

From the recording, we can measure:

• Horizontal position \(x\), relative to launch.
• Vertical position \(y\), relative to launch.
• Time \(t\), measured from release.

Keeping the camera square-on and the measurement grid close to the plane of motion reduces perspective errors.

What should successive video frames show?

If the gravity-only model is correct, the ball should cover equal horizontal distances in equal time intervals. Its vertical displacement during those intervals should shrink towards the apex and then become downward.

Here is an illustrative dataset calculated from the ideal projectile equations. These are model-generated positions, not measurements from a real experiment.

The highest point occurs at \(t=0.80\ \mathrm{s}\).

Every \(0.20\ \mathrm{s}\), the ball moves another \(2.40\ \mathrm{m}\) horizontally.

Therefore:

\[
\begin{aligned}
v_x&=\frac{\Delta x}{\Delta t}\\
&=\frac{2.40}{0.20}\\
&=12.0\ \mathrm{m\,s^{-1}}
\end{aligned}
\]

At the highest point, the speed is still \(12.0\ \mathrm{m\,s^{-1}}\).

This also exposes a measurement trap.

The ball is at the same height at \(0.60\ \mathrm{s}\) and \(1.00\ \mathrm{s}\), but it isn’t stationary vertically at either instant. It’s rising at the first and falling at the second.

The apex lies halfway between those times, at \(0.80\ \mathrm{s}\).

A real video would provide measured positions with some uncertainty. Equal horizontal spacings would support the constant-horizontal-velocity model, while unequal spacings would need investigation. Air resistance, camera perspective, and tracking errors could all affect the result.

The experiment therefore tests an assumption rather than proving that every real projectile follows a perfect parabola.

03Does the launch height change the speed at the top?

Imagine two identical balls launched with the same initial velocity.

One starts at ground level. The other starts from a platform \(12\ \mathrm{m}\) above the ground.

Predict which one has the greater speed at its highest point.

Under the ideal projectile model, neither does. Both have the same initial horizontal velocity, so both have the same speed at their respective apices.

But that doesn’t mean their entire flights are the same.

The raised ball reaches a higher altitude above the ground. It also takes longer to land on the ground below, because it has farther to fall.

This is why you must not use a level-ground range formula automatically for a projectile launched from a building.

The formula for speed at the apex still works because it depends on horizontal velocity, not landing height.

04Which equation should you use?

The fastest method depends on what the question gives you.

These methods assume constant gravitational acceleration and no horizontal acceleration.

Two boundary cases are worth remembering.

A vertically launched object: If \(u_x=0\), its speed really is zero at the apex. Gravity still accelerates it downwards at that instant.

A horizontally launched object: The launch point is already the highest point of its subsequent flight. Its initial vertical velocity is zero, so its speed at that point equals its horizontal launch speed.

For a projectile launched downwards, however, the highest point during the observed flight is the release point. Its vertical velocity isn’t zero there, so you cannot apply the usual apex shortcut.

For an ordinary projectile launched upwards, a reliable decision sequence is to identify the horizontal component, recognise that \(v_y=0\) at the apex, and calculate the remaining speed.

Don’t spend time finding the total flight time unless the question also requires it.

05Questions and solutions

The questions begin with a standard apex calculation, then move towards reconstructing motion from measurements, comparing different launch heights, and testing when the usual shortcut breaks.

Assume \(g=9.8\ \mathrm{m\,s^{-2}}\) throughout. Ignore air resistance unless another horizontal force is explicitly specified.

Question 1

A golf ball is struck at \(30.0\ \mathrm{m\,s^{-1}}\), at \(60.0^\circ\) above the horizontal.

Calculate its speed at the highest point of its path. State the direction and magnitude of its acceleration at that instant.

Solution 1

The ball’s speed at its highest point is \(15.0\ \mathrm{m\,s^{-1}}\). Its acceleration is still \(9.8\ \mathrm{m\,s^{-2}}\) vertically downwards.

First, calculate the horizontal velocity:

\[
\begin{aligned}
u_x&=u\cos\theta\\
&=30.0\cos60.0^\circ\\
&=15.0\ \mathrm{m\,s^{-1}}
\end{aligned}
\]

At the apex, vertical velocity is zero, but horizontal velocity remains \(15.0\ \mathrm{m\,s^{-1}}\).

