What is the relationship between time and height in object free fall?

[BrendanB]

Hello,
I am new to this site. I am currently active duty military (navy) and I am trying to teach myself physics because when I get the time, I would like to take some college physics classes. I downloaded a few physics books online and I am trying to page through them. I am having some trouble figuring out this problem. I feel like a complete idiot because I have the correct answer right in front (teacher solution manual downloaded too) of me and still can't figure it out! Anyways, here is the problem and the correct answer (below). Any help you can offer would be appreciated.

Homework Statement

An object falls freely from height h. It is released at time
zero and strikes the ground at time t. (a) When the object
is at height 0.5h, is the time earlier than 0.5t, equal to 0.5t,
or later than 0.5t? (b) When the time is 0.5t, is the height
of the object greater than 0.5h, equal to 0.5h, or less than
0.5h? Give reasons for your answers.

Homework Equations

The Attempt at a Solution

I understand the formula h=1/2gt^2, but what I don't understand is where they are getting the "0.707t" when they plug in 0.5h.
I also don't understand the second equation where they are getting 0.25h4. Answer
With h=1/2gt^2
(a)0.5h=1/2g(0.707t)^2
(b)The distance fallen is 0.25h=1/2g(0.5t)^2. The elevation is 0.75h, greater than 0.5h.

 

[Chi Meson]

As you do more physics, certain numbers will become familiar. 0.707= 1 over the square root of two. In other words, due to the "t squared" in the formula you mentioned, the height from which an object is released, and the time it takes for the object to hit the ground are not "directly proportional." That is, compared to the first initial test (drop from height H, and record time, t), if you cut the height in half (H/2), the time will not be t/2, but t/√2. If you cut the height to a third (H/3), the time will not be t/3, but t/√3, etc. And if you triple the height (3H), the time will be t*√3.