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Why a projectile travels furthest at 45°

A clean derivation of the range equation, and the one reasoning step that shows 45 degrees gives maximum range on level ground.

18 June 2026 · 2 min read

Most students can quote that 45° gives the maximum range. Far fewer can say why. The "why" is one short step of reasoning, and once you see it you never forget it.

The question

A projectile is launched from level ground at speed vv and angle θ\theta. We ignore air resistance. At what angle is the horizontal range largest?

Set up the two motions

Projectile motion is just two independent motions happening at once. Split the launch velocity into its parts:

  • Horizontal: constant velocity vcos⁡θv\cos\theta.
  • Vertical: starts at vsin⁡θv\sin\theta, pulled down by gravity gg.

The horizontal and vertical positions at time tt are:

x=vcos⁡θ  t,y=vsin⁡θ  t−12gt2x = v\cos\theta \; t, \qquad y = v\sin\theta \; t - \tfrac{1}{2} g t^2

Find the time of flight

The projectile lands when y=0y = 0 again. Solving vsin⁡θ t−12gt2=0v\sin\theta \, t - \tfrac{1}{2} g t^2 = 0 gives t=0t = 0 (the launch) or:

t=2vsin⁡θgt = \frac{2 v \sin\theta}{g}

Find the range

Substitute that flight time into the horizontal equation:

R=vcos⁡θ⋅2vsin⁡θg=2v2sin⁡θcos⁡θgR = v\cos\theta \cdot \frac{2 v \sin\theta}{g} = \frac{2 v^2 \sin\theta \cos\theta}{g}

Using the identity 2sin⁡θcos⁡θ=sin⁡2θ2\sin\theta\cos\theta = \sin 2\theta:

R=v2sin⁡2θgR = \frac{v^2 \sin 2\theta}{g}

The reasoning step

Here is the whole problem in one line. The launch speed vv and gravity gg are fixed. The only thing you control is θ\theta, and it only appears inside sin⁡2θ\sin 2\theta. So the range is largest exactly when sin⁡2θ\sin 2\theta is largest.

The sine function maxes out at 11, when its angle is 90°90°. So we need:

2θ=90°⟹θ=45°2\theta = 90° \quad\Longrightarrow\quad \theta = 45°

What this actually tells you

It is not a coincidence or a rule to memorise. The sin⁡2θ\sin 2\theta is symmetric about 45°45°, which is why a 30° shot and a 60° shot land in the same place: their angles are equal distances either side of the maximum. Understand the shape of sin⁡2θ\sin 2\theta and you have understood every level-ground range question the HSC can ask.

The trace: we did not look up the answer. We followed the motion to a formula, then read the answer off the formula. That path is the point.