Projectile Motion Without Tears: Range, Height, and Hang Time

Published 2026-03-17 · By the CalculatorHive editorial team

Projectile motion looks intimidating because it appears to be one complicated two-dimensional problem. It isn't. It is two simple one-dimensional problems that happen at the same time and share a clock. Once you split the launch velocity into a horizontal part and a vertical part, everything else is substitution.

Why do you split velocity into components?

Gravity pulls straight down. It has no sideways component at all, so it cannot change how fast the projectile moves horizontally. That single fact lets you treat the two directions separately:

• Horizontal: v_x = v·cos(θ), constant for the entire flight. Distance is just v_x × time.
• Vertical: v_y = v·sin(θ) at launch, decreasing by 9.81 m/s every second, reaching zero at the top, then increasing downward at the same rate.

The classic demonstration: a bullet fired horizontally and a bullet dropped from the same height at the same instant hit the ground simultaneously. The fired one travels hundreds of metres downrange, but its vertical motion is identical to a plain drop — see the free fall calculator for the vertical-only case.

The vertical motion is also symmetric on level ground. The time up equals the time down, and the projectile arrives with the same speed it left with, mirrored below the horizontal. That symmetry is where the three standard equations come from.

What are the three projectile motion equations?

For a launch from ground level onto level ground, with launch speed v, angle θ above the horizontal, and g = 9.81 m/s²:

• Flight time: T = 2v·sin(θ) ÷ g
• Maximum height: H = v²·sin²(θ) ÷ (2g)
• Range: R = v²·sin(2θ) ÷ g

The range equation is just the horizontal velocity multiplied by the flight time — v·cos(θ) × 2v·sin(θ)/g — tidied up with the identity 2·sin(θ)·cos(θ) = sin(2θ). Notice what appears and what doesn't. Speed enters squared, so a 10% faster throw goes 21% further. Mass appears nowhere at all.

A worked example: 20 m/s at 35°

Throw a ball at 20 m/s at 35° above the horizontal, releasing and landing at ground level.

1. Components. v_x = 20 × cos(35°) = 20 × 0.8192 = 16.38 m/s. v_y = 20 × sin(35°) = 20 × 0.5736 = 11.47 m/s.
2. Time to the peak. The vertical velocity has to bleed off at 9.81 m/s per second: 11.47 ÷ 9.81 = 1.17 s.
3. Total flight time. Double it by symmetry: 2.34 s.
4. Maximum height. H = 11.47² ÷ (2 × 9.81) = 131.6 ÷ 19.62 = 6.71 m.
5. Range. R = 16.38 × 2.34 = 38.3 m.

Check the range against the direct formula: 20² × sin(70°) ÷ 9.81 = 400 × 0.9397 ÷ 9.81 = 38.3 m. They agree, as they must. The projectile motion calculator runs this in one step and also handles launches from a raised starting height, where the symmetry argument no longer applies.

Two things worth noticing about the landing. The ball arrives at 20 m/s — the same speed it left at, just angled 35° below the horizontal — because gravity gave back exactly the vertical speed it took away. Its kinetic energy at impact therefore equals its kinetic energy at launch; the kinetic energy calculator converts that speed into joules for a given mass.

Why does 45° give the maximum range?

Range depends on sin(2θ), and sine peaks at 90°, so 2θ = 90° means θ = 45°. Physically it is a trade: a shallow angle gives lots of horizontal speed but almost no hang time, while a steep angle gives lots of hang time but little horizontal speed. Range is the product of the two, and 45° balances them exactly.

Because sine is symmetric about 90°, complementary angles produce identical ranges: sin(2 × 30°) = sin(60°) and sin(2 × 60°) = sin(120°) are both 0.866. The 30° shot is fast and flat; the 60° shot is slow and lofted; both land 35.3 m away at 20 m/s. The high one just takes 73% longer to get there and climbs three times as high.

All values at v = 20 m/s, g = 9.81 m/s², level ground, no air resistance.

The most useful column here is the range one, and the useful observation is how flat the curve is near the top. At 40° you still get 40.2 m — 98.5% of the maximum — and the same at 50°. You lose less than 2% of your range for a 5° error either side of optimal. Height and hang time, by contrast, keep climbing all the way to 90°. That is why a punter in football aims high: hang time, not distance, is the objective.

Where does this model break down?

Every number above assumes no air resistance, a flat launch and landing height, constant gravity, and no spin. Three of those matter in practice.

Air resistance is the big one. Drag rises roughly with the square of speed and always opposes motion, which shortens range and — importantly — makes the trajectory asymmetric: the descent is steeper than the ascent. Because drag punishes the longer, slower, higher path more than the flat one, the optimal angle drops. For real sports projectiles it typically lands in the 30–40° range rather than 45°, and for very draggy objects, such as a badminton shuttle, far lower still.

Launch height. If you release above the landing surface — a shot put from shoulder height, a ball thrown off a cliff — the optimal angle is always below 45°, because the extra falling distance is free hang time you don't have to buy with a steep launch. The greater the height advantage relative to the range, the flatter the optimum.

Lift and spin. Backspin on a golf ball or topspin on a tennis ball generates a Magnus force perpendicular to the flight path, which can extend or curtail range dramatically. No ballistic equation covers it.

What the simple model is good for: physics problem sets, order-of-magnitude estimates, dense and slow projectiles where drag is small compared with weight, and building intuition about which variables matter. If you need the force behind the launch rather than its result, the force calculator handles F = ma.

Common questions

Does a heavier ball travel further?

Not in a vacuum — mass cancels out of all three equations, so a marble and a cannonball launched at the same speed and angle land in the same place. In air it does matter, but indirectly: a heavier object of the same size has more momentum relative to the drag force acting on it, so it decelerates less. That is why a bowling ball outflies a beach ball thrown identically.

What if the projectile lands lower than it was launched?

The symmetry breaks and the simple flight-time formula no longer applies. Solve the vertical position equation y = v·sin(θ)·t − ½gt² for the time when y equals the negative drop height (a quadratic in t), then multiply that time by v·cos(θ) for the range. The result is always a longer flight and a longer range than the level-ground formula predicts, with a flatter optimal angle.

Why doesn't the horizontal speed slow down?

In the idealised model, nothing acts horizontally — gravity is purely vertical and there is no air. With no horizontal force there is no horizontal acceleration, so v_x is constant by Newton's first law. In reality drag does act along the flight path and does have a horizontal component, which is exactly why real trajectories fall short of the calculated range.

At what angle should I actually throw or kick something?

For maximum distance with a dense object at modest speed, aim close to 45° — the curve is flat enough near the top that anywhere from 40° to 50° is effectively optimal. Release above ground level, or throw something the air catches, and flatten it toward 35–40°. If you want hang time rather than distance, go steeper: 60° gives 86% of the maximum range but 23% more time in the air.