Question

When a stone is dropped from a height, the time taken to fall, t, is proportional to the square root of the distance fallen, d.

Answer 1

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 = $\sqrt[2]{\frac{2s}{a} }$

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.