How to Derive the Parabolic Projectile Path Equation

A ball rolls off the edge of a table. It keeps moving forwards, but falls further below the tabletop with every passing moment. If its horizontal speed stays constant, why doesn’t it follow a straight diagonal line?

The answer is that two different motions are happening at once. Horizontal displacement increases steadily with time, while vertical displacement changes with the square of time. Eliminate time from those two equations, and you get a quadratic equation in horizontal position. That’s what makes the projectile’s path a parabola.

For a projectile launched from \(x=0\), its trajectory equation is:

\[
\boxed{
\begin{aligned}
y&=y_0+\frac{u_y}{u_x}x\\
&\quad-\frac{g}{2u_x^2}x^2
\end{aligned}
}
\]

Here:

• \(x\) is horizontal displacement from the launch point, in metres.
• \(y\) is vertical position relative to your chosen origin, in metres.
• \(y_0\) is the initial vertical position, in metres.
• \(u_x\) is the initial horizontal velocity, in \(\mathrm{m\,s^{-1}}\).
• \(u_y\) is the initial vertical velocity, in \(\mathrm{m\,s^{-1}}\).
• \(g\) is the positive magnitude of gravitational acceleration, taken as \(9.8\ \mathrm{m\,s^{-2}}\) near Earth’s surface.

We take rightwards and upwards as positive, so vertical acceleration is \(a_y=-g\). We also assume negligible air resistance, constant gravity, and \(u_x\ne0\).

01Why Galileo needed more than a sketch of the trajectory

Before the parabolic model was established, some descriptions of projectile motion used straight sections joined to curved sections. That might look reasonable when watching a thrown object: it appears to travel forwards initially, then turn downwards as it falls.

But a convincing explanation needs more than a curve that looks right. It needs to predict where the projectile will be.

In Two New Sciences (1638), Galileo described projectile motion as the combination of uniform horizontal motion and accelerated vertical motion. His work on inclined planes and falling bodies helped establish the mathematical reasoning behind that description.

The important question was whether the horizontal motion continues independently while gravity changes the vertical motion.

Reconstructing the measurement problem

Imagine a simple experimental arrangement based on the same physical ideas.

On the left, an inclined ramp allows a small metal ball to gain speed. A smooth curved guide leads onto a short, level track at the top of a table.

The ball rolls along that track and leaves its right-hand edge horizontally.

Below the table is the floor, covered with paper that records the ball’s landing position. The horizontal distance is measured from the point directly beneath the table edge to the impact mark. The vertical drop is the height of the edge above the paper.

What should this setup show?

If the ball keeps its horizontal speed after leaving the table, the horizontal distance should increase in direct proportion to flight time. If gravity produces constant downward acceleration, the vertical drop should increase in proportion to the square of flight time.

A single landing mark, however, cannot prove the trajectory is parabolic. A straight line and many different curves can all pass through the launch point and one impact point.

To get a stronger test, we need positions at several points during flight.

Improving the experiment

One approach is to place a catching surface at several different heights and repeat the launch, keeping the initial speed as consistent as possible. Each impact gives another horizontal-distance and vertical-drop pair.

A modern alternative is to record the ball side-on with a stationary camera and a calibrated grid behind its flight path.

The video provides the ball’s horizontal and vertical positions at known time intervals. Unlike the single landing-mark method, it can reveal whether horizontal spacing stays constant and whether downward displacement follows a quadratic pattern.

What the measurements should look like

Suppose the ball leaves horizontally at \(2.00\ \mathrm{m\,s^{-1}}\), with zero initial vertical velocity.

Under the ideal model:

\[
x=2.00t
\]

and

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

where \(t\) is elapsed time in seconds and \(\Delta y\) is vertical displacement from the table edge in metres.

The following values are illustrative predictions, not measurements from a real experiment.

After \(0.10\) seconds, the ball has travelled \(20\) cm horizontally but fallen only \(4.9\) cm.

After \(0.20\) seconds, its horizontal displacement has doubled to \(40\) cm. Its downward displacement has quadrupled to \(19.6\) cm.

