Step-by-step explanation:
We know that in kinematics we can write the formula for uniformly accelerated bodies:
s = 1/2 a t^2 + v0 t + s0
In our case s0 (initial deployment) = 0m and v0 (initial velocity) = 0
So:
s = 1/2 a t^2
Now we isolate t:
t = [tex]\sqrt[2]{\frac{2s}{a} }[/tex]
All proportionalities are obvious.
Note: I forgot to say that a, the acceleration, is the gravitational acceleration. It can be g or -g depending on how you'll choose the reference system.
