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Guest Message by DevFuse
 

final

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#174157

Posted by final on 04 June 2009 - 04:51 AM

ok i was trying to get out of this without doing the math but i guess its hard to convince someone that im right without math (for reference always assume im right and your wrong it will save you alot of time and me alot of work)

I reread the problem as stated and if you are a 2d person that exists entirely in the vertical plane then no drop can hit you if you stand still. So technically the only water that hits you, you have to walk into
so

add all area todethor .56*200=112cubic meters that you walk through
each cubic meter has 200 drops so 112 *200=22400 drops.

you can solve it without ever using speed (or anything dependent on it) so speed doesnt enter into it.

What i think the problem might be is a 3d figure made of two rectangles (one in each plane) I think this because otherwise way too much info is given (even if you ignore the speed because that would be just to trick you)
so...
I have two basic assumptions
these assumptions are the head is in the horizontal plane only. The surface area can be 3d but i think for this problem how much of it lies in each plane needs to be known so i assumed it was the top of the head as any part of it in the vertical plane could have just been added to the body

second the body area is only in the vertical plane so math ho
this can be treated as two different objects and then added togethor so as in the previous problem but only for the .5 of the body

.5*200*200=20000 drops hit body

for the head you have a lil more math
200 drops/cubic meter and the drops move at 9 m/s so 200 drops/cubic meter* 9m/s=1800 drops/(square meter*second)
or 1800 drops per square meter per second (remember this is falling straight down so that is y this doesnt apply to the body)

so 1800 drops/(square meter*second)*.06square meter/head=108 drops per second/head (1 head so per head is stupid but if you think of it as a unit like inch then it might help you follow it more)

anyway for walking 200meter/(1.5meters/second)=133.3333seconds
so drops on head for walking = 108 drops/second *133.3333 seconds= 14400 drops on head
14400+20000=34400


for running body is the same as explained above but the head drops becomes

200/8=25 seconds
25 seconds * 108=2700
2700+20000=22700 drops

so as i explained in my (first?) post
running=(walking-body)ratio of speeds +body
running=(34400-20000)1.5/8+body=22700


the problems and arguements people are having i think come from the two different interpretations but I would offer the thought that either way it is not a direct ratio of the speeds (unless poeple are thinking the rectangle exists entirely in the horizantal plane)

so bushindo will have to explain which way he meant

p.s. i think its hilarious that what bushindo thinks is easy gets this many arguments and different answers

(the only question that has more then one answer on a math test is Name)
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