That’s the key observation. Doubling horizontal displacement quadruples the fall, provided the ball was launched horizontally and the ideal assumptions hold.

The horizontal graph is a straight line because horizontal velocity is constant. The vertical graph curves because the ball gains downward speed.

Now remove time from the measurements.

Since \(x=2.00t\),

\[
t=\frac{x}{2.00}
\]

Substituting into the vertical equation gives:

\[
\begin{aligned}
\Delta y
&=-4.90\left(\frac{x}{2.00}\right)^2\\
&=-1.225x^2
\end{aligned}
\]

Here, both \(x\) and \(\Delta y\) are measured in metres, so the coefficient \(1.225\) has units of \(\mathrm{m^{-1}}\).

This is the equation of a downward-opening parabola.

The experiment’s useful result is not simply that the marks form a curved shape. It’s that one model explains both measurements and predicts the spatial path.

A real ball may experience air resistance, spin-related forces, and measurement error. Those effects limit how closely its recorded positions follow the ideal quadratic equation.

02Deriving the general projectile trajectory equation

The horizontal-launch experiment gives us the simplest version. Now allow the projectile to leave at an angle above the horizontal.

The important change is that the initial velocity has both horizontal and vertical components.

Step 1: Separate horizontal and vertical motion

Suppose the projectile leaves with initial speed \(u\) at angle \(\theta\) above the horizontal.

Its horizontal and vertical velocity components are:

\[
u_x=u\cos\theta
\]

\[
u_y=u\sin\theta
\]

The angle is measured from the positive horizontal direction.

With negligible air resistance, there is no horizontal acceleration:

\[
a_x=0
\]

Gravity provides constant downward acceleration:

\[
a_y=-g
\]

We can therefore use the constant-acceleration equations separately in the two directions.

Step 2: Write the horizontal displacement equation

The horizontal velocity stays constant, so:

\[
\boxed{x=u_xt}
\]

Substituting \(u_x=u\cos\theta\):

\[
x=(u\cos\theta)t
\]

This equation says that equal time intervals produce equal horizontal displacements.

Step 3: Write the vertical displacement equation

Vertically, the projectile begins with velocity \(u_y\) and accelerates downwards at \(g\).

Using the constant-acceleration displacement equation:

\[
y=y_0+u_yt+\frac12a_yt^2
\]

Since \(a_y=-g\):

\[
\boxed{y=y_0+u_yt-\frac12gt^2}
\]

The first term is the launch height. The second describes the vertical displacement that the initial upward velocity would produce if gravity were absent. The third subtracts the effect of downward gravitational acceleration.

Notice that the gravitational term contains \(t^2\). That’s where the curvature comes from.

Step 4: Eliminate time

We have two equations:

\[
\begin{aligned}
x&=u_xt\\
y&=y_0+u_yt-\frac12gt^2
\end{aligned}
\]

The horizontal equation gives:

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

Substitute this into the vertical equation:

\[
\begin{aligned}
y&=y_0+u_y\left(\frac{x}{u_x}\right)\\
&\quad-\frac12g\left(\frac{x}{u_x}\right)^2
\end{aligned}
\]

Simplify:

\[
\boxed{
\begin{aligned}
y&=y_0+\frac{u_y}{u_x}x\\
&\quad-\frac{g}{2u_x^2}x^2
\end{aligned}
}
\]

Time has disappeared. We now have an equation connecting vertical position directly to horizontal position.

That’s the equation of the trajectory, sometimes called the Cartesian equation of projectile motion.

Step 5: Express the equation using launch speed and angle

From the velocity components:

\[
\frac{u_y}{u_x}
=\frac{u\sin\theta}{u\cos\theta}
=\tan\theta
\]

Also:

\[
u_x^2=u^2\cos^2\theta
\]

Substituting both expressions gives:

\[
\boxed{
\begin{aligned}
y&=y_0+x\tan\theta\\
&\quad-\frac{gx^2}{2u^2\cos^2\theta}
\end{aligned}
}
\]

If the launch point is our vertical origin, then \(y_0=0\), and the equation becomes:

\[
\boxed{
\begin{aligned}
y&=x\tan\theta\\
&\quad-\frac{gx^2}{2u^2\cos^2\theta}
\end{aligned}
}
\]

This is the familiar HSC form.