The tempting mistake is to set both velocity and acceleration to zero because the ball momentarily stops rising.

Zero vertical velocity does not mean zero acceleration. Gravity continues acting downwards, causing the ball to begin falling.

Question 2

A fixed camera tracks a ball travelling from left to right. The coordinate origin is its launch position.

At \(t=0.60\ \mathrm{s}\), the ball is at:

• \(x=7.20\ \mathrm{m}\)
• \(y=4.12\ \mathrm{m}\)

At \(t=1.40\ \mathrm{s}\), the ball is at:

• \(x=16.80\ \mathrm{m}\)
• \(y=4.12\ \mathrm{m}\)

The ball experiences constant downward gravitational acceleration and no horizontal acceleration. Both observations belong to the same uninterrupted flight.

Calculate:

a. The time after launch when the ball reaches its highest point.

b. Its speed at that point.

c. Its maximum height above the release position.

d. Whether the ball’s instantaneous speeds at the two recorded positions are equal, and how they compare with its speed at the apex.

Solution 2

The ball reaches its highest point at \(1.00\ \mathrm{s}\), with a speed of \(12.0\ \mathrm{m\,s^{-1}}\) and a height of approximately \(4.90\ \mathrm{m}\) above release. Its speeds at the two recorded positions are equal to each other, but greater than its speed at the top.

Part a: Find the time of the apex.

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

Under constant vertical acceleration, the time from the first crossing to the apex equals the time from the apex to the second crossing.

Therefore:

\[
\begin{aligned}
t_{\text{top}}&=\frac{0.60+1.40}{2}\\
&=1.00\ \mathrm{s}
\end{aligned}
\]

Part b: Find the horizontal speed.

The horizontal displacement between observations is:

\[
\Delta x=16.80-7.20=9.60\ \mathrm{m}
\]

The elapsed time is \(1.40-0.60=0.80\ \mathrm{s}\).

\[
\begin{aligned}
v_x&=\frac{\Delta x}{\Delta t}\\
&=\frac{9.60}{0.80}\\
&=12.0\ \mathrm{m\,s^{-1}}
\end{aligned}
\]

This stays constant, so the apex speed is \(12.0\ \mathrm{m\,s^{-1}}\).

Part c: Find the maximum height.

From \(t=0.60\ \mathrm{s}\) to the apex, the elapsed time is \(0.40\ \mathrm{s}\).

Working backwards from the apex, where vertical velocity is zero, the height difference is:

\[
\begin{aligned}
\Delta y&=\frac12g(0.40)^2\\
&=\frac12(9.8)(0.40)^2\\
&=0.784\ \mathrm{m}
\end{aligned}
\]

Therefore:

\[
H_{\text{max}}=4.12+0.784=4.904\ \mathrm{m}
\]

So the maximum height is approximately \(4.90\ \mathrm{m}\) above release.

Part d: Compare the speeds.

At \(0.40\ \mathrm{s}\) before the apex, the vertical velocity is:

\[
v_y=(9.8)(0.40)=+3.92\ \mathrm{m\,s^{-1}}
\]

At \(0.40\ \mathrm{s}\) after the apex, it is \(-3.92\ \mathrm{m\,s^{-1}}\).

The total speed at either observation is:

\[
\begin{aligned}
v&=\sqrt{12.0^2+3.92^2}\\
&=12.6\ \mathrm{m\,s^{-1}}
\end{aligned}
\]

Both recorded speeds are equal, but both exceed the \(12.0\ \mathrm{m\,s^{-1}}\) apex speed.

The trap is assuming that identical heights mean zero vertical velocity. In fact, the ball has equal and opposite vertical velocities at those positions.

Question 3

Two balls are launched simultaneously to the right from the same vertical line, with unobstructed paths.

Ball A starts at ground level.

Ball B starts from a platform \(14.7\ \mathrm{m}\) above the same level ground.

