Projectile motion

A cricket ball skied for a six, a stone skimmed off a rooftop, a football arcing toward the far post — every one of these traces the same curve once it leaves the hand or foot and gravity takes over. That curve is a projectile's trajectory, and the whole topic reduces to one trick: split the motion into a horizontal part that never changes and a vertical part that behaves exactly like free fall. Nothing about the two directions talks to the other — gravity only ever pulls straight down, so it can only ever change the vertical component of velocity. This article builds on the "Motion in a Plane" content of Unit 2, Kinematics, and stays inside the ISC syllabus: projectiles launched from level ground (or a known height), no incline, no air resistance. We derive the trajectory, time of flight, maximum height and range, work through the two Pass-1 examples in more depth, and flag the traps examiners love to set.

The core insight

A projectile's horizontal motion is constant-velocity and its vertical motion is independent free fall under $g$ , and combining the two gives a parabolic path with time of flight $T = 2u\sin\theta/g$ , maximum height $H = u^2\sin^2\theta/(2g)$ , and range $R = u^2\sin2\theta/g$ .

The core idea

A projectile is any body given an initial velocity and then left to move under gravity alone (air resistance neglected). The defining feature of projectile motion is that the horizontal and vertical motions are completely independent: horizontally there is no acceleration, so velocity stays constant; vertically the body accelerates downward at $g$ , exactly as in one-dimensional free fall. Nothing that happens in one direction affects the other — you can analyse each as a separate 1-D problem and only combine them at the end to describe the actual path. Take a projectile launched from the ground with speed $u$ at angle $\theta$ above the horizontal. Resolving the initial velocity gives a horizontal component that never changes and a vertical component that gravity steadily eats away and then reverses:

u_x = u\cos\theta, \qquad u_y = u\sin\theta.

where: $u$ is the launch speed, $\theta$ is the angle of projection measured from the horizontal, $u_x$ is the (constant) horizontal velocity component, and $u_y$ is the initial vertical velocity component. At any time $t$ after launch, the position follows directly from constant-velocity motion horizontally and free fall vertically:

x = u\cos\theta \cdot t, \qquad y = u\sin\theta \cdot t - \tfrac12 g t^2.

where: $x$ and $y$ are the horizontal and vertical displacements from the launch point, and $g$ is the acceleration due to gravity ( $9.8\ \text{m/s}^2$ unless stated otherwise). Eliminating $t$ between these two equations (using $t = x/(u\cos\theta)$ ) gives the shape of the path itself:

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

This is a parabola in $x$ and $y$ — the reason every projectile's path, ignoring air resistance, is a parabola, not a straight line or a circular arc. The figure below shows this parabolic path together with the three quantities the syllabus asks for: how high it rises, how long it stays up, and how far it travels.

Figure

A parabolic projectile path launched at an angle showing the rising and falling halves symmetric about the peak, with the maximum height marked above the peak and the horizontal range marked along the ground between launch and landing

At the peak the vertical velocity component is momentarily zero, giving the maximum height:

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

where: $H$ is the maximum height above the launch point. The total time the projectile spends in the air, from launch back to the same height ( $y = 0$ ), is the time of flight:

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

where: $T$ is the time of flight. Both follow from ordinary one-dimensional kinematics applied only to the vertical component — the horizontal motion plays no role in either.

Derivation ADeriving the horizontal range formula

The goal is to combine the independent horizontal and vertical motions of a projectile launched from the ground at speed $u$ and angle $\theta$ into a single closed-form expression for how far it travels along the ground before landing back at the same height it was launched from — the range $R$ — using nothing beyond the time of flight already found from the vertical motion and the constant horizontal velocity.

Step 1. Isolate the horizontal motion. Since there is no horizontal acceleration, the horizontal velocity component $u_x = u\cos\theta$ stays fixed for the entire flight, so the horizontal distance covered is simply this constant velocity multiplied by however long the projectile remains airborne.

Step 2. Bring in the time of flight from the vertical motion. The vertical motion alone (free fall under $g$ , starting with upward component $u\sin\theta$ ) fixes how long the projectile is in the air before returning to launch height: $T = 2u\sin\theta/g$ . This is where the two independent 1-D motions reconnect — the horizontal distance depends on a time that only the vertical motion determines.

Step 3. Multiply horizontal velocity by total time. $R = u_x \cdot T = (u\cos\theta)\left(\dfrac{2u\sin\theta}{g}\right) = \dfrac{2u^2\sin\theta\cos\theta}{g}$ .

Step 4. Simplify with the double-angle identity $2\sin\theta\cos\theta = \sin2\theta$ , giving the standard range formula: $R = \dfrac{u^2\sin2\theta}{g}$ . Since $\sin2\theta$ is largest (equal to 1) when $2\theta = 90°$ , range is maximised at $\theta = 45°$ , giving $R_{\max} = u^2/g$ ; and because $\sin2\theta = \sin(180° - 2\theta)$ , complementary angles $\theta$ and $90° - \theta$ always give the same range.

