Help with application

- DarkBlueChocobo

Help with application

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- Future_Einstein

what kind

- Future_Einstein

Job?

- DarkBlueChocobo

I am standing on the upper deck of the football stadium. I have an egg in my hand. I am going to drop it and you are going try to catch it. You are standing the ground. You don't want to stand directly under me. You want to stand as far as possible to one side you can. If you can run at 20 feet per second, and i am at a height of 100 feet, how far away can you stand and still catch the egg if you start running when I let go.

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## More answers

- DarkBlueChocobo

|dw:1443663056043:dw|

- Future_Einstein

you cant calculate this without knowing the amount of time needed for the egg to hit the ground

- Future_Einstein

does it fall at 20 ft per sec. as well?

- DarkBlueChocobo

No doesn't say, but the reason it says you wanna stay so far away is in case of accident

- DarkBlueChocobo

@SolomonZelman @zepdrix

- Future_Einstein

i do not think this is a valid question

- Future_Einstein

@Nichole15 @Loser66 @lovelycharm

- anonymous

Then how far do you think you should stand in order not to get hit?

- DarkBlueChocobo

@freckles

- DarkBlueChocobo

|dw:1443664086583:dw|it kinda looks like we could do something with a triangle

- Future_Einstein

perhaps find the hypotenuse? or slope? thats about all with a triangle that i could think of.

- freckles

I feel like some information is missing :(

- Future_Einstein

thats what i said @freckles

- DarkBlueChocobo

Egh this is like an example from my engineering note book

- Future_Einstein

and we couldnt use a triangle. we only have one plausible side length. the 20 ft per sec. is not a measurement in itself. Right?

- DarkBlueChocobo

So they only give me like 4 key things. 1. I am standing on the upper deck and am dropping an egg. 2. You are standing on the ground and you don't wanna stand direct under it; in fact you wanna stand as far as to one side as you can so that if i accidently release, you wont get hit. 3. Dropping from 100 feet 4. you can run 20 feet per second

- freckles

what book does this come from
and what is the title of the section

- DarkBlueChocobo

Thinking Like an Engineer : An active approach review questions for Chapter 6

- DarkBlueChocobo

Basically am doing questions using the Solvem Method
S: Sketch
O: Objectives/ observations
L: List
V: Variables/Constants
E: Equations and Estimates/Assumptions
M: Manipulation

- Future_Einstein

I am interested in being an architect. And my backup is engineering hopefully. want to fan me and study together. see if there an engineer study group.

- Future_Einstein

there is! i gtg but fan me be on all day tomarrow off and on with my mobile app.

- freckles

i wonder if we can try to look up the speed an egg drops
and use that as our assumption
we may even want to assume the standard football stadium size

- DarkBlueChocobo

wells if we find out how fast the egg drops. We can easily use how fast he can run to see how quick he can still get there in cant we?

- freckles

let's assume the egg falls at 80 ft/sec
|dw:1443666055984:dw|
Let x' be the rate at which the guy is running towards egg drop
Let y' be the rate at which the egg falls
We have
\[x'=-20 \frac{ft}{\sec} \\ y'=-80 \frac{ft}{\sec} \\ \]
The negative just means the distances are decreasing..
For example the distance between the egg and the ground is getting smaller as the egg falls
And the distance between the guy and the egg drop is getting smaller as he gets closer to the egg drop
Let's pretend the runner is x=20 ft away from egg drop
then the guy will get to the egg drop in 1 sec
but in 1 sec we have that egg has only dropped 80 feet by then
Let's pretend the runner is x=40 ft away from egg drop
then the guy will get to the egg drop in 2 sec
but in 2 sec the egg would have dropped 160 feet which means it has already hit the ground since the guy dropping the egg was only 100 ft above
So we know the answer is between x=20 ft and x=40 ft
...
1 sec=20 ft runner=80 feet egg
we want:
? sec=? ft runner=100 feet egg (I guess he can catch it on the ground lol not sure)
so let's divide all sides of
1 sec=20 ft runner=80 feet egg by 80
so we have
1/80 sec=20/80 ft runner=1 feet egg
now multiply both sides by 100
100/80 sec=100*20/80 ft runner=100 feet egg
so this says:
In 1.25 secs if the runner starts 25 ft away from the egg drop he will get there when the egg has landed on the ground lol
but I think we suppose to catch it
...So now I'm beginning to think we should make an assumption about the height of the man running

- freckles

http://marshallbrain.com/science/egg-drop.htm I got my egg speed from this site

- freckles

oh and landed means the egg has cracked :p

- freckles

or broke I mean

- freckles

so... we also need to consider how much the guy is willing to move his hand down to catch the egg
like is he willing to slide ?

- freckles

if we assume the dude is 5 ft and 9.2 inches and he not willing to move his hands lower than his height to catch the egg..
then the egg will only travel 100-(5+9.2/12) ft
which means the velocity will change slightly

- freckles

http://pediatrics.about.com/cs/growthcharts2/f/avg_ht_male.htm
this is the site I used for average male height assuming the dude is from the US

- freckles

\[v=at =\frac{32 ft}{\sec^2} \cdot \sqrt{ \frac{[100-(5+\frac{9.2}{12}) ]ft}{0.5 \cdot \frac{32 ft}{\sec^2}}} \\ v=74.1026...ft/sec\]
is the new velocity of the egg
so we have
\[
\text{ so we have in one sec the egg has moved about } 74.1026 ft \\ \]
\[1 \sec = 20 ft \text{ runner } =74.1026 ft \text{ egg } \\ \text{ divide all sides by } 74.1026 \\ \frac{1}{74.1026} \sec = \frac{20}{74.1026} ft \text{ runner} = 1 ft \text{ egg } \\ \\ \text{ now we are assuming the egg will only travel }[100-(5+\frac{9.2}{12})]ft \]
now we want the egg to travel 100-(5+9.2/12)
so we will multiply all sides by that number
let's go ahead and simplify that number
that number is 94.233333333 which we will just use 94.23
so multiply all sides by this number
\[\frac{94.23}{74.1026} \sec=\frac{20(94.23)}{74.1026} ft \text{ runner } =94.23 ft \text{ egg }\]
so this says in about 1.27 secs we have the runner has ran a distance of 25.4323 so the runner will catch the ball when the egg has fell 94.23 from its starting point

- freckles

you know if the runner started 25.4323 ft away from the egg drop

- freckles

does this make sense

- freckles

I have made a lot of assumptions
but it seems like that is what we must do

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