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DarkBlueChocobo

  • one year ago

Help with application

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  1. Future_Einstein
    • one year ago
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    what kind

  2. Future_Einstein
    • one year ago
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    Job?

  3. DarkBlueChocobo
    • one year ago
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    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.

  4. DarkBlueChocobo
    • one year ago
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    |dw:1443663056043:dw|

  5. Future_Einstein
    • one year ago
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    you cant calculate this without knowing the amount of time needed for the egg to hit the ground

  6. Future_Einstein
    • one year ago
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    does it fall at 20 ft per sec. as well?

  7. DarkBlueChocobo
    • one year ago
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    No doesn't say, but the reason it says you wanna stay so far away is in case of accident

  8. DarkBlueChocobo
    • one year ago
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    @SolomonZelman @zepdrix

  9. Future_Einstein
    • one year ago
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    i do not think this is a valid question

  10. Future_Einstein
    • one year ago
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    @Nichole15 @Loser66 @lovelycharm

  11. anonymous
    • one year ago
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    Then how far do you think you should stand in order not to get hit?

  12. DarkBlueChocobo
    • one year ago
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    @freckles

  13. DarkBlueChocobo
    • one year ago
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    |dw:1443664086583:dw|it kinda looks like we could do something with a triangle

  14. Future_Einstein
    • one year ago
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    perhaps find the hypotenuse? or slope? thats about all with a triangle that i could think of.

  15. freckles
    • one year ago
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    I feel like some information is missing :(

  16. Future_Einstein
    • one year ago
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    thats what i said @freckles

  17. DarkBlueChocobo
    • one year ago
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    Egh this is like an example from my engineering note book

  18. Future_Einstein
    • one year ago
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    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?

  19. DarkBlueChocobo
    • one year ago
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    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

  20. freckles
    • one year ago
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    what book does this come from and what is the title of the section

  21. DarkBlueChocobo
    • one year ago
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    Thinking Like an Engineer : An active approach review questions for Chapter 6

  22. DarkBlueChocobo
    • one year ago
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    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

  23. Future_Einstein
    • one year ago
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    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.

  24. Future_Einstein
    • one year ago
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    there is! i gtg but fan me be on all day tomarrow off and on with my mobile app.

  25. freckles
    • one year ago
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    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

  26. DarkBlueChocobo
    • one year ago
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    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?

  27. freckles
    • one year ago
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    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

  28. freckles
    • one year ago
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    http://marshallbrain.com/science/egg-drop.htm I got my egg speed from this site

  29. freckles
    • one year ago
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    oh and landed means the egg has cracked :p

  30. freckles
    • one year ago
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    or broke I mean

  31. freckles
    • one year ago
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    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 ?

  32. freckles
    • one year ago
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    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

  33. freckles
    • one year ago
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    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

  34. freckles
    • one year ago
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    \[v=at =\frac{32 ft}{\sec^2} \cdot \sqrt{ \frac{[100-(5+\frac{9.2}{12}) ]ft}{0.5 \cdot \frac{32 ft}{\sec^2}}} \\ v=74.1026...ft/sec\] 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

  35. freckles
    • one year ago
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    you know if the runner started 25.4323 ft away from the egg drop

  36. freckles
    • one year ago
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    does this make sense

  37. freckles
    • one year ago
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    I have made a lot of assumptions but it seems like that is what we must do

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