Modeling the motion of a projectile

A projectile is anything launched into the air and then left to move freely — a thrown ball, a kicked football, a shell from a gun. To make the maths tractable, Cambridge asks you to model it. Modelling means replacing the messy real object with a clean idealised one and listing the assumptions you made.

The standard modelling assumptions (learn these — examiners ask you to state them):

• The projectile is a particle — all its mass is at a single point, so it has no size and cannot spin.
• There is no air resistance — the only force acting is gravity.
• Gravity is constant, giving a constant downward acceleration $g$ (we use $g = 10\text{ m s}^{-2}$).
• The ground is flat and horizontal, and we ignore the curvature and rotation of the Earth.

Why each assumption matters. "No air resistance" is the big one — in reality a fast-moving ball is slowed by drag, so the true range is shorter and the path is not a perfect parabola. "Particle" lets us ignore spin (which curves a real football). Stating these tells the examiner you understand the model's limitations.

Cambridge tip. If a question says "state one assumption you have made", a one-line answer such as "air resistance is negligible" earns the mark. Do not write an essay.

This is the single most important principle in the whole topic. Once you launch the projectile, the only force is gravity, and gravity points straight down. So:

• Vertically, gravity provides a constant downward acceleration — the vertical motion is "constant-acceleration" motion (SUVAT).
• Horizontally, there is no force at all — so there is no horizontal acceleration. The horizontal velocity therefore never changes; the projectile travels equal horizontal distances in equal times.

The two directions do not talk to each other. You analyse the vertical motion entirely on its own, and the horizontal motion entirely on its own. The only thing they share is the time $t$ — the clock runs simultaneously for both. This shared time is what links the two when you finally combine them.

Cambridge tip. Set up two clearly labelled columns of work — "Horizontal" and "Vertical" — from the start. This stops you from accidentally putting $g$ into a horizontal equation, the single most common error in the topic.

A projectile is usually launched with speed $V$ at an angle $\theta$ above the horizontal. To use the independence principle, you split this single velocity into its horizontal and vertical parts using basic trigonometry.

The right-angled triangle of velocity. Draw the launch velocity as the hypotenuse of a right-angled triangle:

• Horizontal component (adjacent to $\theta$): $u_x = V\cos\theta$.
• Vertical component (opposite to $\theta$): $u_y = V\sin\theta$.

$u_x = V\cos\theta \qquad u_y = V\sin\theta$

Sign convention — choose it and stick to it. Almost always take upward as positive. Then:

• The horizontal velocity stays $V\cos\theta$ throughout (no acceleration).
• The vertical velocity starts at $+V\sin\theta$, and the acceleration is $-g$ (because gravity pulls down, against the positive direction).

So at the top of the flight the vertical velocity is momentarily $0$; on the way down it becomes negative.

A common trap. If the projectile is launched horizontally (e.g. rolled off a table), then $\theta = 0$, so $u_x = V$ and $u_y = 0$. If it is launched at angle $\theta$ below the horizontal (thrown downward), the vertical component is $-V\sin\theta$.

Cambridge tip. Write "(taking up as positive)" at the top of your solution. It costs one second and protects every sign in the rest of the question.

With the components in hand and a sign convention chosen, you are ready to write down the two independent equations of motion. We will solve them in subtopics 8.2 and 8.3, but the setting up is done here.

Horizontal (constant velocity, $a_x = 0$): initial velocity $V\cos\theta$, so after time $t$, $x = V\cos\theta\,t$

Vertical (constant acceleration $a_y = -g$): initial velocity $V\sin\theta$, so using $s = ut + \tfrac12 a t^2$, $y = V\sin\theta\,t - \tfrac12 g t^2$

Why this is so powerful. Every projectile question — time of flight, range, greatest height, speed at a point, the shape of the path — comes from these two lines plus the SUVAT equations applied to each direction. Identify what you are told (a height? a time? a horizontal distance?) and which direction it belongs to, then pick the right SUVAT equation for that direction.

MF19 status. The SUVAT (constant-acceleration) equations are given in the MF19 formula list, and so is the trajectory equation you will meet in 8.3. But you must decide which equation to use and resolve the velocity yourself — those steps are never given.

Cambridge tip. Before any algebra, sketch the parabola, mark the launch point, the angle, and what you are asked to find. A 10-second diagram prevents the most expensive mistakes.