Time of flight is found from the vertical motion

Projectile Motion — AQA A-Level Physics

At maximum height, the vertical velocityThe rate of change of displacement. A vector quantity. Measured in m s⁻¹. is zero: $v_{y} = 0$.
Using $v = u +$ at with v = 0, u = u sin(theta), $a = -g gives$ the time to maximum height.
For a symmetric trajectory (launch and land at same height), the total time of flight is twice the time to maximum height.

t_{\text{max}} = \frac{u \sin \theta}{g}

$t_max$: time to reach maximum height (s)
$u$: 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 (9.81 m s⁻²)