3.4 Projectile Motion as a Vector-Valued Model

A ball leaves your hands

A basketball leaves a player's hands at an angle, rising in an arc before dropping through the hoop. A soccer ball is kicked across the pitch. A stone is thrown from the top of a cliff into the sea below. All of these are examples of projectile motion --- an object launched into the air, subject only to gravity, with no engine, no wings, and no propulsion after launch.

Here is the surprising claim: you already know everything you need to predict these trajectories. Constant acceleration (Section 2.2) gave you exact formulas for motion under uniform acceleration along a single line. Component independence (this chapter's hero concept) tells you that perpendicular directions don't interfere with each other. Put these two ideas together, and projectile motion falls out automatically. No new physics is required.

Let's see how.

Prediction

Before we derive anything, commit to some guesses.

Before you read on: Two balls are launched from level ground with the same initial speed $v_0$. Ball A is launched at $45°$ above horizontal. Ball B is launched at $60°$ above horizontal. Both land on the same level ground.

Which ball travels farther horizontally (has greater range)?

Which ball reaches a greater maximum height?

Which ball is in the air longer?

Decide on all three before continuing. Write down your answers --- or at least commit mentally. The derivation below will tell you whether your intuition was right.

Many students guess that $60°$ goes farther because "more speed" or "more effort" goes into the launch. Others guess $45°$ because they have heard it is the optimal angle. Both instincts contain a piece of the truth, and sorting out which is correct for which question is the point of this section.

The Setup: Two Problems in One

Consider an object launched from position $(x_0, y_0)$ at time $t = 0$, with initial speed $v_0$ at angle $\theta$ above the horizontal. We choose coordinates with $x$ pointing horizontally in the direction of launch and $y$ pointing vertically upward.

The initial velocity has two components:

$$v_{0x} = v_0 \cos\theta, \qquad v_{0y} = v_0 \sin\theta$$

Now, what forces act on the projectile after launch? Only gravity, pulling straight down. Gravity has no horizontal component. This means:

Horizontal acceleration: $a_x = 0$
Vertical acceleration: $a_y = -g$

And here is the crucial observation: these two statements are independent. The horizontal motion knows nothing about the vertical motion. The vertical motion knows nothing about the horizontal motion. They proceed simultaneously but without communicating.

This is component independence in action --- the hero concept of this chapter. Let's write out the consequences.

Exploration: The Projectile Simulator

Before we do the algebra, build your intuition.

[Interactive: Projectile simulator. The main panel shows the trajectory of a projectile in the $x$-$y$ plane, drawn as a colored arc. Below the main panel, two synchronized graphs appear side by side: $x(t)$ on the left and $y(t)$ on the right, both plotted against the same time axis. Three sliders control the initial speed $v_0$ (from $1$ to $30$ m/s), the launch angle $\theta$ (from $0°$ to $90°$), and the launch height $y_0$ (from $0$ to $20$ m). As the projectile animates, a moving dot on the trajectory and corresponding dots on the $x(t)$ and $y(t)$ graphs advance together. A vertical dashed line on both time-graphs marks the current time.

Guided prompts appear below:

"Set $y_0 = 0$ and $\theta = 45°$. Watch the $x(t)$ graph. What shape is it?"
"Now watch the $y(t)$ graph. What shape is it? How does it compare to the $x(t)$ graph?"
"Why are the two shapes different? What does each shape tell you about the type of motion in that direction?"
"Try $\theta = 0°$ (horizontal launch). What happens to the $y(t)$ graph? What happens to the trajectory?"
"Try $\theta = 90°$ (straight up). What happens to the $x(t)$ graph? Is this still projectile motion?"
"Change $v_0$ while keeping $\theta = 45°$. What changes in the trajectory? What stays the same about the shapes of $x(t)$ and $y(t)$?"]

Spend time with this. The shapes you see --- a straight line for $x(t)$ and a parabola for $y(t)$ --- are the entire story of projectile motion, and everything else follows from them.

Concept Reveal: Two Independent Constant-Acceleration Problems

Here is what the simulator showed you, expressed in equations.

