Module 3: Kinematics · Theory · Prerequisites: K11, Foundation 4
One motion, two equations, one clock. The clock is what makes them a single problem rather than two separate ones.
Mahavakya
Projectile motion is two independent one-dimensional motions sharing the same clock. The components don’t know about each other. Time does.
On this page
Module 1 established the independence of perpendicular components and demonstrated it with dropped and projected balls — showing the horizontal motion continuing unchanged as the vertical motion added below it. This part uses that result without repeating it. The independence is K11’s; what follows is what it produces.
1. Setting up
A particle is projected from the origin at speed u at angle θ above the horizontal. Gravity acts downward with magnitude g. Take rightward and upward as positive.
horizontal: ax = 0, vx = u cos θ (constant)
vertical: ay = −g, vy = u sin θ − gt
This is K7 in the horizontal direction and K8 in the vertical direction, with one shared t. That shared clock is the only coupling between the two components, and it is what makes this a two-dimensional problem rather than two one-dimensional ones.
Horizontal: constant velocity. Vertical: constant acceleration. Same t couples them.
2. The four standard results
Applying K8’s equations to each component, then connecting via t:
Time to maximum height. Set vy = 0:
tH = u sin θ / g
Maximum height. Substitute into y(t):
H = u² sin² θ / (2g)
Time of flight. The trajectory is symmetric, so the total time is twice the rise time — or set y = 0:
T = 2u sin θ / g
Range. Horizontal component over the full flight time:
R = u² sin 2θ / g
T = 2u sin θ/g and R = u² sin 2θ/g hold only for equal launch and landing heights. Both are obtained by setting y = 0, which is only true when the particle returns to the level it started from. Launch from a cliff, or land on a slope, and these formulas do not apply. Return to the equations.
3. The trajectory equation
Eliminate t between x = u cos θ · t and the y equation:
y = x tan θ − gx² / (2u² cos² θ)
This is y as a function of x — the trajectory shape, not the motion in time. It is a parabola opening downward.
Foundation 7’s Contemplation noted that y(x) and y(t) are both parabolas and entirely different objects — one is the path, the other is the motion. Their slopes carry different units: dy/dx is dimensionless (a slope on the path), while dy/dt is m/s (the vertical velocity). They are not interchangeable.
Worked example
A ball is thrown at 20 m/s at 30° above horizontal (g = 10). Find the speed at the highest point, the maximum height, and the range.
Components: ux = 20 cos 30° = 10√3 m/s, uy = 20 sin 30° = 10 m/s.
Speed at top: only the horizontal component survives, so v = 10√3 ≈ 17.3 m/s. Not zero.
H = 100 / 20 = 5 m T = 2 s R = 10√3 × 2 = 20√3 ≈ 34.6 m
Always state the speed at the apex, not the velocity. The velocity is purely horizontal; the speed is its magnitude.
4. Complementary angles
Since sin 2θ = sin(180° − 2θ) = sin 2(90° − θ), angles θ and 90° − θ give the same range.
Equal range does not mean identical motion.
The maximum heights are u² sin² θ/(2g) and u² cos² θ/(2g). Their ratio is tan² θ, and for 30° and 60° that is exactly 3. Same launch speed, same range — but one goes three times higher and takes longer to do it.
The figure makes this visible in a way the algebra does not: the 60° trajectory is noticeably narrower and taller, while the 30° one is wide and low. Equal range is a one-dimensional comparison; the motions differ in every other respect.
5. The graphs of projectile motion
K6 argued that a problem becomes easier when you choose the graph that makes the relevant feature visible. Projectile motion has two natural graphs.
vx–t graph — a horizontal line at u cos θ. The horizontal motion is uniform and the line never moves. Its area gives the horizontal displacement.
vy–t graph — a straight line with slope −g, crossing zero at tH. This is a standard K8 graph. Its signed area gives the vertical displacement; the crossing is the apex; and the signed area over the whole flight is zero, which says the landing height equals the launch height.
