A ball thrown horizontally and a ball dropped vertically from the same height hit the ground at the same time. This single fact is the entire engine of projectile motion. Here is why it is true, and how to use it.

Projectile Motion: One Idea, Not Two

Fire a bullet horizontally. Drop a bullet from the same height at the same instant. Which hits the ground first?

Most people guess the dropped bullet — it has a head start, it's going straight down. But both bullets hit the ground at exactly the same time. The horizontal motion of the fired bullet has no effect whatsoever on how fast it falls.

This is the entire engine of projectile motion. Horizontal and vertical motions are completely independent of each other. They share only one thing: time. Once you see this clearly, projectile motion is not a new topic — it is K02 applied twice, simultaneously, in two perpendicular directions.

Why independence holds

The fired bullet travels horizontally. Gravity acts vertically. These two are perpendicular — at right angles to each other.

Force in the vertical direction changes only the vertical velocity. It cannot reach across and touch the horizontal velocity. This is exactly what vectors mean: components along perpendicular axes do not interfere with each other.

Galileo demonstrated this experimentally in the early 1600s — the simultaneous drop result above. But the reason it works is the vector structure of Newton's second law: F = ma is a vector equation, and a force in one direction produces acceleration only in that direction.

There is no horizontal force on a projectile (ignoring air resistance). So horizontal acceleration is zero. Horizontal velocity stays constant throughout the flight — the same at launch, at the peak, at landing. Gravity is the only force, and it acts purely vertically. The vertical motion is free fall — constant downward acceleration g.

These two motions run in parallel, linked only by the clock.

Setting up the equations

Positive direction: rightward for horizontal (x), upward for vertical (y). A projectile is launched from the origin with speed u at angle θ above the horizontal.

Initial conditions:

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

Horizontal motion (a = 0, uniform motion):

$x = u_x t = (u\cos\theta)\, t$

Vertical motion (a = −g, uniform acceleration downward):

$y = u_y t - \frac{1}{2}gt^2 = (u\sin\theta)\, t - \frac{1}{2}gt^2$

$v_y = u_y - gt = u\sin\theta - gt$

The horizontal equation is just distance = speed × time. The vertical equation is K02's s = ut + ½at² with a = −g.

That is the complete description. Everything else — time of flight, maximum height, range, trajectory shape — is extracted from these two equations.

Time of flight

The projectile lands when y = 0 again (back at the launch height):

$(u\sin\theta)\, t - \frac{1}{2}gt^2 = 0$

$t\left(u\sin\theta - \frac{g}{2}t\right) = 0$

Two solutions: t = 0 (the launch moment) and:

T = \frac{2u\sin\theta}{g}

The time of flight is determined entirely by the vertical component of launch velocity. The horizontal component plays no role. A projectile launched at a steeper angle stays airborne longer — because it has more upward velocity to lose before falling back.

Maximum height

At the peak, the vertical velocity is zero — the projectile is moving purely horizontally at that instant. Set v_y = 0:

$u\sin\theta - gt_{peak} = 0 \implies t_{peak} = \frac{u\sin\theta}{g} = \frac{T}{2}$

The peak is exactly halfway through the flight — by symmetry, which holds because the vertical motion is symmetric under constant acceleration. Substitute into y:

H = \frac{u^2\sin^2\theta}{2g}

Maximum height depends only on the vertical component squared. A ball thrown straight up (θ = 90°) reaches maximum height — all the launch speed goes into fighting gravity. A ball thrown horizontally (θ = 0°) has zero maximum height — it starts falling immediately.

Range

Range R is the horizontal distance covered during the full flight time T:

$R = u_x \cdot T = (u\cos\theta) \cdot \frac{2u\sin\theta}{g}$

$R = \frac{u^2 \cdot 2\sin\theta\cos\theta}{g}$

Using the identity 2sinθcosθ = sin2θ:

R = \frac{u^2\sin 2\theta}{g}

Range is maximised when sin2θ = 1, i.e., 2θ = 90°, i.e., θ = 45°. This is not a fact to memorise — it falls directly from the formula. At 45°, the launch speed is split equally between horizontal and vertical, giving the best compromise between time in the air and horizontal speed.

Two angles give the same range: θ and (90° − θ). For example, 30° and 60° give identical ranges. This follows from sin2θ = sin(180° − 2θ). The parabolic paths are different — one flatter, one steeper — but they land at the same point.

The trajectory shape

Eliminate t from the x and y equations to get y as a function of x — the shape of the path.

From the horizontal equation: $t = \frac{x}{u\cos\theta}$ . Substitute into the vertical:

$y = (u\sin\theta) \cdot \frac{x}{u\cos\theta} - \frac{1}{2}g\left(\frac{x}{u\cos\theta}\right)^2$

$y = x\tan\theta - \frac{g}{2u^2\cos^2\theta}\, x^2$

This is a parabola: y = Ax − Bx², where A and B are constants. The path is symmetric, peaks at x = R/2, and returns to y = 0 at x = R. The parabolic shape is a direct consequence of constant acceleration in one direction and zero acceleration in the other — one linear, one quadratic, combined.