Horizontal motion: constant velocity

There is no horizontal acceleration, so the horizontal motion is the simplest possible case from Section 2.2 --- constant velocity:

$$x(t) = x_0 + v_{0x}\,t = x_0 + (v_0 \cos\theta)\,t$$

The $x(t)$ graph is a straight line. The horizontal velocity never changes:

$$v_x(t) = v_{0x} = v_0 \cos\theta$$

Vertical motion: constant acceleration

The vertical acceleration is $-g$ (downward), so this is exactly the free-fall problem from Section 2.2:

$$y(t) = y_0 + v_{0y}\,t - \tfrac{1}{2}g\,t^2 = y_0 + (v_0 \sin\theta)\,t - \tfrac{1}{2}g\,t^2$$

$$v_y(t) = v_{0y} - g\,t = v_0 \sin\theta - g\,t$$

The $y(t)$ graph is a downward-opening parabola. The vertical velocity starts positive (if $\theta > 0$), decreases, passes through zero at the peak, and becomes increasingly negative.

The key insight

Projectile motion is not a new type of problem. It is two old problems --- one for each direction --- running simultaneously and independently. The horizontal problem is trivial (constant velocity). The vertical problem is one you have already solved (constant acceleration under gravity). Component independence means you can solve them separately and combine the results.

This is Section 2.2 applied independently in two directions. The only new idea is that the two directions don't talk to each other.

Connection: Galileo's Insight

Galileo figured out the parabolic shape of projectile trajectories around 1608 --- by arguing exactly this component-independence idea, centuries before calculus existed. He reasoned that a ball rolling off a table continues its horizontal motion uniformly (no force to change it) while simultaneously falling vertically under gravity. The combination of uniform horizontal motion and accelerated vertical motion produces a parabolic path.

Galileo did not have the equations we just wrote down. He had the idea --- that the horizontal and vertical motions are independent --- and from that idea alone, he deduced the shape of the trajectory. The equations are a cleaner way of expressing the same argument.

Why the Trajectory is a Parabola

You might have been told that projectile trajectories are parabolas. But why a parabola and not some other curve? This is not a fact to memorize --- it is something you can prove in two lines.

Start with the parametric equations:

$$x = x_0 + (v_0\cos\theta)\,t, \qquad y = y_0 + (v_0\sin\theta)\,t - \tfrac{1}{2}g\,t^2$$

Solve the first equation for $t$:

$$t = \frac{x - x_0}{v_0 \cos\theta}$$

Substitute into the second equation:

$$y = y_0 + (v_0\sin\theta)\cdot\frac{x - x_0}{v_0\cos\theta} - \frac{1}{2}g\left(\frac{x - x_0}{v_0\cos\theta}\right)^2$$

$$y = y_0 + (\tan\theta)(x - x_0) - \frac{g}{2v_0^2\cos^2\theta}(x - x_0)^2$$

This is of the form $y = A + B(x - x_0) + C(x - x_0)^2$ --- a quadratic function of $x$. A quadratic function is a parabola. The trajectory must be parabolic because $x$ grows linearly in time while $y$ grows quadratically. Eliminating the linear parameter $t$ between a linear and a quadratic function always produces a quadratic relationship.

Before you read on: Look at the eliminated equation above. What would change if there were a horizontal acceleration (say, a wind pushing the ball forward)? Would the trajectory still be a parabola?

Check your answer

If there were a constant horizontal acceleration $a_x$, then $x(t) = x_0 + v_{0x}t + \frac{1}{2}a_x t^2$, which is quadratic in $t$. Eliminating $t$ between two quadratics is more complicated --- the trajectory would generally *not* be a parabola. The parabolic shape depends on one component being linear in $t$ (constant velocity) and the other being quadratic (constant acceleration). Remove the linearity, and the special shape is lost.

Solving Projectile Problems: The Key Quantities

For a projectile launched from ground level ($y_0 = 0$) on level ground, three quantities appear again and again: the time of flight, the maximum height, and the range. Let's derive each one from the component equations.

Time of flight

The projectile returns to ground level when $y(t) = 0$:

$$0 = (v_0\sin\theta)\,t - \tfrac{1}{2}g\,t^2 = t\bigl(v_0\sin\theta - \tfrac{1}{2}g\,t\bigr)$$

One solution is $t = 0$ (the launch). The other is:

$$t_{\text{flight}} = \frac{2v_0\sin\theta}{g}$$

The flight time depends on the vertical component of the initial velocity. A steeper launch angle means more initial vertical speed, which means more time in the air.

Maximum height

At the peak, the vertical velocity is zero: $v_y = 0$. Using $v_y = v_0\sin\theta - gt$:

$$t_{\text{peak}} = \frac{v_0\sin\theta}{g}$$

This is exactly half the total flight time --- the trajectory is symmetric in time. Substituting back into $y(t)$:

$$h_{\max} = \frac{v_0^2\sin^2\theta}{2g}$$

The maximum height depends on $\sin^2\theta$. The steeper the angle, the higher the peak.

