Projectile Motion: Complete Guide with Worked Examples

What Is Projectile Motion?

A projectile is any object that, once launched, moves only under the influence of gravity — no engine, no thrust, no air resistance (in the idealised model). A thrown ball, a bullet fired horizontally, a long-jumper in the air, a droplet of water from a fountain: all projectiles.

What makes projectile motion so elegant — and so important as a teaching topic — is that it's the simplest possible example of two-dimensional kinematics. The horizontal and vertical components of motion are completely independent of each other. Gravity only acts downward; it doesn't touch the horizontal velocity at all. That independence is the insight that unlocks everything.

Galileo Galilei was the first to analyse projectile motion mathematically, around 1590. He showed that the path of a projectile (ignoring air resistance) is a parabola — a result that contradicted centuries of Aristotelian physics, which held that projectiles followed a straight-line path before suddenly dropping. See our Classical Mechanics guide for the full historical context and Newton's Laws, which provide the underlying framework.

The Assumptions — and When They Break Down

The standard projectile motion model makes three assumptions that are worth being explicit about, because real-world projectiles violate all three:

1. No air resistance. In reality, drag forces slow projectiles, especially at high speeds. A real golf ball travels far less than the ideal model predicts. Air resistance makes the maths significantly harder and the trajectory asymmetric.
2. Constant gravitational acceleration. We assume g = 9.8 m/s² throughout the flight. This is valid for objects that don't reach significant altitude. For rockets or ballistic missiles, g varies with height.
3. Flat Earth approximation. We treat the ground as flat. For very long-range projectiles (intercontinental ballistic missiles, for example), the curvature of the Earth matters.

For almost every physics exam problem and most real-world engineering estimates, these assumptions hold well enough. A cricket ball bowled at 140 km/h travels about 20 metres — the ideal model gives an excellent approximation. It's when you extend to extreme cases that you need the more complete picture.

The Core Equations — Derived from Scratch

We start from the two kinematic equations of uniform acceleration (derived in our Classical Mechanics guide):

For a projectile launched at speed v₀ and angle θ above the horizontal, we split the initial velocity into components:

Now we apply the kinematic equations to each direction independently:

Horizontal Motion (a = 0)

Vertical Motion (a = −g)

The Three Key Results Derived

1. Time of Flight

The projectile lands when y = 0 again (assuming it returns to the same height it was launched from):

This is a satisfying result: the time in the air is entirely determined by the vertical component. If you fire two projectiles at the same angle but one goes twice as fast, the faster one spends twice as long in the air.

2. Maximum Height

At maximum height, the vertical velocity is zero: vᵧ = 0. Substituting into vᵧ = v₀ sin θ − gt:

3. Range

Substitute the time of flight T back into the horizontal position equation:

Why Does 45° Give Maximum Range?

This is one of the most satisfying results in introductory physics. The range formula is R = v₀² sin 2θ / g. For fixed v₀ and g, the range is maximised when sin 2θ is maximised — and sin achieves its maximum value of 1 when its argument equals 90°.

So: 2θ = 90° ⟹ θ = 45°. Maximum range at 45°.

But here's the deeper geometric insight: at 45°, the horizontal and vertical components of initial velocity are equal (v₀ cos 45° = v₀ sin 45° = v₀/√2). Any angle lower than 45° means too little vertical velocity — the projectile doesn't spend enough time in the air to travel far horizontally. Any angle higher than 45° means too much vertical velocity — the projectile goes high but comes straight down without much horizontal distance.

A beautiful symmetry: complementary angles give equal range. A shot at 30° and a shot at 60° (both launched at the same speed) travel exactly the same horizontal distance — but the 60° shot goes higher and takes longer. This follows directly from sin 2(30°) = sin 60° and sin 2(60°) = sin 120° = sin 60°.

Six Worked Examples

Common Mistakes and Misconceptions

• Mixing up v₀ sin θ and v₀ cos θ. Remember: sin gives the component in the direction of θ (vertical for launch angle), cos gives the adjacent component (horizontal). Drawing a right-angled velocity triangle every time eliminates this error.
• Using g = 9.8 m/s² with the wrong sign. Choose a sign convention (up positive is standard) and stick to it. In y(t) = v₀ sin θ · t − ½gt², the minus sign is already built in for upward-positive convention.
• Assuming the range formula works for non-level ground. R = v₀² sin 2θ / g only applies when the launch and landing heights are equal. For elevated or depressed launches, go back to y(t) = 0 and solve the quadratic (as in Example 5).
• Forgetting the two solutions when finding angle from range. Because sin 2θ = sin(180° − 2θ), there are always two launch angles (below and above 45°) that produce the same range. Both are physically valid; choose based on context.
• Treating maximum height as occurring at the halfway point in time. This only holds for level-ground launches. For elevated launches, it occurs at t = v₀ sin θ / g regardless of ground height.

Frequently Asked Questions