{/* DIAGRAM: Parabolic trajectory with ux, uy components shown at launch, velocity vector at peak showing only horizontal component, symmetric path */}

Velocity at any point

At any time t, the velocity has two components:

$v_x = u\cos\theta \quad \text{(constant throughout)}$ $v_y = u\sin\theta - gt$

Speed at that instant:

$v = \sqrt{v_x^2 + v_y^2}$

Direction (angle with horizontal):

$\phi = \tan^{-1}\left(\frac{v_y}{v_x}\right)$

At the peak: v_y = 0, so v = v_x = u cosθ. The minimum speed during flight is at the peak, and it equals the horizontal component of launch velocity. Not zero — a common error.

Note: 1. Resolve the launch velocity into horizontal and vertical components: ux = u cosθ, uy = u sinθ. 2. Write the two equations separately. Horizontal: x = uxt. Vertical: y = uyt − ½gt², vy = uy − gt. 3. Identify what is asked. Time of flight? Range? Height? Velocity at a point? 4. Use the vertical equation to find time in most cases — it contains g and is usually where the constraint lives. 5. Use time in the horizontal equation to find horizontal distance. 6. Never mix horizontal and vertical in the same equation. They are solved in parallel, not together. 7. At the peak: vy = 0. Speed at peak = ux = u cosθ. 8. For landing on uneven ground: y ≠ 0 at landing. Set y equal to the landing height and solve the quadratic for t.

Warning: "At the peak, velocity is zero." Only the vertical component is zero at the peak. The horizontal component is u cosθ throughout — including at the peak. Speed at the peak is u cosθ, not zero.

> > **"Heavier objects fall faster in projectile motion."** No. In the absence of air resistance, g is the same for all masses. The horizontal component of initial velocity has no effect on how fast the object falls vertically. > > **"The range formula R = u²sin2θ/g always applies."** Only when the projectile lands at the same height it was launched from. If the landing height is different — a cliff, a slope, a table — derive from first principles using the two equations. > > **Forgetting to check which root of the quadratic to use.** When solving for time, you often get two solutions. t = 0 is always the launch. The other is the landing. Always identify which is which before using the value. > > **Using g = +10 m/s² in the vertical equation without checking the sign convention.** If upward is positive, a = −g = −10 m/s². If downward is positive, a = +g = +10 m/s² and all signs flip. Pick one convention and apply it everywhere.

A worked example

A ball is launched from the ground at 20 m/s at 30° above the horizontal. Take g = 10 m/s². Find: (a) time of flight, (b) maximum height, (c) range, (d) speed at the highest point.

Initial components:

$u_x = 20\cos30° = 20 \times \frac{\sqrt{3}}{2} = 10\sqrt{3} \approx 17.3 \text{ m/s}$

$u_y = 20\sin30° = 20 \times \frac{1}{2} = 10 \text{ m/s}$

Part (a) — time of flight:

$T = \frac{2u_y}{g} = \frac{2 \times 10}{10} = 2 \text{ s}$

Part (b) — maximum height:

$H = \frac{u_y^2}{2g} = \frac{100}{20} = 5 \text{ m}$

Part (c) — range:

$R = u_x \times T = 10\sqrt{3} \times 2 = 20\sqrt{3} \approx 34.6 \text{ m}$

Or using the formula: R = u²sin2θ/g = 400 × sin60°/10 = 40 × (√3/2) = 20√3 ✓

Part (d) — speed at highest point:

At the peak, vy = 0. Speed = vx = ux = 10√3 ≈ 17.3 m/s horizontally.

Twist variant

Same launch, but now the ball is launched from the edge of a cliff 20 m above the ground below. How far from the base of the cliff does it land?

The range formula no longer applies — the landing height is −20 m (below the launch point). Work from first principles.

Vertical equation with y = −20 m at landing:

$-20 = 10t - \frac{1}{2}(10)t^2 = 10t - 5t^2$

$5t^2 - 10t - 20 = 0 \implies t^2 - 2t - 4 = 0$

$t = \frac{2 \pm \sqrt{4 + 16}}{2} = \frac{2 \pm \sqrt{20}}{2} = 1 \pm \sqrt{5}$

Taking the positive root: t = 1 + √5 ≈ 1 + 2.236 = 3.236 s

Horizontal distance from launch point: x = ux × t = 10√3 × 3.236 ≈ 56.0 m

Compare with the flat-ground range of 34.6 m — the extra height gives the ball an additional 21.4 m. The range formula would have given the wrong answer here; the two-equation approach always works.

What to remember

Horizontal and vertical motions are independent. They share only time.

Horizontal: constant velocity, zero acceleration. Vertical: constant acceleration g downward, identical to free fall.

At the peak: vertical velocity is zero, horizontal velocity is unchanged.

The three derived results — T, H, R — are valid only for launch and landing at the same height. For any other geometry, return to the two equations and solve from scratch.

K04 extends this idea of independent components to relative motion — what velocity looks like from a moving reference frame. The river-boat and rain-man problems are vector addition problems in disguise, and the tools from M01 and K01 are everything you need.