Both boxed equations describe the same physical relationship. The first uses velocity components; the second uses launch speed and angle.

For an exam problem, choose whichever form matches the information provided.

03Why the equation describes a parabola

A quadratic equation has the general structure:

\[
y=Ax^2+Bx+C
\]

Our trajectory equation can be written as:

\[
y=-\frac{g}{2u_x^2}x^2
+\frac{u_y}{u_x}x+y_0
\]

So its coefficients are:

\[
A=-\frac{g}{2u_x^2}
\]

\[
B=\frac{u_y}{u_x}
\]

\[
C=y_0
\]

Each coefficient has a physical meaning.

Because \(g>0\) and \(u_x^2>0\), the coefficient of \(x^2\) is negative.

The parabola therefore opens downwards.

What controls the curvature?

Imagine launching two balls horizontally from the same height. One has twice the horizontal speed of the other.

Would they fall through different vertical distances in the first second?

No. Both begin with zero vertical velocity and experience the same downward acceleration. After one second, their vertical displacement is the same.

But the faster ball has travelled twice as far horizontally.

That means its trajectory looks flatter.

Mathematically, doubling \(u_x\) makes the magnitude of the quadratic coefficient four times smaller:

\[
\frac{g}{2(2u_x)^2}
=\frac14\frac{g}{2u_x^2}
\]

The curvature coefficient becomes one-quarter as large.

Be careful with the language here. The faster ball does not experience weaker gravity. It simply travels farther horizontally during the same fall.

05Launching from a height: don’t confuse two coordinate systems

The equation

\[
y=x\tan\theta-
\frac{gx^2}{2u^2\cos^2\theta}
\]

is often introduced with the launch point at \((0,0)\).

That does not mean the projectile must be launched from ground level. It means the vertical position is being measured relative to the launch point.

If a ball is launched from a balcony, you have two useful coordinate choices.

Here \(h\) is the height of the launch point above the ground, in metres.

Both choices are correct. What matters is using one consistently.

The mistake is to measure height from the launch point and then automatically set \(y=0\) at ground impact.

06A quick method for exam questions

When a question asks you to derive, use, or interpret a projectile path equation, make these decisions in order.

1. Choose an origin. Decide whether \(y=0\) means the launch point or the ground.
2. Resolve the initial velocity. Find \(u_x\) and \(u_y\), with signs consistent with your axes.
3. Write both motion equations. Use \(x=u_xt\) and \(y=y_0+u_yt-\frac12gt^2\).
4. Eliminate time only when you need the path. Substitute \(t=x/u_x\) to obtain \(y(x)\).
5. Apply the actual target or landing condition. A landing below launch height has a different vertical coordinate from a return to launch height.
6. Check the result physically. A path below ground, a negative flight time, or a shortcut used at the wrong height should make you reconsider the setup.

If the question asks for the position at a particular time, there’s no benefit in eliminating time first. Use the component equations directly.

Also remember the limits of the method. The usual Cartesian derivation assumes a non-zero, constant horizontal velocity. A purely vertical launch has \(x=0\) throughout the flight, so division by \(u_x\) is impossible; its spatial path is a vertical line. If a horizontal force acts during flight, \(x=u_xt\) may no longer hold.

07Questions and solutions

The first question checks the derivation. The others ask you to work backwards from observations, test competing explanations, or recognise when one projectile path must be treated as more than a single uninterrupted motion.

Assume negligible air resistance and constant \(g=9.80\ \mathrm{m\,s^{-2}}\) unless stated otherwise.

Question 1

A ball is launched from level ground with an initial horizontal velocity of \(7.00\ \mathrm{m\,s^{-1}}\) and an initial vertical velocity of \(4.90\ \mathrm{m\,s^{-1}}\).

Taking the launch point as \((0,0)\):

a) Derive the Cartesian equation of its trajectory.

b) Calculate its height after travelling \(3.50\) m horizontally.

