Fine, lets do a retry of this.
Δd = -0.9m
v₁ = 0
v₂ = ?
a = -9.8 m/s²
Δt = ?
We can use the following kinematic equation and solve for Δt.
Δd = v₁Δt + 0.5(a)(Δt)²
Δd = 0.5(a)(Δt)²
2Δd = a(Δt)²
√2Δd/a = Δt
√2(-0.9m)/(-9.8 m/s²) = Δt
0.428571428574048 = Δt
Therefore, it takes 0.4 seconds for the glass to hit the ground, or 0.43s as you said (even though I don't believe it follows significant digit rules)