Several problems become graphical. The signed area crossing zero at tH tells you the trajectory is symmetric without solving anything. And the total signed area being zero immediately gives you the time of flight: the two triangles must be equal, so T = 2tH.
6. Elevated launches
When the landing height differs from the launch height, the standard formulas break. The procedure does not.
Worked example
A ball is projected horizontally at 15 m/s from the top of a cliff 45 m high. How far from the base does it land? (g = 10)
Horizontal: ux = 15, ax = 0. Vertical: uy = 0, ay = −10. Landing at y = −45.
−45 = 0 − ½(10)t² → t = 3 s
x = 15 × 3 = 45 m
The horizontal projection formula R = u² sin 2θ/g would give zero for θ = 0. Ignore it and work from the component equations directly.
7. The method
Every projectile problem reduces to the same sequence, in the same order.
- Choose axes and sign convention — horizontal positive right, vertical positive up, unless something else makes the constraint simpler.
- Resolve the initial velocity — ux = u cos θ, uy = u sin θ.
- Write the component equations — two equations with one shared t.
- State what the problem requires — a height, a range, a time, an angle at impact.
- Apply the condition — set vy = 0 for the apex, set y = 0 for equal-height landing, or set both coordinates to the target position.
- Check physical admissibility — is t positive? Does the position lie on the intended part of the trajectory?
Steps 1 to 3 are setup and take under a minute. Step 6 is K7’s lesson applied to two-dimensional geometry. Steps 4 and 5 are the part students skip — knowing what condition to apply, and applying it before starting algebra, saves more time than any shortcut formula.
Contemplation
The trajectory parabola is a geometrical object — a curve in space. The motion is an event in time. They are not the same thing, and the trajectory equation hides the time entirely.
That is often useful: the trajectory tells you where the ball will be without telling you when, and for questions about aim and clearance the when is irrelevant. But the flight time comes from the motion, not the trajectory, and no amount of reading the parabola will produce it.
The clock is what you carry between the two component equations. The parabola is what you see when you throw the clock away. Knowing which description answers which question is the real skill in this topic — not the formulas.
Common misconceptions
1. The ball stops at the apex
Only vy = 0 at the apex. The horizontal velocity is untouched, and the speed there is u cos θ, which is not zero for any θ < 90°.
2. T = 2u sin θ/g always
Only when landing and launch heights are equal. Set y equal to the actual landing height and solve the resulting quadratic.
3. The range formula gives the horizontal distance for any launch
It gives the horizontal displacement when the vertical displacement is zero. For elevated launches or landings on slopes, use the component equations directly.
4. Equal range means identical motion
Complementary angles give the same range with completely different heights, times of flight and intermediate positions.
5. The trajectory equation gives the velocity
The slope dy/dx is a geometric gradient with no units. The velocity involves dy/dt, which has units of m/s. They are different derivatives of different functions.
Reflection
- Maximum range is at θ = 45°. What does the trajectory look like for very small θ, and what happens as θ approaches 90°?
- A projectile and a ball dropped from the apex at the same instant: do they land at the same time? Why?
- At what point on a projectile’s path is the speed minimum? Write it in terms of u and θ.
Key takeaways
- Horizontal: constant velocity u cos θ. Vertical: constant acceleration −g. Shared t.
- H = u² sin² θ/(2g); T = 2u sin θ/g; R = u² sin 2θ/g — all for equal-height launch and landing only.
- The trajectory y(x) shows the path. The motion y(t) shows how it is traversed. Their gradients differ in both meaning and units.
- Complementary angles give the same range but differ in height by tan² θ.
- For non-level launches: ignore the formulas, write the component equations, apply the appropriate geometric condition.
- Speed at apex = u cos θ; never zero for θ < 90°.
What comes next
The trajectory was described relative to the ground. K13 asks what it looks like from a different frame — a frame that is also moving. The subtraction that produced relative velocity in K7 extends directly here, and several problems that look complicated in the ground frame become simple in a moving one.
Prerequisites: K11 and Foundation 4 · Reading: 16 min · Practice: 40 min · Difficulty: Intermediate