Puzzle/Math Problem [Closed]

Kick_04

I recieved this email from a co-worker of mine to try and assist him in solving the question.

p.s. - A few days ago, I was reminded of something I hadn't thought about in a long time so I looked up the derivation of the relevant formula and, amazingly, it simplifies down to X=(Dg)^1/2 or X equals the square root of the product Dg, where X=desired value, D=Diameter, and g=acceleration due to gravity.
So, for extra credit: Be the first to reply directly to me with the correct identity of "X" and I will pay for your dinner on 8-25.
Hint: I was reminded of this exactly one week ago, today.

I have never met this person, all I have heard about him is he has identic memory, in his early 70s, and is considered very cheap. So for him to offer buying someones dinner left the people astonished.


The message was sent out on July 23rd 2014. So the exactly one week ago, today would of been Wednesday July 16th.

I know that X= speed (diameter could be meters, gravity would be meters per second squared. Multiplied and square root of this would leave meters per second)... I just can not figure out what calculation you would ever use diameter for or what is relative to July 16.

I found "Fat Boy" (atomic bomb) was first tested on July 16th but that was 69 years ago, so unlikely to be in the news. Also 45 years ago on that day is when Apollo 11 launched from Flordia to put first man on moon, which was on the news. Co-worker seems to think it has to do with the 2004 tsunami.

Any ideas/help on the "identity of X"?


E: It was escape velocity

Hirocon

Escape velocity

Kick_04

I'll read through it in more detail later (at work), but unless I glanced over it those calculations still do not use Diameter; they use radius/height.

Actually as I typed this I realised that 2r would be diameter.

Thank you.

Daenris

July 16th, 1969 was the launch of Apollo 11.

khain

Hirocon wrote: »

Escape velocity

This is correct. The equation for escape velocity is ve = sqrt(2GM/r), and the equation to calculate gravity at a certain radius is g = GM/r^2. Substitution results in ve = sqrt(2rg).

Kick_04

Yeah I emailed the person and waiting for a response back.

I have no idea what else it could be. It was just annoyning for him replacing the "2r" with "d", was driving me mad because nothing uses diameter for calculations. (that I have dealt with anyway)

My answer to them.

The identity of X is “Escape Velocity calculation for Apollo 11 launch on July 16th, 1969”.

The true equation is… Ve = (2rg)^1/2 … however 2r is the same as d, so I am unsure if you changed it to make it more complicated to find the answer or you were being completely literal by simplifying.