Solution 1

The trajectory is \(y=0.700x-0.100x^2\), and the ball is \(1.23\) m above its launch point at \(x=3.50\) m.

Part a)

The horizontal equation is:

\[
x=7.00t
\]

The vertical equation is:

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

Eliminate time using:

\[
t=\frac{x}{7.00}
\]

Therefore:

\[
\begin{aligned}
y&=4.90\left(\frac{x}{7.00}\right)\\
&\quad-4.90\left(\frac{x}{7.00}\right)^2
\end{aligned}
\]

Simplifying:

\[
\boxed{y=0.700x-0.100x^2}
\]

Here \(x\) and \(y\) are in metres.

Part b)

Substitute \(x=3.50\) m:

\[
\begin{aligned}
y&=0.700(3.50)\\
&\quad-0.100(3.50)^2\\
&=2.450-1.225\\
&=1.225\ \mathrm{m}
\end{aligned}
\]

To three significant figures:

\[
\boxed{y=1.23\ \mathrm{m}}
\]

A common mistake is to use \(x=u_xt\) as though it were the vertical displacement equation too. The horizontal motion is uniform, but the vertical motion is accelerated. That difference produces the quadratic term.

Question 2

A small ball is fired from a platform \(6.00\) m above level ground towards a target on the ground \(20.0\) m horizontally from the launch point.

Its horizontal launch velocity is fixed at \(10.0\ \mathrm{m\,s^{-1}}\), but its initial vertical velocity can be adjusted.

A thin vertical wall stands \(10.0\) m from the launch point. The top of the wall is \(8.00\) m above the ground.

Treat the ball as a particle.

a) Calculate the initial vertical velocity required for the ball to hit the target.

b) Determine whether that trajectory clears the wall.

c) Can any adjustment to the initial vertical velocity make the ball both clear the wall and hit the target while keeping \(u_x=10.0\ \mathrm{m\,s^{-1}}\)? Explain.

Solution 2

The ball needs \(u_y=6.80\ \mathrm{m\,s^{-1}}\) to hit the target, but that trajectory passes \(0.100\) m below the top of the wall. With horizontal velocity fixed at \(10.0\ \mathrm{m\,s^{-1}}\), no choice of vertical launch velocity can satisfy both conditions.

Part a)

The horizontal flight time to the target is:

\[
\begin{aligned}
t&=\frac{x}{u_x}\\
&=\frac{20.0}{10.0}\\
&=2.00\ \mathrm{s}
\end{aligned}
\]

Using ground level as \(y=0\):

\[
y=6.00+u_yt-4.90t^2
\]

At the target, \(y=0\) and \(t=2.00\) s.

\[
\begin{aligned}
0&=6.00+2.00u_y-19.6\\
2.00u_y&=13.6\\
u_y&=6.80\ \mathrm{m\,s^{-1}}
\end{aligned}
\]

Part b)

The wall is halfway to the target horizontally.

Because horizontal speed is constant:

\[
t_{\mathrm{wall}}=\frac{10.0}{10.0}
=1.00\ \mathrm{s}
\]

The ball’s height there would be:

\[
\begin{aligned}
y&=6.00+6.80(1.00)\\
&\quad-4.90(1.00)^2\\
&=7.90\ \mathrm{m}
\end{aligned}
\]

The wall is \(8.00\) m tall.

The ball is \(0.100\) m too low and would strike the wall instead of reaching the target.

Part c)

To pass strictly above the wall, the ball would need:

\[
6.00+u_y-4.90>8.00
\]

Therefore:

\[
u_y>6.90\ \mathrm{m\,s^{-1}}
\]

But hitting the target requires exactly \(6.80\ \mathrm{m\,s^{-1}}\).

Those requirements are incompatible.

For example, with \(u_y=6.90\ \mathrm{m\,s^{-1}}\), the ball’s height at the target’s horizontal position would be:

\[
\begin{aligned}
y&=6.00+6.90(2.00)-19.6\\
&=0.200\ \mathrm{m}
\end{aligned}
\]

It would pass above the target.

