3.4.1.4MechanicsCore

Projectile Motion

A projectile is any object moving under gravity alone, once launched. The key insight is that horizontal and vertical motion are $independent$ : they share the same time, but otherwise you treat them as two separate SUVAT problems.

40 min Video by Science Shorts 3.4.1 Force, energy and momentum

Projectile Motion — AS/A-Level PhysicsWatch the full walkthrough before the notes below.

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What you'll be able to do

Explain why horizontal and vertical motion can be treated independently
Resolve a launch velocity into horizontal and vertical components
Use SUVAT separately in each direction, linked by a common time
Calculate time of flight, maximum height and horizontal range
Explain (qualitatively) the effect of air resistance on a real projectile

The big idea: independence of motion

Gravity acts only vertically, so it changes the vertical velocity but never the horizontal velocity. That means the $horizontal$ motion is at constant velocity ( $a = 0$ ), while the $vertical$ motion is uniform acceleration with $a = g$ downward.

The one thing the two directions share is $time$ : an object is in the air for the same length of time whether you look at its horizontal or vertical motion. Time is the bridge between the two SUVAT calculations.

Tip — Set out two columns — horizontal and vertical — and list u, v, a, s, t separately for each. The shared value of t connects them.

Horizontally launched projectiles

If an object is launched horizontally (e.g. rolling off a table), its initial vertical velocity is zero. The vertical drop sets the time in the air, and that same time fixes how far it travels horizontally.

s_{x} = u_{x} t

Horizontal: constant velocity, so distance = speed × time.

s_{y} = \frac{1}{2} g t^{2}

Vertical drop from rest, taking down as positive.

1Vertical: find the time to fall

1.25

m from rest.

s_{y} = \frac{1}{2} g t^{2}

gives

1.25 = \frac{1}{2} (9.81) t^{2}

2So

t^{2} = 0.255

and

t = 0.50

3Horizontal:

s_{x} = u_{x} t = 2.0 \times 0.50 = 1.0

Answer

s_{x} = 1.0

Projectiles launched at an angle

When a projectile is launched at angle $θ$ with speed $u$ , first resolve the launch velocity: the horizontal component is $u cos θ$ and the vertical component is $u sin θ$ .

At maximum height the vertical velocity is momentarily zero (the horizontal velocity is unchanged). By symmetry over level ground, the time to the top is half the total time of flight.

u_{x} = u cos θ, u_{y} = u sin θ

Resolve the launch velocity first.

1Vertical component:

u_{y} = 20 sin 3 0^{\circ} = 10

m/s.

2At the top

v_{y} = 0

, so time up:

v = u + a t

gives

0 = 10 - 9.81 t

t = 1.02

3Total flight (up and back down)

= 2 \times 1.02 = 2.04

AnswerTime of flight

= 2.0

Tip — Take one direction as positive and be consistent. If up is positive, the vertical acceleration is −9.81 m/s² for the whole flight, including on the way up.

The effect of air resistance

The calculations above ignore air resistance. In reality, drag acts opposite to the velocity, reducing both the range and the maximum height, and making the path $asymmetric$ — the descent is steeper than the ascent. Exam answers should state this when asked to compare an ideal projectile with a real one.

Equation recap

u_{x} = u cos θ

Horizontal component of launch velocity.

u_{y} = u sin θ

Vertical component of launch velocity.

s_{x} = u_{x} t

Horizontal: constant velocity.

s_{y} = u_{y} t - \frac{1}{2} g t^{2}

Vertical displacement (up positive).

Common mistakes to avoid

Applying g to the horizontal motion.

Gravity is vertical only — horizontal velocity stays constant throughout the flight.

Forgetting to resolve an angled launch velocity into components.

Always split u into u cosθ (horizontal) and u sinθ (vertical) before using SUVAT.

Thinking the vertical velocity is zero throughout at the top.

Only the vertical velocity is zero at the highest point; the horizontal velocity is unchanged.

Key takeaways

A projectile moves under gravity alone; horizontal and vertical motion are independent.
Horizontal motion is at constant velocity; vertical motion has a = g.
Time is shared between the two directions — it links the two SUVAT calculations.
Resolve an angled launch into u cosθ and u sinθ before solving.

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