Anyone good at physics?

Question

Suppose the first rock in Conceptual Checkpoint 2-5 drops through a height h before the second rock is released from rest. Show that the separation between the rocks, S, is given by the following expression:
S=h+(square root of (2gh))t
In this result, the time t is measured from the time the second rock is dropped.

Answer 1

The time elapsed between releasing of both rocks is
t₀ = √(2g/h).

The speed of the first rock at the moment of releasing the second rock is
v₀ = g t₀ = g √(2g/h) = √(2gh).

The distance of the first rock from the Checkpoint at time t after releasing the second rock is
h + v₀ t + ½ g t² = h + √(2gh) t + ½ g t².

The distance of the second rock from the Checkpoint is
½ g t².

The difference of both distances, i.e. the distance between the two rocks is
S = h + √(2gh) t
Q.E.D.
-

Answer 2

In time t:
The distance traveled by the first rock is
h + vt + 05gt^2,
[since the initial velocity is v for the first rock at time t = 0] and it has traveled a distance h in this time t.]

The distance traveled by the second rock is 05gt^2

In time't' Distance of separation, s = h + vt. [Subtracting the second from the first]
Using v^2 -u^2 = 2gh, for the first rock, v = √ [2gh] since u = 0.
Distance of separation in time t = h + t *√ [2gh]
===============================================