The angle that throws furthest, and why nobody notices

Throw a ball as far as possible and the answer is forty-five degrees. It is one of the first genuinely quantitative results anybody meets in physics, it is exactly right, and almost nobody throwing a ball has ever used it.

That is not because people are bad at physics. It is because the maximum is extraordinarily flat, and the physics that produces the flatness is more interesting than the physics that produces the answer.

The trick that makes it easy #

The whole of projectile motion rests on one observation, and it is a strange one: the horizontal and vertical motions do not know about each other.

Gravity pulls straight down. It has no horizontal component at all, so it cannot change the horizontal velocity — and if nothing changes the horizontal velocity, it is constant. Meanwhile the vertical velocity changes at a steady rate, exactly as it would for an object dropped from rest at the same moment.

That second half is the part that reliably surprises. A ball fired horizontally from a cannon and a ball dropped from the muzzle at the same instant hit the ground together. The fired ball travels much further, and it falls at the same rate while doing so.

So a hard problem — motion under a force, in two dimensions — becomes two easy problems that happen to share a clock. The horizontal problem is distance equals speed times time. The vertical problem is the one-dimensional fall, which was solved before Newton.

This decomposition is worth naming, because it is not a trick specific to projectiles. It is the reason vectors are useful at all: a vector equation is several scalar equations that can be solved separately and reassembled at the end. The same move appears in the forces on an inclined plane, where the weight is split into a component along the slope and one across it, and neither half is affected by the other.

Where forty-five comes from #

Launch at speed $v$ and angle $\theta$. The vertical component is $v\sin\theta$ and the horizontal is $v\cos\theta$.

The flight lasts until the vertical motion has come back to its starting height, which takes

$$t = \frac{2v\sin\theta}{g},$$

and during that time the projectile has travelled horizontally at the unchanging speed $v\cos\theta$. The range is the product:

$$R = \frac{2v^2\sin\theta\cos\theta}{g} = \frac{v^2\sin 2\theta}{g}.$$

The last step uses the double-angle identity, and it makes the answer immediate. The range depends on the angle only through $\sin 2\theta$, and a sine is largest when its argument is a right angle. So $2\theta = 90°$, and the best launch angle is $45°$.

The same expression explains the pairing visible in the first figure. Since $\sin 2\theta = \sin(180° - 2\theta)$, the angles $\theta$ and $90° - \theta$ give identical ranges. Thirty degrees and sixty degrees land together; twenty and seventy land together. One is a flat, fast, short-lived shot and the other is a high lob, and they arrive at the same place by two entirely different routes.

Why the answer does not matter much #

Here is the part that the derivation hides. $\sin 2\theta$ near its maximum is very nearly flat, because every smooth function is flat near its maximum. That is not a fact about projectiles; it is the reason maxima are hard to locate experimentally and easy to live with.

At forty degrees, $\sin 80° = 0.985$. At thirty-five degrees, $\sin 70° = 0.940$. A throw ten degrees off the optimum loses about one and a half percent of its range — well below the variation in how hard a person throws from one attempt to the next.

So the practical content of the forty-five degree result is not aim at forty-five degrees. It is anything between about thirty and sixty is fine, which is exactly the range of angles a person throwing hard produces naturally. The theory and the instinct agree, and the theory adds a tolerance rather than a correction.

The flatness has a second consequence that matters more. Because the maximum is flat, a measurement of the optimum angle is a bad way to test the theory: the data would need absurd precision to distinguish forty-five from forty-two. The sharp, testable prediction is the pairing — that $\theta$ and $90° - \theta$ land together — because that is a statement about two trajectories being equal, and equalities are easy to check.

Energy, arriving at the same answer sideways #

Nothing above mentioned energy, and the whole thing can be redone without mentioning time.

At the top of the arc the vertical velocity is zero but the horizontal velocity is untouched, so the projectile is still moving — at $v\cos\theta$, which for a forty-five degree launch is about seventy percent of its launch speed. The kinetic energy at the top is therefore not zero, and the height reached is set by only the vertical part of the launch energy.