The tempting solution is to keep increasing the launch angle until the wall is cleared. But the target position already fixes the required vertical velocity once the horizontal velocity is specified.

A projectile’s trajectory must satisfy every spatial constraint, not just the final one.

Question 3

A projectile is launched from a platform \(5.00\) m above level ground.

Two position sensors detect it at the following locations, measured relative to the ground:

Assume the projectile travels to the right under the standard projectile model and that its launch point is at \(x=0\).

a) Determine its full Cartesian trajectory equation.

b) Calculate the horizontal and vertical components of its initial velocity.

c) Find the horizontal position and height of the highest point.

d) Determine where it lands on the ground, and explain why simply doubling the horizontal position of the highest point would give the wrong landing distance.

Solution 3

The trajectory is \(y=5.00+\frac23x-\frac{x^2}{36}\). Its initial velocity components are approximately \(13.3\ \mathrm{m\,s^{-1}}\) horizontally and \(8.85\ \mathrm{m\,s^{-1}}\) upwards. The highest point is \((12.0,\ 9.00)\) m, and the ball lands at \(x=30.0\) m.

Part a)

The general trajectory equation is:

\[
y=y_0+bx-cx^2
\]

where:

\[
b=\frac{u_y}{u_x}
\]

and

\[
c=\frac{g}{2u_x^2}
\]

Since \(y_0=5.00\) m:

\[
y=5.00+bx-cx^2
\]

Use the first sensor, at \((6.00,8.00)\):

\[
8=5+6b-36c
\]

So:

\[
6b-36c=3
\]

Use the second sensor, at \((18.0,8.00)\):

\[
18b-324c=3
\]

Subtracting the first equation from the second gives:

\[
12b-288c=0
\]

Therefore:

\[
b=24c
\]

Substitute into \(6b-36c=3\):

\[
\begin{aligned}
6(24c)-36c&=3\\
108c&=3\\
c&=\frac1{36}
\end{aligned}
\]

Then:

\[
b=24\left(\frac1{36}\right)
=\frac23
\]

The trajectory is:

\[
\boxed{y=5.00+\frac23x-\frac{x^2}{36}}
\]

All positions are in metres. The coefficient \(1/36\) has units of \(\mathrm{m^{-1}}\).

Part b)

Because:

\[
c=\frac{g}{2u_x^2}
\]

we have:

\[
\frac1{36}=\frac{9.80}{2u_x^2}
\]

Rearranging:

\[
\begin{aligned}
u_x^2&=176.4\\
u_x&=13.3\ \mathrm{m\,s^{-1}}
\end{aligned}
\]

We choose the positive root because the projectile travels to the right.

Now:

\[
b=\frac{u_y}{u_x}=\frac23
\]

Thus:

\[
\begin{aligned}
u_y&=\frac23u_x\\
&=\frac23\sqrt{176.4}\\
&=8.85\ \mathrm{m\,s^{-1}}
\end{aligned}
\]

Part c)

A parabola is symmetric about the vertical line through its highest point.

The ball passes through the same height at \(x=6.00\) m and \(x=18.0\) m.

Therefore the highest point lies halfway between those positions:

\[
x_{\mathrm{top}}=\frac{6.00+18.0}{2}
=12.0\ \mathrm{m}
\]

Substitute \(x=12.0\) m:

\[
\begin{aligned}
y_{\mathrm{top}}
&=5.00+\frac23(12.0)\\
&\quad-\frac{(12.0)^2}{36}\\
&=5.00+8.00-4.00\\
&=9.00\ \mathrm{m}
\end{aligned}
\]

So the highest point is:

\[
\boxed{(12.0,\ 9.00)\ \mathrm{m}}
\]

Part d)

At ground impact, \(y=0\):

\[
0=5+\frac23x-\frac{x^2}{36}
\]

Multiply by \(36\):

\[
0=180+24x-x^2
\]

Rearrange:

\[
x^2-24x-180=0
\]

Factorise:

\[
(x-30)(x+6)=0
\]

The physical landing position is:

\[
\boxed{x=30.0\ \mathrm{m}}
\]

The negative root belongs to the mathematical extension of the path before launch.

