Projectile Motion Explained: A-Level Physics (9702)

Short answer: Projectile motion is just two separate motions happening at once. The horizontal velocity stays constant because no horizontal force acts, while the vertical motion accelerates under gravity. Split the launch velocity into horizontal and vertical parts, apply the suvat equations to each, and the whole problem becomes straightforward.

The one idea that unlocks everything

A projectile is any object moving only under gravity after launch, with air resistance ignored. The key insight for 9702 is that the horizontal and vertical motions are independent.

• Horizontal: no force acts, so velocity is constant.
• Vertical: gravity acts, so the object accelerates downwards at $g = 9.81 \text{ m s}^{-2}$.

Time is the only quantity shared between the two directions. Once you find the time of flight, you can solve for range, height, or final velocity.

Splitting the launch velocity

If an object is launched with speed $u$ at angle $\theta$ above the horizontal, resolve it into components:

• Horizontal: $u_x = u\cos\theta$
• Vertical: $u_y = u\sin\theta$

The horizontal component never changes. The vertical component decreases as the projectile rises, reaches zero at the top, then increases downwards.

The equations you actually use

These are the standard suvat equations, applied separately to each direction.

Quantity	Horizontal	Vertical
Acceleration	$0$	$g$ (downwards)
Velocity	$v_x = u\cos\theta$ (constant)	$v_y = u\sin\theta - gt$
Displacement	$x = u\cos\theta \cdot t$	$y = u\sin\theta \cdot t - \frac{1}{2}gt^2$

Two more that save time:

• Time to reach maximum height: $t = \frac{u\sin\theta}{g}$
• Maximum height: $H = \frac{(u\sin\theta)^2}{2g}$

A full worked example

A ball is launched at $u = 20 \text{ m s}^{-1}$ at $\theta = 30^\circ$ above the horizontal. Take $g = 9.81 \text{ m s}^{-2}$ and ignore air resistance.

Step 1: Resolve the velocity.

$u_x = 20\cos 30^\circ = 17.3 \text{ m s}^{-1}$

$u_y = 20\sin 30^\circ = 10.0 \text{ m s}^{-1}$

Step 2: Time to maximum height. At the top, vertical velocity is zero.

$t = \frac{u_y}{g} = \frac{10.0}{9.81} = 1.02 \text{ s}$

Step 3: Maximum height.

$H = \frac{u_y^2}{2g} = \frac{10.0^2}{2 \times 9.81} = 5.10 \text{ m}$

Step 4: Total time of flight. By symmetry, the time up equals the time down (for a level launch and landing), so:

$T = 2t = 2.04 \text{ s}$

Step 5: Horizontal range. Horizontal velocity is constant, so:

$R = u_x \times T = 17.3 \times 2.04 = 35.3 \text{ m}$

Every number here comes from treating the two directions separately and sharing only the time. That is the whole method.

Sign conventions that catch people out

• Pick a positive direction for the vertical axis and stick to it. If up is positive, then $g$ is negative in your equations.
• Horizontal displacement uses constant velocity, so never put $g$ into the horizontal calculation.
• The speed at any instant is the vector sum: $v = \sqrt{v_x^2 + v_y^2}$, and the direction is $\tan\theta = \frac{v_y}{v_x}$.

What changes when air resistance is included

The 9702 syllabus expects you to describe the effect qualitatively. With air resistance:

• The range is shorter.
• The maximum height is lower.
• The path is no longer symmetrical; the descent is steeper than the ascent.
• Horizontal velocity is no longer constant, because a drag force now acts backwards.

You will not usually be asked to calculate this, but you may need to explain it, so link the drag force to a reduced horizontal velocity.

Quick revision checklist

• Resolve the launch velocity into $u\cos\theta$ and $u\sin\theta$.
• Horizontal velocity is constant; vertical velocity changes by $g$ each second.
• Find the time first, then use it in the other direction.
• At maximum height, vertical velocity is zero, not the whole velocity.

If you want to check your working, ExamPal is an AI tutor that already knows the 9702 syllabus and marks your projectile answers against the Cambridge mark scheme, so you can confirm your method and significant figures are exam-ready.