This is the same accounting that governs a pendulum swinging, where potential energy peaks at the ends of the swing instead of the middle, and the same accounting that decides which collisions lose energy and which do not. What changes between those cases is where the potential sits and whether anything drains the total. What does not change is that the sum is a constant, and that solving for the constant is nearly always faster than following the motion.

The velocity, resolved a second way #

The decomposition into horizontal and vertical is one choice among several, and it is worth seeing that the choice was made for convenience rather than forced.

Resolving along and across the velocity instead — into the component that changes the speed and the component that changes the direction — gives a different and equally valid description, and it is the natural one for anything moving in a circle. On a slope, the useful axes are the ones aligned with the constraint rather than with the force. The rule in all three cases is the same: pick the axes that make one of the two equations trivial, and accept whatever the other one turns out to be.

For a projectile the horizontal equation is the trivial one, which is why this decomposition wins. Nothing acts horizontally, so that component is a constant, and a constant is the easiest object in mechanics to work with.

Where the model stops #

The trajectory drawn above is a parabola. Real projectiles do not follow parabolas, and the discrepancy is not small.

The figure assumes exactly one force acts: a constant downward pull. Four things are missing.

Air resistance is the big one, and it is the point at which the model stops being an approximation and starts being a different problem — there is no closed-form trajectory with quadratic drag, only a numerical one. Drag grows roughly as the square of speed, so it is fiercest at launch and it acts backwards along the path, which means it removes horizontal speed that nothing restores. A real trajectory is asymmetric — steeper coming down than going up — and its optimum launch angle drops. For a golf ball the best angle is well under forty-five degrees; for a shot put, which is heavy and slow enough for drag to be nearly irrelevant, forty-five is close to right, and the actual optimum is a degree or two lower simply because the shot is released above the ground it lands on.

The release height breaks the symmetry even in vacuum. The range formula assumed the landing height equals the launch height. Throwing from shoulder height onto flat ground gives the projectile a small bonus of extra fall, which slightly favours a flatter, faster shot.

Constant gravity is an approximation that a thrown ball never notices and an artillery shell does. Over a few hundred metres the direction of “down” changes measurably, and the parabola’s true shape is a small arc of an ellipse around the centre of the Earth.

A non-rotating Earth is assumed too. The Coriolis deflection is imperceptible for a cricket ball and decisive for a naval gun, which is the reason long-range gunnery tables carry a correction that depends on which way the ship is facing.

Naming these is not a disclaimer. It is the point: the parabola is the answer to a well-posed question — constant force, no medium, flat ground — and knowing which question a picture answers is what makes it possible to tell when the picture applies. Every idealisation on this site has the same status, from the frictionless pendulum to the ideal gas, and each is worth exactly as much as the range of conditions over which its neglected terms stay small.

Two versions of the same shot #

The pairing of complementary angles has a use that survives the corrections. Given a target within range, there are always two ways to hit it: a flat shot and a lob. Artillery calls them the low and high trajectories, and the choice between them is tactical rather than physical.

The flat shot arrives sooner and is harder to intercept. The lob arrives steeply, which lets it clear an obstacle and drop nearly vertically onto something the flat shot would skid off — and mortars exist entirely to exploit that end of the pair. Both solutions come out of the same quadratic, and the existence of exactly two is the same statement as the existence of exactly one maximum between them.

At the maximum the two merge. A target at the very limit of range can be reached in only one way, at forty-five degrees, and that coincidence — two solutions collapsing into one at the boundary of the possible — is a shape that recurs all over physics, from critical angles in optics to the separatrix in a pendulum’s phase portrait, where the curve dividing swinging from spinning belongs to neither family.

The ladder from here #

Later rungs on this anchor: the trajectory with quadratic drag, which has no closed-form solution and has to be integrated. The optimum angle as a function of drag, which falls from forty-five toward the low thirties. Launching onto sloping ground, where the optimum shifts by half the slope angle. The Magnus force on a spinning ball, which is the reason a curveball curves and the reason a golf ball is dimpled. The envelope of all trajectories at a fixed speed — itself a parabola, and the boundary of everywhere that can be reached. And the orbital limit, where the range formula fails completely because the ground curves away as fast as the projectile falls.

Galileo worked out the parabola and also knew it was wrong in air; he wrote that the deviation was “so great that it cannot be reduced to any rule”. He was right about that too.