Projectile Motion: Analysing Parabolic Trajectories

What is Projectile Motion?

Projectile motion occurs when an object is launched into the air and moves freely under the sole influence of gravity (ignoring air resistance).

The resultant path is a predictable curve called a parabola. The genius of Newton's mechanics allows us to solve these complex 2D arcs by splitting them into two completely distinct 1D motions: horizontal and vertical.

Learning Goals: By the end of this guide, you should be able to:

Understand the independence of horizontal and vertical motion.

Decompose a launch velocity into initial $v_x$ and $v_y$ components.

Use the kinematic equations to calculate the time of flight, maximum height, and horizontal range.

Solve projectile motion problems logically across different launch scenarios.

The Core Principle: Independence of Motion

The most fundamental rule of projectile motion is: Perpendicular components of motion are entirely independent of each other.

1. Horizontal Motion (Constant Velocity)

Since gravity acts downwards, there is no horizontal acceleration ( $a_x = 0$ ).
Thus, the horizontal velocity ( $v_x$ ) remains constant throughout the flight.
Equation: $\Delta x = v_x t$

2. Vertical Motion (Constant Acceleration)

Gravity acts constantly downwards to accelerate the object ( $a_y \approx -9.81\ \text{m/s}^2$ ).
Thus, the vertical velocity ( $v_y$ ) changes steadily. Positive when going up, zero at the peak, and negative when coming down.
Equations: The standard kinematic (SUVAT) equations apply here.

The only variable that links these two separate dimensions together is time ( $t$ ). The time it takes for the object to complete its vertical arc is the exact same time it has to travel horizontally.

The Three Key Projectile Formulas

Given an initial velocity $\vec{v}_0$ at launch angle $\theta$ :

Initial Horizontal Velocity: $v_{0x} = v_0 \cos\theta$
Initial Vertical Velocity: $v_{0y} = v_0 \sin\theta$

If the projectile lands at the same height from which it was launched, we can derive three extremely useful formulas:

Time of Flight ( $T$ )

The time to reach the apex is $t = \frac{v_0 \sin\theta}{g}$ . The total time of flight is double that:

T = \frac{2v_0 \sin\theta}{g}

Maximum Height ( $H$ )

Using $v_y^2 = v_{0y}^2 + 2a_y \Delta y$ , and setting final $v_y = 0$ at the apex:

H = \frac{(v_0 \sin\theta)^2}{2g}

Horizontal Range ( $R$ )

Using $R = v_x \times T = (v_0 \cos\theta) \times (\frac{2v_0 \sin\theta}{g})$ . Because $2\sin\theta\cos\theta = \sin(2\theta)$ :

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

(Note: Maximum range occurs at $\theta = 45^\circ$ because $\sin(90^\circ) = 1$ .)

Worked Examples

Example 1: Kicked Football

Question: A football is kicked from the ground with an initial velocity of $20\ \text{m/s}$ at an angle of $40^\circ$ . Calculate its maximum height and how far away it lands. ( $g = 9.8\ \text{m/s}^2$ )

Step 1: Maximum Height ( $H$ )

H = \frac{(20 \sin40^\circ)^2}{2 \times 9.8} = \frac{(12.86)^2}{19.6} = \frac{165.23}{19.6} = 8.43\ \text{m}

Step 2: Horizontal Range ( $R$ )

R = \frac{(20)^2 \sin(2 \times 40^\circ)}{9.8} = \frac{400 \sin(80^\circ)}{9.8} = \frac{400 \times 0.985}{9.8} = 40.2\ \text{m}

Example 2: Horizontal Launch (Off a Cliff)

Question: A stone is thrown horizontally off a $50\ \text{m}$ cliff at $15\ \text{m/s}$ . How far from the base does it land?

(Note: The derived formulas above do not apply here because it doesn't land at the same height. We must use the basic equations).

Step 1: Find Time ( $t$ ) from vertical motion. $v_{0y} = 0$ (horizontally launched), $\Delta y = -50\ \text{m}$ .

\Delta y = v_{0y}t + \frac{1}{2}a_yt^2 \implies -50 = 0 - 4.9t^2 \implies t^2 = 10.2 \implies t = 3.19\ \text{s}

Step 2: Find Distance ( $\Delta x$ ) from horizontal motion. $v_x = 15\ \text{m/s}$ .

\Delta x = v_x \times t = 15 \times 3.19 = 47.9\ \text{m}

The stone lands $47.9\ \text{m}$ from the base of the cliff.

Common Mistakes

Using Range/Max Height formulas blindly — The $R$ , $H$ , and $T$ formulas only work if the projectile takes off and lands at the very same elevation. For launching off a cliff or shooting a basketball into a hoop, you must separate $x$ and $y$ motions and solve manually.
Mixing x and y variables — Never, ever plug a horizontal velocity into a vertical acceleration equation. Treat them as completely separate problems that share only time ( $t$ ).
Puttings a non-zero $a_x$ — Once an object is in the air, there is no horizontal force (ignoring air resistance) so $a_x = 0$ . Velocity $v_x$ does not change.
Sign errors — If you define "up" as positive, $g$ must be $-9.8\ \text{m/s}^2$ and downward displacement is negative. Be consistent.

Exam Tips (A-Level / AP / IB)

Always set up an $x/y$ table listing your known and unknown kinematic variables before calculating anything.
Remember that at the highest point of an arc, vertical velocity $v_y = 0$ , but the horizontal velocity $v_x$ is still there! So the kinetic energy is not zero at the apex.
If an AP/IB question asks "what angle provides the same range as $30^\circ$ ?", the answer is its complement: $90^\circ - 30^\circ = 60^\circ$ . Complementary angles give identical ranges.

Frequently Asked Questions

What happens if we include air resistance?

Everything changes! The parabola becomes asymmetrical. The projectile peaks earlier, its maximum height is lower, its range is much shorter, and it falls at a steeper angle than it went up.

Why is 45° the optimal angle for maximum range?

Because Range depends on $\sin(2\theta)$ . The maximum value of the sine function is 1, which happens when the angle is $90^\circ$ . Setting $2\theta = 90^\circ$ gives $\theta = 45^\circ$ . Mathematically, this perfectly balances vertical hang-time against horizontal speed.

Vector Decomposition — Learn how to split the initial launch velocity into its $v_x$ and $v_y$ components.
Motion Graphs — Visualise how velocity and position change smoothly through time.