A tempting answer is \(24.0\) m, twice the apex’s horizontal position.

But symmetry around \(x=12.0\) means the ball returns to its launch height of \(5.00\) m at \(x=24.0\) m.

It still has another \(5.00\) m vertically to fall before reaching ground level.

The trajectory is symmetric about its highest point, but the actual flight between launch and landing is not symmetric when those endpoints have different heights.

Question 4

A small charged bead leaves an experimental launcher horizontally. Gravity acts downwards, while a uniform electric field produces an additional constant horizontal force.

A side-on recording gives the following illustrative positions relative to the launch point:

Assume the bead remains within the field throughout the recorded interval, the field does not affect vertical motion, and air resistance is negligible.

a) Show that the vertical data agree with constant gravitational acceleration.

b) Calculate the bead’s initial horizontal velocity and horizontal acceleration.

c) A student uses the standard horizontal-projectile equation:

\[
y=-\frac{gx^2}{2u_x^2}
\]

Using the initial horizontal velocity found in part b), calculate the height this equation predicts at \(x=0.72\) m. Compare it with the recorded position.

d) Explain precisely which step of the usual parabolic-path derivation is invalid. Does observing a curved path alone establish that horizontal acceleration is zero?

Solution 4

The vertical motion follows gravitational acceleration, but the bead has an initial horizontal velocity of \(2.20\ \mathrm{m\,s^{-1}}\) and a horizontal acceleration of \(-2.00\ \mathrm{m\,s^{-2}}\). The standard horizontal-projectile equation therefore gives an incorrect height.

Part a)

Because the bead starts with zero vertical velocity, its vertical equation is:

\[
y=-\frac12gt^2
\]

At \(t=0.20\) s:

\[
\begin{aligned}
y&=-4.90(0.20)^2\\
&=-0.196\ \mathrm{m}
\end{aligned}
\]

At \(t=0.40\) s:

\[
y=-4.90(0.40)^2
=-0.784\ \mathrm{m}
\]

At \(t=0.60\) s:

\[
y=-4.90(0.60)^2
=-1.764\ \mathrm{m}
\]

All three positions agree with the predicted vertical motion under constant gravity.

Part b)

Because horizontal acceleration may be non-zero, use:

\[
x=u_{x0}t+\frac12a_xt^2
\]

Here \(u_{x0}\) is the initial horizontal velocity and \(a_x\) is horizontal acceleration.

At \(t=0.20\) s:

\[
0.40=0.20u_{x0}+0.020a_x
\]

At \(t=0.40\) s:

\[
0.72=0.40u_{x0}+0.080a_x
\]

Double the first equation:

\[
0.80=0.40u_{x0}+0.040a_x
\]

Subtract it from the second:

\[
-0.08=0.040a_x
\]

Therefore:

\[
\boxed{a_x=-2.00\ \mathrm{m\,s^{-2}}}
\]

Substitute back:

\[
0.40=0.20u_{x0}-0.040
\]

Hence:

\[
\boxed{u_{x0}=2.20\ \mathrm{m\,s^{-1}}}
\]

The negative horizontal acceleration means the bead slows while travelling to the right.

Part c)

The standard equation would predict:

\[
y=-\frac{9.80x^2}{2(2.20)^2}
\]

At \(x=0.72\) m:

\[
\begin{aligned}
y&=-\frac{9.80(0.72)^2}{9.68}\\
&=-0.525\ \mathrm{m}
\end{aligned}
\]

But the recorded vertical position is:

\[
y=-0.784\ \mathrm{m}
\]

The standard formula underestimates the downward displacement by:

\[
0.784-0.525=0.259\ \mathrm{m}
\]

The bead has fallen farther because horizontal slowing means it takes longer to reach \(x=0.72\) m than the constant-horizontal-speed model predicts.

Part d)

The failed step is:

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

That expression assumes \(u_x\) is constant.

In this experiment:

\[
x=u_{x0}t+\frac12a_xt^2
\]

So horizontal displacement is not proportional to time.

The presence of a curved path does not by itself prove zero horizontal acceleration.

