3.1MechanicsCore

Projectile Motion

Fire a bullet perfectly horizontally at the exact same instant you simply drop one from the same height, and — ignoring air resistance — both hit the ground at exactly the same time. That surprising result falls straight out of treating horizontal and vertical motion as two completely independent problems happening simultaneously.

40 min Video by Physics Online Module 3.1 Motion

Projectile Motion (of a Bullet) — 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 in projectile motion
Use SUVAT equations separately for the horizontal and vertical components of projectile motion
Calculate the time of flight, range and maximum height of a projectile
Solve problems involving a projectile launched at an angle

Independence of horizontal and vertical motion

A projectile’s motion can always be split into two completely independent components: a horizontal component with $constant velocity$ (assuming no air resistance, since gravity acts only vertically), and a vertical component with $constant acceleration$ $g$ downward — behaving exactly like an object simply dropped or thrown straight up. Neither component affects the other at all.

Tip — This is why a bullet fired horizontally and one simply dropped from the same height land at the same time — their vertical motion is identical in both cases; only the horizontal motion differs.

Horizontally launched projectiles

For an object launched purely horizontally, the initial vertical velocity is zero, so the SUVAT equations apply independently and simply to each direction.

Horizontal: x = v_{x} t Vertical: y = \frac{1}{2} g t^{2}

Horizontal (constant velocity) and vertical (constant acceleration g) motion.

1Vertical motion only:

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

, so

20 = 0.5 \times 9.81 \times t^{2}

t^{2} = \frac{20}{4.905} = 4.08

, so

t = 4.08

3Time of flight

t \approx 2.02

4Horizontal distance:

x = v_{x} t = 12 \times 2.02

AnswerTime of flight ≈ 2.02 s; horizontal distance ≈ 24.2 m.

Tip — The time of flight is found ENTIRELY from the vertical motion — the horizontal velocity has no effect whatsoever on how long the fall takes.

Projectiles launched at an angle

When a projectile is launched at an angle $θ$ above the horizontal with speed $u$ , resolve the initial velocity into horizontal and vertical components first ( $u cos θ$ and $u sin θ$ ), then treat each component with its own SUVAT equations exactly as before.

1Vertical component of initial velocity:

u_{y} = u sin θ = 18 sin 40° = 11.57

m s⁻¹.

2Maximum height (where vertical velocity = 0):

v^{2} = u_{y}^{2} - 2 g h \Rightarrow h = \frac{u _{y}^{2}}{2 g} = \frac{11.5 7 ^{2}}{2 \times 9.81}

h \approx 6.82

4Time of flight (up and back down to the same height):

t = \frac{2 u _{y}}{g} = \frac{2 \times 11.57}{9.81}

AnswerMaximum height ≈ 6.82 m; time of flight ≈ 2.36 s.

Tip — Always resolve the launch velocity into components FIRST — trying to work with the single launch speed and angle directly in a SUVAT equation is the most common source of errors in this topic.

Equation recap

x = v_{x} t

Horizontal displacement (constant velocity).

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

Vertical displacement for a horizontally-launched projectile.

u_{x} = u cos θ, u_{y} = u sin θ

Resolving a launch velocity at angle θ.

Common mistakes to avoid

Using the full launch speed (rather than its vertical component) to find time of flight for an angled projectile.

Always resolve the launch velocity first — only the VERTICAL component determines the time of flight.

Believing horizontal velocity affects how long a projectile stays in the air.

Horizontal and vertical motion are completely independent — changing horizontal speed alone never changes the time of flight.

Forgetting gravity acts throughout the entire flight, including a projectile’s upward stage.

Gravity decelerates a projectile on the way up and accelerates it on the way down — it acts continuously and unchanged throughout the whole flight.

Key takeaways

Horizontal motion (constant velocity) and vertical motion (constant acceleration g) are entirely independent in projectile motion.
For a horizontally-launched projectile, time of flight comes purely from the vertical drop; horizontal distance = horizontal velocity × time of flight.
For a projectile launched at angle θ, resolve the launch velocity into ux=ucosθ and uy=usinθ before applying SUVAT to each component separately.
Maximum height and time of flight depend only on the vertical component of velocity; range depends on both components.

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Projectile Motion

40 min Video by Physics Online Module 3.1 Motion