Both have identical initial velocity components:

\[
u_x=12.0\ \mathrm{m\,s^{-1}}
\]

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

Calculate:

a. The speed of each ball at its respective highest point.

b. The time at which each ball lands, and its horizontal landing distance from the launch line.

c. The impact speed of each ball.

Explain why treating both flights as level-ground projectile motion would give an incorrect result.

Solution 3

Both balls have an apex speed of \(12.0\ \mathrm{m\,s^{-1}}\). Ball A lands after \(2.00\ \mathrm{s}\), travelling \(24.0\ \mathrm{m}\) horizontally. Ball B lands after \(3.00\ \mathrm{s}\), travelling \(36.0\ \mathrm{m}\). Ball B also has the greater impact speed.

Part a: Apex speeds.

The balls have identical horizontal velocity components, and neither experiences horizontal acceleration.

Therefore:

\[
v_{\text{top},A}=v_{\text{top},B}
=12.0\ \mathrm{m\,s^{-1}}
\]

Their different launch heights don’t change the result.

Part b: Landing times and distances.

For Ball A, the landing height equals its launch height.

Its upward and downward vertical motions are symmetric, so:

\[
\begin{aligned}
t_A&=\frac{2u_y}{g}\\
&=\frac{2(9.8)}{9.8}\\
&=2.00\ \mathrm{s}
\end{aligned}
\]

Its horizontal distance is:

\[
x_A=12.0(2.00)=24.0\ \mathrm{m}
\]

Ball B lands \(14.7\ \mathrm{m}\) below its launch position.

Using ground as the vertical origin:

\[
0=14.7+9.8t-4.9t^2
\]

Divide by \(4.9\):

\[
0=3+2t-t^2
\]

Rearranging:

\[
(t-3)(t+1)=0
\]

The physically relevant root is \(t=3.00\ \mathrm{s}\), because negative time is not part of the flight after launch.

Therefore:

\[
x_B=12.0(3.00)=36.0\ \mathrm{m}
\]

Part c: Compare impact speeds.

Both balls retain their horizontal velocity of \(12.0\ \mathrm{m\,s^{-1}}\).

Vertically:

\[
v_y=u_y-gt
\]

For Ball A:

\[
v_{y,A}=9.8-9.8(2.00)
=-9.8\ \mathrm{m\,s^{-1}}
\]

For Ball B:

\[
v_{y,B}=9.8-9.8(3.00)
=-19.6\ \mathrm{m\,s^{-1}}
\]

Using the magnitudes of the perpendicular components:

\[
\begin{aligned}
v_{\text{land},A}
&=\sqrt{12.0^2+9.8^2}\\
&=15.5\ \mathrm{m\,s^{-1}}\\
v_{\text{land},B}
&=\sqrt{12.0^2+19.6^2}\\
&=23.0\ \mathrm{m\,s^{-1}}
\end{aligned}
\]

Ball B hits the ground faster because it falls through a greater vertical distance.

The level-ground flight-time shortcut \(t=2u_y/g\) works for Ball A, not Ball B.

The surprising point is that two balls can have identical speeds at their apices but different landing times, ranges, and impact speeds. Apex speed and landing speed are different questions.

Question 4

A small bead is launched from the origin with:

• Initial horizontal velocity \(u_x=+8.0\ \mathrm{m\,s^{-1}}\).
• Initial vertical velocity \(u_y=+9.8\ \mathrm{m\,s^{-1}}\).

Right and up are positive.

Unlike an ideal projectile, the bead experiences an additional constant horizontal force throughout its flight. Its accelerations are:

\[
a_x=-12.0\ \mathrm{m\,s^{-2}}
\]

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

Assume these accelerations remain constant and ignore other forces.

a. Find the time when the bead reaches its greatest height, and its speed and direction at that instant.

b. Find the time and horizontal coordinate when the bead is furthest to the right.

c. Explain why the highest point and furthest-right point are different, and why \(v_{\text{top}}=u_x\) is no longer valid.

This is an extension of the standard gravity-only projectile model, using the same two-dimensional kinematics.

Solution 4

The bead reaches its highest point after \(1.00\ \mathrm{s}\), moving left at \(4.0\ \mathrm{m\,s^{-1}}\). It reaches its furthest-right position earlier, after \(0.667\ \mathrm{s}\), at \(x=2.67\ \mathrm{m}\).