There is a useful mathematical subtlety here. Motion with a constant acceleration vector can still trace a geometric parabola when viewed in suitably rotated axes. But with horizontal acceleration present, the standard downward-opening quadratic \(y(x)\) derived from \(x=u_xt\) does not generally apply in the original horizontal and vertical axes.

The governing issue is not whether a curve looks parabolic. It’s whether the motion equations used to derive it match the forces acting on the object.

Question 5

A ball is launched from level ground with initial velocity components:

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

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

It moves under gravity, with no air resistance.

An experimental device can give the ball a brief, purely horizontal impulse, instantly reversing its horizontal velocity from \(+12.0\ \mathrm{m\,s^{-1}}\) to \(-12.0\ \mathrm{m\,s^{-1}}\). The impulse does not change its vertical velocity.

First, suppose the impulse occurs exactly at the highest point.

a) Calculate the time and coordinates of the highest point before the impulse.

b) Derive the Cartesian equation of the ball’s trajectory after the impulse.

c) Does the ball follow a different curved path, or does it retrace its original path in reverse? Prove your conclusion and determine where it lands.

Now suppose the same horizontal velocity reversal occurs \(0.50\) s after launch instead of at the highest point.

d) Determine the ball’s position and vertical velocity at the moment of this earlier impulse.

e) Will the ball still retrace its original trajectory? Determine its eventual landing position and explain the physical condition that makes the two cases behave differently.

Assume each impulse is effectively instantaneous and that gravity is the only force acting between impulses.

Solution 5

When the horizontal velocity is reversed exactly at the highest point, the ball retraces its original parabolic path and lands back at the launch point. Reversing it \(0.50\) s after launch does not produce retracing: the ball continues rising while moving left and eventually lands \(12.0\) m to the left of the launch point.

The crucial difference is whether the impulse reverses the ball’s entire velocity vector, not merely its horizontal component.

Part a)

At the highest point, the vertical velocity is zero:

\[
v_y=u_y-gt
\]

So:

\[
\begin{aligned}
0&=9.80-9.80t\\
t&=1.00\ \mathrm{s}
\end{aligned}
\]

The horizontal position is:

\[
x=(12.0)(1.00)=12.0\ \mathrm{m}
\]

The vertical position is:

\[
\begin{aligned}
y&=9.80(1.00)-4.90(1.00)^2\\
&=4.90\ \mathrm{m}
\end{aligned}
\]

The highest point is:

\[
\boxed{(12.0,\ 4.90)\ \mathrm{m}}
\]

Part b)

Immediately after the impulse:

• The ball is at \((12.0,\ 4.90)\) m.
• Its horizontal velocity is \(-12.0\ \mathrm{m\,s^{-1}}\).
• Its vertical velocity is zero.
• Its vertical acceleration remains \(-9.80\ \mathrm{m\,s^{-2}}\).

Let \(\tau\) be the time elapsed since the impulse, in seconds.

The new horizontal equation is:

\[
x=12.0-12.0\tau
\]

The vertical equation is:

\[
y=4.90-4.90\tau^2
\]

From horizontal motion:

\[
\tau=\frac{12.0-x}{12.0}
\]

Substitute into the vertical equation:

\[
\boxed{
y=4.90-\frac{4.90}{144}(12.0-x)^2
}
\]

This is the post-impulse trajectory, with positions measured in metres.

Part c)

The post-impulse path is exactly the same geometric parabola that the ball followed before reaching the highest point.

Before the impulse:

\[
x=12.0t
\]

\[
y=9.80t-4.90t^2
\]

Eliminating time:

\[
y=\frac{9.80}{12.0}x-\frac{4.90}{144}x^2
\]

Completing the square gives:

\[
\boxed{
y=4.90-\frac{4.90}{144}(x-12.0)^2
}
\]

Compare this with the post-impulse equation:

\[
y=4.90-\frac{4.90}{144}(12.0-x)^2
\]

Because:

\[
(12.0-x)^2=(x-12.0)^2
\]

the two equations describe exactly the same curve.

The ball therefore descends along the same positions it occupied while ascending, but in reverse order.