Range

The horizontal distance traveled during the flight time:

$$R = v_{0x}\cdot t_{\text{flight}} = (v_0\cos\theta)\cdot\frac{2v_0\sin\theta}{g} = \frac{v_0^2 \cdot 2\sin\theta\cos\theta}{g}$$

Using the double-angle identity $2\sin\theta\cos\theta = \sin 2\theta$:

$$R = \frac{v_0^2\sin 2\theta}{g}$$

This is the range formula. It has a beautiful structure: the range depends on $\sin 2\theta$, which is maximized when $2\theta = 90°$, i.e., $\theta = 45°$.

Returning to the Prediction

Now we can answer the opening questions. For two balls launched from level ground with the same speed $v_0$, one at $\theta_A = 45°$ and the other at $\theta_B = 60°$:

1. Which goes farther? The range is $R = \frac{v_0^2\sin 2\theta}{g}$. - $R_A = \frac{v_0^2\sin 90°}{g} = \frac{v_0^2}{g}$ - $R_B = \frac{v_0^2\sin 120°}{g} = \frac{v_0^2(\sqrt{3}/2)}{g} \approx 0.87\,\frac{v_0^2}{g}$

Ball A ($45°$) goes farther. The $45°$ angle maximizes range.

2. Which goes higher? The max height is $h = \frac{v_0^2\sin^2\theta}{2g}$. - $h_A = \frac{v_0^2(1/2)}{2g} = \frac{v_0^2}{4g}$ - $h_B = \frac{v_0^2(3/4)}{2g} = \frac{3v_0^2}{8g}$

Ball B ($60°$) goes higher. More of its initial velocity is directed upward.

3. Which is in the air longer? The flight time is $T = \frac{2v_0\sin\theta}{g}$. - $T_A = \frac{2v_0\sin 45°}{g} = \frac{v_0\sqrt{2}}{g}$ - $T_B = \frac{2v_0\sin 60°}{g} = \frac{v_0\sqrt{3}}{g}$

Ball B ($60°$) is in the air longer. A steeper launch puts more velocity into the vertical component, and the vertical component alone determines flight time.

Pause and reflect: Each answer is governed by a different component. Range involves both components (horizontal speed times flight time). Height depends only on the vertical component. Flight time depends only on the vertical component. Component independence lets you isolate exactly which part of the initial velocity controls each quantity.

The Non-Example: When the Model Breaks

Everything we have done rests on a critical assumption: the only force after launch is gravity. This means no air resistance, no wind, no spin effects. How good is this assumption?

For a dense, slowly-moving object --- a shot put, a ball thrown across a room --- the assumption is excellent. Air resistance is negligible, and the trajectory is indistinguishable from a parabola.

For a fast-moving or lightweight object --- a golf ball, a baseball pitched at 90 mph, a badminton shuttlecock --- air resistance is not negligible. And here is the important point: air resistance does not just "slow things down a little." It couples the components together, which violates the foundation of our entire analysis.

Why? Air drag opposes the velocity vector. If the projectile moves at an angle, the drag force has both horizontal and vertical components. And those components depend on the speed, which involves both $v_x$ and $v_y$. The horizontal drag depends on how fast the object is moving vertically, and vice versa. The two directions are no longer independent.

$$F_{\text{drag}} \propto v^2 = v_x^2 + v_y^2$$

The drag in the $x$-direction depends on $v_y$ (through the total speed), and the drag in the $y$-direction depends on $v_x$. The components talk to each other. Component independence is lost, and the trajectory is no longer parabolic --- it is shorter, steeper, and asymmetric.

The lesson: The projectile motion model breaks precisely when the forces in one direction depend on the motion in the other. This is what "component independence" means physically --- it is not just a mathematical convenience but a physical condition that holds only when the forces in each direction are independent of the motion in other directions.

Practice

Layer 1: Concrete

Problem 1. A ball is launched from ground level at $30°$ above the horizontal with an initial speed of $20$ m/s. Take $g = 10$ m/s$^2$.

(a) Find the time of flight.

(b) Find the maximum height.

Check your answer

The components of initial velocity are: - $v_{0x} = 20\cos 30° = 20 \cdot \frac{\sqrt{3}}{2} \approx 17.3$ m/s - $v_{0y} = 20\sin 30° = 20 \cdot \frac{1}{2} = 10$ m/s (a) Time of flight: $T = \frac{2v_{0y}}{g} = \frac{2(10)}{10} = 2$ s. (b) Maximum height: $h = \frac{v_{0y}^2}{2g} = \frac{100}{20} = 5$ m. (c) Range: $R = v_{0x}\cdot T = 17.3 \times 2 \approx 34.6$ m. Or using the formula: $R = \frac{v_0^2\sin 2\theta}{g} = \frac{400\sin 60°}{10} = 40 \cdot \frac{\sqrt{3}}{2} \approx 34.6$ m.

