A basketball arcing toward the hoop, a ball thrown off a cliff, a stream of water from a garden hose — all follow the same curved path through the air. That path is a parabola, and the physics behind it is projectile motion. Once an object leaves the ground, only one force acts on it: gravity, pulling vertically downward at 9.8 m/s². Everything else — the speed, the angle, the shape of the arc — follows from that single fact.
Below, every key equation is derived from first principles — range, maximum height, time of flight, and the parabolic trajectory itself — with five worked examples, common mistakes corrected, and a full treatment of launches from height.
What Is Projectile Motion?
Projectile motion is the motion of an object launched into the air that subsequently moves under the influence of gravity alone — no engine thrust, no aerodynamic lift, no significant air resistance. Once launched, the only force acting on the object is its weight: F = mg directed vertically downward.
A basketball arcing toward the hoop, a ball rolling off a table, a stream of water from a hose, a long jumper leaving the ground — all are examples of projectile motion. The mathematics that describes them is identical, because the underlying physics is the same in every case: a constant downward force producing a constant downward acceleration, while the horizontal motion continues undisturbed.
Galileo understood the key insight in the early 17th century. Newton’s laws explain why it works.
The Key Assumption: Independent Components
The idea that makes projectile motion tractable is that horizontal and vertical motions are completely independent of each other.
Gravity acts only vertically. It has no horizontal component. Therefore it affects only the vertical velocity and cannot change the horizontal velocity at all. This follows directly from Newton’s second law applied component by component:
- Horizontal: Net horizontal force = 0, so horizontal acceleration = 0, so horizontal velocity is constant
- Vertical: Net vertical force = mg downward, so vertical acceleration = g downward, so vertical velocity changes continuously
A famous demonstration makes this vivid: drop one ball and fire another horizontally from the same height at the same instant. Both hit the ground at exactly the same time — because both have identical vertical motion. The horizontal velocity of the second ball is irrelevant to its vertical fall.
Setting Up the Equations
For an object launched with initial speed v0 at angle θ above the horizontal:
Initial velocity components:
v0x = v0 cos θ v0y = v0 sin θ
Position at time t:
x(t) = v0 cos θ · t y(t) = v0 sin θ · t − (1/2)gt²
Velocity at time t:
vx(t) = v0 cos θ (constant — no horizontal force)
vy(t) = v0 sin θ − gt
The horizontal velocity never changes. The vertical velocity decreases at rate g = 9.8 m/s² on the way up, reaches zero at the peak, then increases downward on the descent. At the peak, vy = 0 — but the object still moves horizontally at v0 cos θ. It is never completely stationary unless launched straight up.
The Four Key Results
These apply when the projectile lands at the same height it was launched from.
Time to maximum height:
t_top = v0 sin θ / g
Maximum height:
H = v0² sin²θ / (2g)
Total time of flight:
T = 2v0 sin θ / g
Horizontal range:
R = v0² sin(2θ) / g
The range equation reveals a key insight: sin(2θ) is maximised when θ = 45°. A 45° launch gives the maximum range for any given speed. Angles equally above and below 45° produce identical range — a 30° launch travels exactly as far as a 60° launch, though the 60° trajectory is higher and takes longer.
Equations Summary
| Quantity | Formula | Notes |
|---|---|---|
| Horizontal velocity | vx = v0 cos θ | Constant throughout flight |
| Vertical velocity | vy = v0 sin θ − gt | Zero at peak |
| Horizontal position | x = v0 cos θ · t | Linear in time |
| Vertical position | y = v0 sin θ · t − ½gt² | Quadratic in time |
| Time to peak | t_top = v0 sin θ / g | Same height launch only |
| Maximum height | H = v0² sin²θ / (2g) | Same height launch only |
| Total flight time | T = 2v0 sin θ / g | Same height launch only |
| Horizontal range | R = v0² sin(2θ) / g | Same height launch only |
The Parabolic Trajectory
From x(t): t = x / (v0 cos θ). Substituting into y(t):
y = x tan θ − [g / (2v0² cos²θ)] · x²
This is quadratic in x — a parabola. The trajectory of a projectile under uniform gravity is always exactly parabolic, regardless of launch angle or initial speed. This was Galileo’s insight, and it follows from constant horizontal velocity combined with constant vertical acceleration. The parabola opens downward and its shape depends only on launch angle and initial speed.
Energy Perspective on Projectile Motion
Conservation of energy gives an alternative route to the same results. At any point in the trajectory, total mechanical energy is conserved:
(1/2)mv0² = (1/2)mv² + mgy
This gives speed at any height y directly, without tracking time:
v = √(v0² − 2gy)
At the peak (height H), only horizontal velocity remains. The energy method is fastest when you need the speed at a given height but do not need the time or direction — one equation replaces several kinematic steps.
