Breaking everything into components is the standard approach to analyzing projectile motion questions, but sometimes looking at whole vectors can also get you to an answer.

For instance, look at the whole vectors for position and let the simulation go until it stops. You get a right-angled triangle where simple geometry will get you an equation for the time-of-flight.

Calling the launch angle (measured from the horizontal) q, we get:

sinq

=

0.5 a t2 vo t

=

0.5 a t vo

Setting the acceleration equal to g, and solving for the time, which we call the time-of-flight, gives:

t

=

2 vo sinq g

=

2 voy g

If you look at the whole vectors for velocity and go to the maximum height position, the geometry of right-angled triangles gets you the time to reach maximum height:

sinq

=

a t vo

So,   t

=

vo sinq g

=

voy g

The time to reach maximum height is half the time-of-flight, which makes sense in this situation where the object lands at the same level it was launched from.