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anonymous
 5 years ago
suppose a molecule of water is propelled from a point 6 inches below floor level at an angle of 60 with an initial velocity of 21.3 feet per second. write modeling equations that describe the position of the molecule in the water stream for any time t
anonymous
 5 years ago
suppose a molecule of water is propelled from a point 6 inches below floor level at an angle of 60 with an initial velocity of 21.3 feet per second. write modeling equations that describe the position of the molecule in the water stream for any time t

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TuringTest
 5 years ago
Best ResponseYou've already chosen the best response.0this problem is wordfor word for sure? like "below floor level" and all that weirdsounding stuff?

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0yes that's what it said word for word i tried plugging this into my calculator: x = 21.3T cos (60) and y = 21.3T sin (60)16t^2  0.5 and the graph goes downward which isn't right

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.01) Convert units to be the same. 2) Draw a diagram (on paper) of what's going on. 3) Use trig relationships to figure out missing parts of your diagram. 4) Integrate the velocity formula to find the position function. That should get you pointed in the right direction.

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0tried graphing it and it points downward which isn't right..

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0You stated you tried graphing it with a calculator graphing utility. I suggest you draw it out on paper and begin thinking about it from there.

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0tried that too and it seems like it should point upward

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0Okay, label all the information you have on your diagram. Can you use Trig to find the missing information you need to solve for the position function?

TuringTest
 5 years ago
Best ResponseYou've already chosen the best response.0There is no way it should be pointing upward with the angle at 60 deg

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0well this if for parametric equations so yes you can

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0not straight upward but to some degree yes it should be

TuringTest
 5 years ago
Best ResponseYou've already chosen the best response.0your calculator isn't in radians or anything is it? and what do you mean above? how can the fact that is is parametric change the fact that it angles upwards if initiated at 60 deg

TuringTest
 5 years ago
Best ResponseYou've already chosen the best response.0my calculator has it pointing upwards

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0no its in degree mode. it should be in quadrant 1 and slanted at a 60 degree angle, i mean above the x axis basically, when i graphed it, it appeared below the x axis which cant be right

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0what did you use as the equations then?

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0Are you translating by 6 inches? That would be below the x axis pointed up.

TuringTest
 5 years ago
Best ResponseYou've already chosen the best response.0same as you x=21.3Tcos60 y=21.3sin6016t^2.5 wait, I did something when I tried to zoom and it changed downward, weird...

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0i put 0.5 so thats like 6 inches. yeah this doesnt make much sense....the next question is how long is the water molecule in the air before coming down to floor level..

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0but if you look at the table it appears to be positive for just a little bit

TuringTest
 5 years ago
Best ResponseYou've already chosen the best response.0for the next part we should be solving\[h(t)=21.3t \sin(60)16t^20.5=0\]to find the time in the air

TuringTest
 5 years ago
Best ResponseYou've already chosen the best response.0of course we will get two answers and we want the larger one

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0Best wishes with this. My time's up for today. Bye guys.

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0but couldnt you just look at the table? i would get between 1.1 and 1.15 if i do it that way?

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0i would get 1.3 seconds

TuringTest
 5 years ago
Best ResponseYou've already chosen the best response.0I keep getting different answers... I must be tired. did I put this into wolfram right? http://www.wolframalpha.com/input/?i=t%3D%2821.3sin60%2Bsqrt%28%2821.3sin60%29%5E24*16*.5%29%29%29%2F%282*16%29

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0im not really familiar with that site....but the next part was to figure out the maximum height and i got 4.816 feet from the table which doesnt make sense either

TuringTest
 5 years ago
Best ResponseYou've already chosen the best response.0look at the link you should get to learn the site, it is very useful for this type of thing you can see if the formula entered is the same as yours I have to go, sorry. Good luck though.

phi
 5 years ago
Best ResponseYou've already chosen the best response.1Here's wolfram's plot http://www.wolframalpha.com/input/?i=plot+x+%3D+21.3*t*cos+%2860%29+%2C+y%3D+21.3*t*sin%2860%29++16t%5E2+0.5

TuringTest
 5 years ago
Best ResponseYou've already chosen the best response.0thanks phi, I couldn't figure how to put parametric equations in wolfram. "plot", huh?

phi
 5 years ago
Best ResponseYou've already chosen the best response.1and here's t for y=0 http://www.wolframalpha.com/input/?i=solve++0%3D+21.3*t*sin%2860%29++16t%5E2+0.5

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0oh wow yeah my graph looks nothing like that..dont know why

TuringTest
 5 years ago
Best ResponseYou've already chosen the best response.0oh, I forgot to use the negative root to make for the larger time, because the denominator is negative...

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0well i guess it sort of does match my answers, confused about the next part though

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0d.) it is 12 feet from the point at which the water is propelled to the center of the receptacle in the floor where it needs to land. if the receptacle is circular with a diameter of 3 inches, will the water molecule land in the receptacle?

TuringTest
 5 years ago
Best ResponseYou've already chosen the best response.0well you need to find r you have the time t of travel in the air, so the distance it travels is v_x=21.3cos(60)t=21.3cos(60)(1.125) now is this x within \[12\pm1.5\]where 1.5 is the radius of the receptacle?

TuringTest
 5 years ago
Best ResponseYou've already chosen the best response.0by r I mean the distance travelled

TuringTest
 5 years ago
Best ResponseYou've already chosen the best response.0*should be r=(v_x)t=21.3cos(60)t=21.3cos(60)(1.125)

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0okay based on that i think it will make it to the receptacle...

phi
 5 years ago
Best ResponseYou've already chosen the best response.1double check it's 1.5 inches so 1.5/12 feet

TuringTest
 5 years ago
Best ResponseYou've already chosen the best response.0right, because \[21.3\cos(60)(1.125)=11.98\] which is in \[12\pm1.5\]

TuringTest
 5 years ago
Best ResponseYou've already chosen the best response.0oh, I forgot conversions...

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0i did that at first too...

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0okay good, so last thing is e.) an arc of water needs to land in a 3inch circular receptacle in the floor whose center is 15 feet from the point at which the water is propelled. suppose the water is propelled from a point 6 inches below floor level at an angle of 55. at approximately what intitial velocity must the water be propelled in order to land in the receptacle?

TuringTest
 5 years ago
Best ResponseYou've already chosen the best response.0sorry, I'm going to bed, hopefully phi can help you. g'night!

phi
 5 years ago
Best ResponseYou've already chosen the best response.12 equations and 2 unknowns. I'd use the x one to find v in terms of t. sub into the y equation, and solve for t when y=0

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0and how exactly would you set that up?

phi
 5 years ago
Best ResponseYou've already chosen the best response.1x= v t cos(55), with x= 15 so 15= v t cos(55) v= 15/(t cos(55) y= v t sin(55)  16 t^2 0.5, with v as above. solve for t when y= 0 ( the water hits the floor)

phi
 5 years ago
Best ResponseYou've already chosen the best response.1plug into the first equation to find v

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0sorry but how do you solve for t when y = 0...im at 0 = (15/t cos (55)) t sin (55)  16t^2  0.5

phi
 5 years ago
Best ResponseYou've already chosen the best response.1For starter's, the t cancels in the first term. The sin, cos are constants so it's essentially a t^2 = b for t= sqrt(b/a)

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0so what would the answer for that be?
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