Kinematics gives positions and velocities at specific times. Energy methods give speeds at specific heights without needing time. Use whichever the problem calls for — often both together.
Projectile Launched from a Height
When a projectile launches from height h above the landing point, the symmetry assumptions break down. The four key results no longer apply. Return to the full kinematic equations.
For a horizontal launch (θ = 0°) from height h:
x(t) = v0 t y(t) = h − (1/2)gt²
Landing when y = 0:
t_land = √(2h/g)
Horizontal distance at landing:
R = v0 · √(2h/g)
The time of flight depends only on h and g — not on horizontal speed. A ball thrown at 5 m/s and one at 50 m/s from the same cliff hit the ground at the same time. Horizontal speed determines only where they land.
For an angled launch from height h, set y(t) = 0 in the full equation and solve the resulting quadratic for t. Take the positive root.
Common Mistakes
Mistake 1: Thinking velocity is zero at the peak. Only vy is zero. The horizontal speed v0 cos θ continues unchanged throughout — including at the highest point.
Mistake 2: Applying the range formula when heights differ. R = v0² sin(2θ)/g only works when launch and landing heights are equal. For cliff launches or elevated targets, use the full equations.
Mistake 3: Thinking gravity switches off at the peak. At the peak, vy = 0 but acceleration is still 9.8 m/s² downward. Gravity does not pause. The vertical velocity is momentarily zero but changing at rate g at every instant.
Mistake 4: Assuming 45° is always optimal. For launches and landings at different heights, or with air resistance, the optimal angle shifts — often below 45°.
Mistake 5: Applying g to the full initial speed instead of only the vertical component. Always split v0 into v0x = v0 cos θ and v0y = v0 sin θ first. Gravity acts only on the vertical component.
Real-World Applications
Sports science — in basketball, the optimal release angle depends on release height and distance to the hoop. Angles between 45° and 55° minimise sensitivity to small errors in release speed and give a larger effective target because the ball enters the hoop from above rather than at a shallow angle. Projectile motion analysis defines this trade-off precisely.
Ballistics and forensics — the range equation allows forensic scientists to reconstruct launch conditions from an impact site. Given the horizontal distance and the impact angle, the initial speed and launch point can be calculated. This is used in accident reconstruction and criminal investigations.
Aerospace — during coast phases between engine burns, a spacecraft follows a ballistic arc closely approximated by projectile motion equations. Mission planners use them to estimate downrange distance and plan the timing of subsequent burns.
Civil engineering — the parabolic shape of projectile trajectories appears in suspension bridge cable design. Under distributed traffic loading, a cable’s shape approaches a parabola. Engineers use the same quadratic mathematics to analyse cable tension and deflection.
Worked Examples
Example 1 — Maximum Height and Range
Problem: A ball is launched at 20 m/s at 35°. Find maximum height and range.
v0x = 20 cos 35° ≈ 16.4 m/s v0y = 20 sin 35° ≈ 11.5 m/s
H = (11.5)² / (2 × 9.8) = 132.25 / 19.6 ≈ 6.7 m
R = (20)² × sin 70° / 9.8 = 400 × 0.940 / 9.8 ≈ 38.4 m
Example 2 — Time of Flight
Problem: A projectile launches at 15 m/s at 60°. How long is it in the air?
T = 2v0 sin θ / g = 2 × 15 × sin 60° / 9.8 = 2 × 15 × 0.866 / 9.8 ≈ 2.65 s
Example 3 — Horizontal Launch from a Cliff
Problem: A ball is thrown horizontally at 12 m/s from a 45 m cliff. Find time of flight and horizontal distance.
t = √(2h/g) = √(90/9.8) = √9.18 ≈ 3.03 s
x = v0 · t = 12 × 3.03 ≈ 36.4 m
Example 4 — Speed at a Given Height
Problem: A projectile launches at 25 m/s at 40°. Find its speed at 8 m above the launch point.
v = √(v0² − 2gy) = √(625 − 2 × 9.8 × 8) = √(625 − 156.8) = √468.2 ≈ 21.6 m/s
Example 5 — Finding Both Launch Angles for a Target
Problem: A ball must reach a target 80 m away at the same height. Initial speed is 30 m/s. Find both possible angles.
sin(2θ) = Rg / v0² = (80 × 9.8) / 900 = 784 / 900 ≈ 0.871
2θ = 60.5° → θ1 ≈ 30.3° 2θ = 119.5° → θ2 ≈ 59.7°
Both angles give the same range. The lower angle produces a flatter, faster trajectory. The higher angle produces a steeper arc that takes longer.