Projectile Motion Explained: Range, Height, Flight Time — and Why 45° Is Not Always Best

Two independent motions glued together by one clock

Split the launch velocity into components: horizontally v₀ cos θ, vertically v₀ sin θ. Nothing pushes the projectile sideways once it is in the air, so the horizontal component never changes and the horizontal distance is simply speed times time. Vertically it is a free-fall problem with an upward start. The two axes never talk to each other except through the shared elapsed time, and that single fact generates all three formulas. Time of flight comes from the vertical axis returning to zero: T = 2v₀ sinθ/g. Feed that time into the horizontal axis and you get the range, v₀ cos θ × T, which trigonometry compresses to v₀² sin(2θ)/g.

The independence of the two axes has a consequence people find counter-intuitive: a ball fired horizontally and a ball simply dropped from the same height hit the ground at the same moment. The horizontal speed buys distance, never hang time. It also explains why the vertical component alone sets both the peak and the duration, and the horizontal component alone converts that duration into distance. When a problem mixes the two — substituting 20 m/s where only the 10.0 m/s vertical component of a 30° launch belongs — every subsequent number is wrong by the same trigonometric factor.

Why 45° maximises range, and why complementary angles tie

The range formula contains the angle only through sin(2θ), and a sine reaches its largest value of 1 when its argument is 90 degrees — that is, when θ is 45°. There the range collapses to v₀²/g, which for 20 m/s is 40.79 m. The same expression explains the ties: sin(2θ) takes the same value at θ and at 90° − θ, because 2(90° − θ) = 180° − 2θ and the sine of an angle equals the sine of its supplement. So 30° and 60° both land at 35.32 m, and 15° and 75° both at 20.39 m, from identical launch speeds.

Equal range does not mean equal trajectory, and choosing between two tied angles is a real decision. The 60° shot reaches 15.30 m and hangs for 3.53 s; the 30° shot peaks at 5.10 m and is down in 2.04 s. A fielder throwing to a base wants the flat option, because time in the air is time for the runner; an artillery piece firing over a ridge wants the steep one. It is also worth knowing how flat the maximum is: 40° and 50° both give 40.17 m, only 1.5 percent short of the best possible 40.79 m, so a couple of degrees of aiming error costs almost nothing.

The assumption almost every real example violates

R = v₀² sin(2θ)/g is derived by asking when the projectile comes back to its starting height. Almost nothing does. A ball leaves a hand at shoulder height and lands on the grass; a shot is put from above the ground; a stone is thrown off a cliff. With a launch height h the flight time becomes t = (v₀ sinθ + √(v₀² sin²θ + 2gh))/g and the range is v₀ cos θ × t. For 20 m/s at 45° released from 2.00 m, that is 42.70 m rather than 40.79 m — 4.7 percent further, and the level-ground formula understates it every time.

The optimal angle moves too. Launching from a height, the best angle satisfies sin²θ = v₀²/(2v₀² + 2gh), which for this example gives 43.66° and a range of 42.74 m. The gain over simply keeping 45° is four centimetres, so in practice the height matters far more than re-optimising the angle. Air resistance is the other omission, and it is the larger one for anything light or fast: a real thrown ball loses range and its best angle drops well below 45°, which is why the formulas here are labelled as the drag-free idealisation rather than as a description of a real throw.

Frequently asked questions

Is 45 degrees always the best launch angle?

Only when launch and landing are at the same height and air resistance is ignored. Raise the launch point and the optimum drops below 45°: from 2.00 m at 20 m/s the best angle is 43.66°. Add air resistance and it drops further, because a flatter trajectory spends less time being slowed. Aim below the target and the optimum rises above 45° instead. The 45° result is a clean consequence of a clean assumption, not a universal rule of thumb.

Does a heavier projectile travel less far?

Not in the idealised formulas: mass appears nowhere in R, H or T, so two objects launched at the same speed and angle follow identical paths regardless of what they weigh. In reality drag breaks the tie, and it breaks it in favour of the heavier object — the retarding force depends on size and shape, while the inertia resisting it depends on mass, so a dense body of the same shape is slowed less. That is why a steel ball outranges a plastic one of the same diameter thrown identically, even though the drag-free formulas say they should tie.

What is the speed at the highest point of the trajectory?

Not zero — that is the classic slip. At the apex only the vertical component has vanished; the horizontal component v₀ cos θ is untouched, because nothing has acted horizontally. A 20 m/s launch at 30° is still moving at 20 × cos 30° = 17.32 m/s when it peaks, and at 60° it is still doing 10.0 m/s. The acceleration at the apex is not zero either: it is the full 9.80665 m/s² pointing down, which is exactly why the vertical velocity does not stay at zero.