Problem 2. A stone is thrown horizontally from the top of a $45$ m cliff at $15$ m/s. Take $g = 10$ m/s$^2$.

(a) How long does it take to hit the ground?

(b) How far from the base of the cliff does it land?

Check your answer

Horizontal launch means $\theta = 0°$, so $v_{0x} = 15$ m/s and $v_{0y} = 0$. (a) The vertical motion is free fall from rest: $y_0 - \frac{1}{2}gt^2 = 0$ gives $t = \sqrt{\frac{2y_0}{g}} = \sqrt{\frac{2(45)}{10}} = 3$ s. (b) Horizontal distance: $x = v_{0x}\cdot t = 15 \times 3 = 45$ m from the base of the cliff. (c) At impact: $v_x = 15$ m/s (unchanged), $v_y = gt = 10 \times 3 = 30$ m/s (downward). Speed: $v = \sqrt{v_x^2 + v_y^2} = \sqrt{225 + 900} = \sqrt{1125} \approx 33.5$ m/s. Notice that the horizontal and vertical motions were solved completely independently. The time came from the vertical problem alone; the horizontal distance then followed.

Layer 2: Pattern

Problem 3. Using the range formula $R = \frac{v_0^2 \sin 2\theta}{g}$:

(a) Find two different launch angles that give the same range for a fixed $v_0$.

(b) What is the relationship between these two angles? Verify your answer with a specific pair.

Check your answer

(a) Since $R$ depends on $\sin 2\theta$, two angles give the same range whenever they produce the same value of $\sin 2\theta$. We need $\sin 2\theta_1 = \sin 2\theta_2$. This happens when $2\theta_2 = 180° - 2\theta_1$, i.e., $\theta_2 = 90° - \theta_1$. The paired angles are *complementary* --- they add up to $90°$. (b) For example, $\theta = 30°$ and $\theta = 60°$ are complementary. Check: $\sin 60° = \sin 120° = \frac{\sqrt{3}}{2}$. Both give the same range. (c) The pairing exists because complementary angles swap the horizontal and vertical components. At $30°$, the ball has more horizontal speed and less vertical speed --- it goes fast and low. At $60°$, the ball has less horizontal speed and more vertical speed --- it goes slow and high. The product $v_{0x}\cdot T = (v_0\cos\theta)(2v_0\sin\theta/g)$ contains $\cos\theta\sin\theta$, which is symmetric under $\theta \to 90° - \theta$. The two trajectories have the same range but very different shapes: the low-angle trajectory is flat and fast; the high-angle trajectory is tall and slow.

Layer 3: Structure

Problem 4. We showed that the trajectory is a parabola by eliminating $t$ from the parametric equations. But this derivation assumed that $x(t)$ is linear in $t$, which requires $a_x = 0$.

(a) Starting from the component equations, derive the trajectory equation $y(x)$ explicitly.

(b) Identify the coefficients of the linear and quadratic terms. What physical quantities control the curvature of the parabola?

Check your answer

(a) With $x_0 = y_0 = 0$ for simplicity: $$t = \frac{x}{v_0\cos\theta}$$ $$y = (\tan\theta)\,x - \frac{g}{2v_0^2\cos^2\theta}\,x^2$$ (b) The linear coefficient is $\tan\theta$ --- this is the initial slope of the trajectory at launch. The quadratic coefficient is $-\frac{g}{2v_0^2\cos^2\theta}$. The curvature of the parabola depends on: - $g$: stronger gravity means more curvature (tighter arc). - $v_0$: faster launch means less curvature (flatter arc) --- the ball "outruns" gravity. - $\cos^2\theta$: at steep angles, $\cos\theta$ is small, so the denominator is small and the curvature is large. This makes sense: a steeply launched ball spends a long time at each horizontal position, giving gravity more time to pull it down per unit of horizontal distance. (c) When $g \to 0$, the quadratic term vanishes and $y = (\tan\theta)\,x$ --- a straight line in the launch direction. With no gravity, the projectile travels in a straight line forever. This is Newton's first law.

Layer 4: Debug

Problem 5. A student is asked: "A ball is launched at $30°$ from the edge of a $20$ m cliff with a speed of $25$ m/s. How far from the base of the cliff does it land?"

The student writes:

$$R = \frac{v_0^2\sin 2\theta}{g} = \frac{625\sin 60°}{10} \approx 54.1 \text{ m}$$

What went wrong?