This happens because the bead’s horizontal and vertical velocities now both change with time.

Part a: Find the highest point.

Vertical velocity is:

\[
v_y=u_y+a_yt
\]

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

\[
\begin{aligned}
0&=9.8-9.8t\\
t&=1.00\ \mathrm{s}
\end{aligned}
\]

Horizontal velocity at this time is:

\[
\begin{aligned}
v_x&=u_x+a_xt\\
&=8.0-12.0(1.00)\\
&=-4.0\ \mathrm{m\,s^{-1}}
\end{aligned}
\]

Since \(v_y=0\), the total speed is \(4.0\ \mathrm{m\,s^{-1}}\), directed left.

So an object can be moving left at its highest point even though it was initially launched to the right.

Part b: Find the furthest-right position.

The bead stops moving right when its horizontal velocity becomes zero.

\[
\begin{aligned}
0&=8.0-12.0t\\
t&=\frac{8.0}{12.0}\\
&=0.667\ \mathrm{s}
\end{aligned}
\]

Horizontal displacement is:

\[
x=u_xt+\frac12a_xt^2
\]

Substitute \(t=2/3\ \mathrm{s}\):

\[
\begin{aligned}
x&=8.0\left(\frac23\right)\\
&\quad-6.0\left(\frac23\right)^2\\
&=2.67\ \mathrm{m}
\end{aligned}
\]

At that instant, the vertical velocity is:

\[
\begin{aligned}
v_y&=9.8-9.8\left(\frac23\right)\\
&=+3.27\ \mathrm{m\,s^{-1}}
\end{aligned}
\]

The bead is still rising when it reaches its furthest-right point.

Part c: Explain the difference.

Two separate conditions determine these turning points:

• Greatest height occurs when \(v_y=0\).
• Furthest-right position occurs when \(v_x=0\).

There is no reason for those conditions to happen simultaneously when both accelerations are non-zero.

The usual result \(v_{\text{top}}=|u_x|\) assumes horizontal acceleration is zero. That assumption fails here.

But the more fundamental result survives: at a genuine vertical apex, \(v_y=0\), so the speed is still the magnitude of the horizontal velocity at that instant.

We must calculate the current \(v_x\), not reuse its initial value.

Question 5

Two ball launchers are mounted on the same vertical wall, pointing to the right.

Launcher A is mounted above Launcher B.

Both launch balls at exactly \(19.6\ \mathrm{m\,s^{-1}}\), but at different angles:

• Launcher A fires at \(30^\circ\) above the horizontal.
• Launcher B fires at \(60^\circ\) above the horizontal.

You can choose the vertical distance between the launchers and the time delay between their launches.

Is it possible for the two balls to reach exactly the same point in space at exactly the same instant, with both balls at the highest point of their respective trajectories?

If it is possible:

a. Find the required vertical separation between the launchers, and state which must be higher.

b. Determine which launcher must fire first, and by how much time.

c. Find the horizontal position of the meeting point relative to the wall.

d. Calculate both balls’ speeds immediately before they meet. Explain why those speeds are or are not equal.

Assume constant gravity, negligible air resistance, unobstructed motion, and that both balls travel freely until the instant they meet.

Solution 5

Yes, the two balls can meet exactly at their apices. Launcher A must be \(9.80\ \mathrm{m}\) above Launcher B, and B must fire approximately \(0.732\ \mathrm{s}\) earlier.

The shared apex will be approximately \(17.0\ \mathrm{m}\) horizontally from the wall. Immediately before meeting, A will be moving at \(17.0\ \mathrm{m\,s^{-1}}\) to the right, while B will be moving at \(9.80\ \mathrm{m\,s^{-1}}\) to the right.

The key is that the angles \(30^\circ\) and \(60^\circ\) are complementary.

Part a: Compare their maximum vertical rises.

