Range is found from horizontal velocity multiplied by total time

Projectile Motion — AQA A-Level Physics

Horizontal velocityThe rate of change of displacement. A vector quantity. Measured in m s⁻¹. is constant (no horizontal accelerationThe rate of change of velocityThe rate of change of displacement. A vector quantity. Measured in m s⁻¹.. A vector quantity. Measured in m s⁻².), so: $Range = u \cos(\theta) times$ total time of flight.
For a symmetric trajectory, substituting the total time gives $R = u^2 \sin(2theta) / g$.
Maximum range occurs at $\theta = 45 degrees (since \sin 90 = 1)$.

R = \frac{u^2 \sin 2\theta}{g}

$R$: range (m)
$u$: initial speed (m s⁻¹)
$θ$: angle of projection
$g$: accelerationThe rate of change of velocityThe rate of change of displacement. A vector quantity. Measured in m s⁻¹.. A vector quantity. Measured in m s⁻². due to gravity (m s⁻²)

Examiner Tips and Tricks

The range equation $R = u^2 \sin(2theta)/g$ is only valid when the launch and landing heights are the same.
For asymmetric cases (e.g. launched from a cliff), solve horizontal and vertical components separately.