Check your answer

The range formula $R = \frac{v_0^2\sin 2\theta}{g}$ applies **only** to projectiles launched and landing at the same height. The student used it for a projectile launched from a cliff, where the landing height is different from the launch height. The range formula was derived by setting $y(t_{\text{flight}}) = 0$ with $y_0 = 0$. When $y_0 \neq 0$, the flight time is longer (the ball has an extra $20$ m to fall), so the horizontal distance is greater than $54.1$ m. The correct approach: use the component equations with $y_0 = 20$ m, solve $0 = 20 + (v_0\sin 30°)t - \frac{1}{2}gt^2$ for $t$ (which gives a quadratic equation), and then compute $x = (v_0\cos 30°)t$. The lesson: the range formula is a *derived shortcut* that carries hidden assumptions (launch and landing at the same height). When those assumptions fail, you must return to the underlying component equations. This is why understanding *where* formulas come from matters more than memorizing them.

Layer 5: Creation

Problem 6. A target is located at the point $(40, 10)$ meters from the launch point (i.e., $40$ m downrange and $10$ m above launch height). The launch speed is $v_0 = 25$ m/s. Find a launch angle $\theta$ that makes the projectile pass through this point.

Hint: Substitute the target coordinates into the trajectory equation and solve for $\theta$. You may find it helpful to write $\tan\theta = u$ and express $\cos^2\theta$ in terms of $u$.

Check your answer

Using the trajectory equation with $x_0 = y_0 = 0$: $$y = (\tan\theta)\,x - \frac{g}{2v_0^2\cos^2\theta}\,x^2$$ Substitute $(x, y) = (40, 10)$, $g = 10$, $v_0 = 25$: $$10 = 40\tan\theta - \frac{10}{2(625)\cos^2\theta}(1600)$$ $$10 = 40\tan\theta - \frac{16000}{1250\cos^2\theta}$$ $$10 = 40\tan\theta - \frac{12.8}{\cos^2\theta}$$ Using the identity $\frac{1}{\cos^2\theta} = 1 + \tan^2\theta$ and letting $u = \tan\theta$: $$10 = 40u - 12.8(1 + u^2)$$ $$10 = 40u - 12.8 - 12.8u^2$$ $$12.8u^2 - 40u + 22.8 = 0$$ Using the quadratic formula: $$u = \frac{40 \pm \sqrt{1600 - 4(12.8)(22.8)}}{2(12.8)} = \frac{40 \pm \sqrt{1600 - 1167.36}}{25.6} = \frac{40 \pm \sqrt{432.64}}{25.6}$$ $$u = \frac{40 \pm 20.8}{25.6}$$ This gives $u_1 \approx 2.375$ or $u_2 \approx 0.75$, corresponding to $\theta_1 \approx 67.1°$ and $\theta_2 \approx 36.9°$. There are *two* valid launch angles. The lower angle produces a flatter trajectory that passes through the point on the way up (or along a shallow arc). The higher angle produces a tall, arcing trajectory that passes through the same point on the way down. This is another manifestation of the complementary-angle pairing, now generalized to an elevated target.

Reflection

Reflection: What made projectile motion simpler than you expected?

If you look back at this section, the entire analysis reduced to two independent one-dimensional problems --- both of which you had already solved in Chapter 2. The horizontal problem was trivial (constant velocity). The vertical problem was free fall. No new equations were needed. No new techniques. The only new idea was component independence: the recognition that horizontal and vertical motions proceed without influencing each other.

Ask yourself: what assumption is doing all the work? It is the assumption that the only force is gravity, which acts purely vertically. This is what guarantees $a_x = 0$ and makes the components independent. The moment a force has components that depend on the full velocity vector --- like air drag --- the independence breaks, and the simplicity evaporates.

Projectile motion is not important because of its formulas. It is important because it is the first example of a strategy: decompose a vector problem into independent scalar problems, solve each one separately, and combine the results. You will use this strategy again and again --- in circular motion, in multi-body problems, and eventually in every branch of physics where symmetry allows you to separate directions.

Looking Ahead

Projectile motion was the cleanest possible application of component independence: one component had zero acceleration, and the other had constant acceleration. The two directions were completely decoupled.

In Section 3.5, we turn to relative motion and moving reference frames. There, you will add velocity vectors that belong to different observers --- the boat and the river, the plane and the wind. The independence idea will still be at work (you decompose into components to add vectors), but now the question becomes: whose components? Choosing the right reference frame will be as important as choosing the right coordinate axes.