At the apex:

\[
v_y=0
\]

The vertical rise above each launch point is:

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

Since \(u_y=u\sin\theta\):

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

Both balls have initial speed \(19.6\ \mathrm{m\,s^{-1}}\), giving:

\[
\frac{u^2}{2g}
=\frac{19.6^2}{2(9.8)}
=19.6\ \mathrm{m}
\]

For Ball A:

\[
\begin{aligned}
H_A&=19.6\sin^230^\circ\\
&=4.90\ \mathrm{m}
\end{aligned}
\]

For Ball B:

\[
\begin{aligned}
H_B&=19.6\sin^260^\circ\\
&=14.7\ \mathrm{m}
\end{aligned}
\]

Ball B rises \(9.80\ \mathrm{m}\) farther above its launch position.

To reach the same absolute height, B must therefore start \(9.80\ \mathrm{m}\) below A.

Part b: Compare their times to the apex.

For a rising projectile:

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

For A:

\[
\begin{aligned}
t_A&=\frac{19.6\sin30^\circ}{9.8}\\
&=1.00\ \mathrm{s}
\end{aligned}
\]

For B:

\[
\begin{aligned}
t_B&=\frac{19.6\sin60^\circ}{9.8}\\
&=1.732\ \mathrm{s}
\end{aligned}
\]

B takes longer to reach its apex.

For both balls to arrive at their apices simultaneously, B must launch first.

The required delay is:

\[
\Delta t=1.732-1.00
=0.732\ \mathrm{s}
\]

Part c: Check whether the apex positions line up horizontally.

Matching heights and times isn’t enough. The balls must also reach the same horizontal position.

Horizontal displacement to the apex is:

\[
x_{\text{top}}=u_xt_{\text{top}}
\]

Substitute the horizontal and vertical launch components:

\[
\begin{aligned}
x_{\text{top}}
&=(u\cos\theta)\frac{u\sin\theta}{g}\\
&=\frac{u^2\sin(2\theta)}{2g}
\end{aligned}
\]

Here is the crucial relationship:

\[
\sin(2\times30^\circ)
=\sin(2\times60^\circ)
\]

because:

\[
\sin60^\circ=\sin120^\circ
\]

Therefore, both balls travel exactly the same horizontal distance from the wall before reaching their apices.

Numerically:

\[
\begin{aligned}
x_{\text{top}}&=19.6\sin60^\circ\\
&=16.97\ \mathrm{m}
\end{aligned}
\]

So their apex positions can coincide horizontally as well as vertically.

Combined with the launch separation and timing, this gives a meeting at the same point and instant.

Part d: Compare their speeds at the meeting point.

Both balls have zero vertical velocity at their apices, but their horizontal velocities differ.

For Ball A:

\[
\begin{aligned}
v_A&=19.6\cos30^\circ\\
&=16.97\ \mathrm{m\,s^{-1}}
\end{aligned}
\]

For Ball B:

\[
\begin{aligned}
v_B&=19.6\cos60^\circ\\
&=9.80\ \mathrm{m\,s^{-1}}
\end{aligned}
\]

Both are moving to the right immediately before they meet, but A is moving faster.

Their relative horizontal speed is approximately:

\[
16.97-9.80=7.17\ \mathrm{m\,s^{-1}}
\]

A tempting prediction is that two balls with equal initial speeds, meeting at their highest points, must also have equal speeds there.

That ignores the different amounts of vertical rise.

Ball B converts more of its initial kinetic energy into gravitational potential energy as it rises \(14.7\ \mathrm{m}\) above its launch point. Ball A rises only \(4.90\ \mathrm{m}\) above its launch point.

Under the gravity-only model, each ball keeps its own horizontal velocity unchanged. So their speeds at the shared apex are different, even though they began with equal speeds.

Equal initial speeds and equal apex positions do not force equal apex velocities.

06What changes after the highest point?

At the apex, the vertical velocity is zero, so the speed of an ideal projectile is entirely horizontal. It is also the minimum speed during a complete rising-and-falling flight.

Immediately afterwards, gravity makes the vertical velocity negative. The projectile still moves horizontally, but now it also gains downward speed.

That means the simple apex rule is no longer enough. At a later point, you need to calculate the horizontal and vertical components and combine them.

The final velocity of a projectile guide develops that next step, including why landing height changes the final result even when the horizontal component stays constant.