It takes another \(1.00\) s to fall from the highest point to ground level.

Its horizontal displacement during that second is:

\[
\Delta x=(-12.0)(1.00)
=-12.0\ \mathrm{m}
\]

Starting from \(x=12.0\) m, it lands at:

\[
\boxed{x=0}
\]

It returns to the launch point after a total elapsed time of \(2.00\) s.

This is not because gravity stops acting or reverses direction.

At the highest point, the ball has no vertical velocity. Its velocity is entirely horizontal.

Reversing that horizontal velocity therefore reverses the complete instantaneous velocity vector. Under the same subsequent gravitational acceleration, the ball follows its earlier motion backwards in space.

A photograph showing only the positions visited between launch and apex would not, by itself, reveal that the ball had travelled along that same arc twice. Time and direction information are needed.

Part d)

Now the impulse occurs after only \(0.50\) s.

Before the impulse, the horizontal position is:

\[
\begin{aligned}
x&=12.0(0.50)\\
&=6.00\ \mathrm{m}
\end{aligned}
\]

The vertical position is:

\[
\begin{aligned}
y&=9.80(0.50)-4.90(0.50)^2\\
&=4.90-1.225\\
&=3.675\ \mathrm{m}
\end{aligned}
\]

The vertical velocity is:

\[
\begin{aligned}
v_y&=9.80-9.80(0.50)\\
&=+4.90\ \mathrm{m\,s^{-1}}
\end{aligned}
\]

So the impulse occurs at:

\[
\boxed{(6.00,\ 3.675)\ \mathrm{m}}
\]

The ball is still moving upwards at \(4.90\ \mathrm{m\,s^{-1}}\).

Part e)

The ball will not retrace its earlier path.

Just before the impulse, its velocity components are:

\[
(12.0,\ +4.90)\ \mathrm{m\,s^{-1}}
\]

Just after the impulse, they are:

\[
(-12.0,\ +4.90)\ \mathrm{m\,s^{-1}}
\]

Only the horizontal component has reversed. The vertical component still points upwards.

The ball therefore starts moving left while continuing to rise.

Its vertical motion is unchanged by the horizontal impulse, so it still reaches ground level \(2.00\) s after the original launch.

The remaining flight time after the impulse is:

\[
\Delta t=2.00-0.50
=1.50\ \mathrm{s}
\]

The horizontal velocity during that interval is \(-12.0\ \mathrm{m\,s^{-1}}\).

Thus:

\[
\begin{aligned}
x_{\mathrm{land}}
&=6.00+(-12.0)(1.50)\\
&=6.00-18.0\\
&=-12.0\ \mathrm{m}
\end{aligned}
\]

The ball lands:

\[
\boxed{12.0\ \mathrm{m}}
\]

to the left of its original launch point.

There is another way to expose the misconception.

Half a second after the early impulse, the ball reaches \(x=0\) again. But its vertical motion is then at the highest point, \(y=4.90\) m.

So it passes directly above the launch point rather than returning to it.

The two impulse cases differ because exact retracing requires the entire velocity vector to reverse at the impulse. A purely horizontal impulse can achieve that only when vertical velocity is already zero, which occurs at the highest point.

The ideal projectile equations still work between impulses, but you must start a new set of motion equations whenever an impulse changes the velocity.

08What the parabolic equation tells you – and what it doesn’t

The Cartesian equation tells you where a projectile can be. It describes the shape of its path without requiring time as an explicit variable.

That makes it useful for checking whether a ball clears an obstacle, finding the height at a particular horizontal position, interpreting recorded trajectory data, or reconstructing launch conditions.

But removing time also removes information.

Two movements can occupy the same geometric curve without reaching each point at the same time. An impulse can even make a projectile retrace an earlier path.

If the next question involves a moving target, simultaneous arrivals, or a sudden change in velocity, keep the time equations alongside the Cartesian equation.

And before relying on any of them, consider whether the model fits the physical situation. The separate discussion of Projectile Motion Assumptions in HSC Physics Explained covers when constant gravity and negligible air resistance are appropriate – and when the real trajectory stops following the parabola you’ve derived.