Throw a ball, fire a cannon, or jump off a cliff (don't do that). Gravity pulls everything down at the same rate (-9.8 m/s²). The path that results is always a Parabola.
The secret that unlocks every projectile problem: horizontal and vertical motion are completely independent. Gravity only acts vertically, so the sideways speed never changes, while the up-and-down speed steadily loses 9.8 m/s every second. A bullet fired horizontally and a bullet dropped from the same height hit the ground at the same instant — the horizontal motion doesn't buy any extra airtime.
The Equations
x(t) = v_x * t (Horizontal motion is constant)
y(t) = v_y * t - 0.5 * g * t² (Vertical motion slows down)
You can graph these quadratics to see exactly where the object will land. If the launch speed is v at angle θ, split it first: v_x = v·cos(θ) and v_y = v·sin(θ). From there, three questions answer everything: When does it land (set y = 0)? How high does it go (vertical speed hits 0 at the top)? How far does it travel (plug landing time into x)?
Worked Example: A Kicked Football
A football is kicked at 20 m/s at 30° above the ground. Find the flight time, maximum height, and range. (Use g = 9.8 m/s².)
- Split the velocity. v_x = 20·cos(30°) = 17.3 m/s; v_y = 20·sin(30°) = 10 m/s.
- Flight time. The ball lands when y = 0: t(10 − 4.9t) = 0, giving t = 0 (launch) or t = 10/4.9 ≈ 2.04 s.
- Maximum height. The peak comes at half the flight time (t ≈ 1.02 s), when vertical speed reaches zero: y = 10(1.02) − 4.9(1.02)² ≈ 5.1 m.
- Range. Horizontal distance = v_x × flight time = 17.3 × 2.04 ≈ 35.3 m.
- Sanity check. The trajectory is symmetric: up in 1.02 s, down in 1.02 s, peak exactly halfway along the 35 m range. Everything is consistent.
Every projectile problem — cannonballs, basketball shots, water fountains — is this same recipe with different numbers.
Graphing It Yourself
A rewarding exercise: plot the football's trajectory as y against x. Using our numbers, the path is y = 0.577x − 0.0164x², a downward parabola you can enter directly into a graphing tool. You'll see the peak at about (17.6, 5.1) and the landing at x ≈ 35.3 — the same answers we calculated, now visible as geometry. Changing the launch angle in the equation and watching the arc flatten or steepen builds more intuition in five minutes than an hour of formula drills.
Why Always a Parabola?
Substitute t = x/v_x into the height equation and y becomes a quadratic function of x. Constant horizontal speed plus constant downward acceleration mathematically forces a parabolic arc — the same curve as y = ax² + bx from algebra class, just drawn in the air.
Common Mistakes to Avoid
- Applying gravity to horizontal motion. Gravity is strictly vertical. The horizontal velocity at landing is identical to the horizontal velocity at launch (ignoring air resistance).
- Swapping sine and cosine. The vertical component takes sin(θ), the horizontal takes cos(θ). Swapping them gives a flat, short trajectory when you expected a high, long one — or vice versa.
- Sign confusion with g. Pick a convention (up = positive, so acceleration = −9.8 m/s²) and hold it for the entire problem. Mixing signs mid-solution is the most common source of impossible answers.
- Assuming the speed at the peak is zero. Only the vertical velocity vanishes at the top. The projectile is still moving horizontally at full v_x — that's why it keeps traveling forward.
Frequently Asked Questions
What launch angle gives the maximum range?
45° — on flat ground with no air resistance, it perfectly balances airtime against forward speed. The range formula R = v²·sin(2θ)/g peaks when 2θ = 90°. Complementary angles (30° and 60°) land in exactly the same spot, just via different arcs.
Does the mass of the projectile change its path?
Not in the ideal (vacuum) model — a bowling ball and a marble launched identically follow identical parabolas, because gravity accelerates all masses equally. In real air, lighter and less aerodynamic objects feel drag more strongly, which is why a feather misbehaves.
How does air resistance change the real trajectory?
Drag steals speed continuously, so real projectiles fall shorter and steeper than the ideal parabola — the descent is more cramped than the ascent. Real-world optimal launch angles drop below 45° (long jumpers and shot putters use roughly 35–42°). Physics courses ignore drag first because the clean model captures the essential behavior.