Worked example

A ball is thrown with a speed of $20\ \text{m/s}$ at an angle of $30°$ above the horizontal from level ground. Take $g = 9.8\ \text{m/s}^2$ . Find (a) the time of flight, (b) the maximum height, and (c) the horizontal range.

Step 1. Resolve the given data: $u = 20\ \text{m/s}$ , $\theta = 30°$ , so $\sin30° = 0.5$ and $\sin60° = 0.866$ .

Step 2. Time of flight: $T = \dfrac{2u\sin\theta}{g} = \dfrac{2(20)(0.5)}{9.8} = \dfrac{20}{9.8} \approx 2.04\ \text{s}$ .

Step 3. Maximum height: $H = \dfrac{u^2\sin^2\theta}{2g} = \dfrac{(20)^2(0.5)^2}{2(9.8)} = \dfrac{100}{19.6} \approx 5.10\ \text{m}$ .

Step 4. Horizontal range: $R = \dfrac{u^2\sin2\theta}{g} = \dfrac{(400)(0.866)}{9.8} \approx \dfrac{346.4}{9.8} \approx 35.35\ \text{m}$ . The ball is airborne for about 2.04 s, rises to about 5.1 m, and lands about 35.35 m from the launch point.

Competitive stretch

The same ball is now thrown horizontally ( $\theta = 0°$ ) from the top of a $20\ \text{m}$ tower at $15\ \text{m/s}$ , the Pass-1 companion problem. Here $u_y = 0$ , so vertical motion is plain free fall from $20\ \text{m}$ : $20 = \tfrac12(9.8)t^2$ gives $t \approx 2.02\ \text{s}$ , independent of the horizontal speed. The horizontal distance covered in that time is $x = u_x t = 15 \times 2.02 \approx 30.3\ \text{m}$ . The stone lands about 30.3 m from the base of the tower, about 2.02 s after release — a bullet fired horizontally and one simply dropped from the same height would hit the ground at exactly the same instant, since their vertical motions are identical.

Common traps

Using the initial speed $u$ directly instead of resolving it into $u\cos\theta$ and $u\sin\theta$ first — every formula here needs the components, not the raw speed.
Applying $R = u^2\sin2\theta/g$ when launch and landing heights differ (e.g., off a tower) — this range formula only holds when the projectile lands back at the launch height.
Forgetting horizontal velocity stays constant throughout the flight — a common slip is applying $g$ to the horizontal motion as well as the vertical.
Assuming a bigger launch angle always means a bigger range — range peaks at exactly $45°$ , and equal ranges appear for complementary angles like $30°$ and $60°$ .

Ask the tutor: if the trajectory, time-of-flight, height and range formulas keep blurring together, ask the AI tutor to walk through a few mixed problems with you.

Why does horizontal velocity not change even though the ball is clearly slowing down and speeding up as it goes up and comes down?

What you are seeing change is the vertical component, not the horizontal one. Gravity acts straight down, so it can only alter the vertical part of the velocity — the horizontal part has nothing pulling on it and stays exactly $u\cos\theta$ from launch to landing. The ball's overall speed does change (it is slowest near the peak), but that is entirely because the vertical component shrinks to zero and grows again; the horizontal component is a silent constant riding underneath the whole time.

Can I use the range formula R = u squared sin2θ over g for a ball thrown off a building?

No, not directly. That formula was derived assuming the projectile lands at the same height it was launched from, so the time of flight is $T = 2u\sin\theta/g$ . If the launch and landing heights differ — thrown from a tower, or landing in a pit below ground level — you must go back to the basic equation $y = u\sin\theta\, t - \tfrac12 g t^2$ , set $y$ to the actual vertical drop, solve for $t$ , and then multiply by $u\cos\theta$ to get the horizontal distance, exactly as in the tower example above.

Why is the trajectory a parabola and not, say, a straight line or a circle?

The trajectory equation $y = x\tan\theta - gx^2/(2u^2\cos^2\theta)$ has $y$ depending on $x^2$ , which is exactly the algebraic signature of a parabola. Physically this comes from combining a motion that is linear in time ( $x = u\cos\theta\, t$ ) with one that is quadratic in time ( $y$ involves $t^2$ because of constant acceleration). Any time you eliminate $t$ between a linear-in- $t$ expression and a quadratic-in- $t$ expression, you get a $y$ versus $x^2$ relationship — a parabola. This is why the shape never changes qualitatively, only its width and height, however $u$ or $\theta$ are chosen.

Projectile motion

The core idea

Worked example

Competitive stretch

Common traps

Before this

Where this leads