anonymous
  • anonymous
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
Mathematics
  • Stacey Warren - Expert brainly.com
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SOLVED
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jamiebookeater
  • jamiebookeater
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TuringTest
  • TuringTest
this problem is word-for word for sure? like "below floor level" and all that weird-sounding stuff?
anonymous
  • anonymous
yes 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
  • anonymous
1) 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.

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anonymous
  • anonymous
tried graphing it and it points downward which isn't right..
anonymous
  • anonymous
You 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
  • anonymous
tried that too and it seems like it should point upward
anonymous
  • anonymous
Okay, 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
  • TuringTest
There is no way it should be pointing upward with the angle at 60 deg
anonymous
  • anonymous
well this if for parametric equations so yes you can
anonymous
  • anonymous
not straight upward but to some degree yes it should be
TuringTest
  • TuringTest
your 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
  • TuringTest
my calculator has it pointing upwards
anonymous
  • anonymous
no 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
  • anonymous
what did you use as the equations then?
anonymous
  • anonymous
Are you translating by 6 inches? That would be below the x axis pointed up.
TuringTest
  • TuringTest
same as you x=21.3Tcos60 y=21.3sin60-16t^2-.5 wait, I did something when I tried to zoom and it changed downward, weird...
anonymous
  • anonymous
i 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
  • anonymous
but if you look at the table it appears to be positive for just a little bit
TuringTest
  • TuringTest
for the next part we should be solving\[h(t)=21.3t \sin(60)-16t^2-0.5=0\]to find the time in the air
TuringTest
  • TuringTest
of course we will get two answers and we want the larger one
anonymous
  • anonymous
Best wishes with this. My time's up for today. Bye guys.
TuringTest
  • TuringTest
later :)
anonymous
  • anonymous
but couldnt you just look at the table? i would get between 1.1 and 1.15 if i do it that way?
TuringTest
  • TuringTest
what table?
anonymous
  • anonymous
hit 2nd and graph
anonymous
  • anonymous
i would get 1.3 seconds
anonymous
  • anonymous
sorry 1.13
TuringTest
  • TuringTest
I keep getting different answers... I must be tired. did I put this into wolfram right? http://www.wolframalpha.com/input/?i=t%3D%28-21.3sin60%2Bsqrt%28%2821.3sin60%29%5E2-4*-16*-.5%29%29%29%2F%282*-16%29
anonymous
  • anonymous
im 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
  • TuringTest
look 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
  • phi
Here'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
  • TuringTest
thanks phi, I couldn't figure how to put parametric equations in wolfram. "plot", huh?
phi
  • phi
and 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
  • anonymous
oh wow yeah my graph looks nothing like that..dont know why
TuringTest
  • TuringTest
oh, I forgot to use the negative root to make for the larger time, because the denominator is negative...
anonymous
  • anonymous
well i guess it sort of does match my answers, confused about the next part though
TuringTest
  • TuringTest
which part?
anonymous
  • anonymous
d.) 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
  • TuringTest
well 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
  • TuringTest
by r I mean the distance travelled
TuringTest
  • TuringTest
*should be r=(v_x)t=21.3cos(60)t=21.3cos(60)(1.125)
anonymous
  • anonymous
okay based on that i think it will make it to the receptacle...
phi
  • phi
double check it's 1.5 inches so 1.5/12 feet
TuringTest
  • TuringTest
right, because \[21.3\cos(60)(1.125)=11.98\] which is in \[12\pm1.5\]
TuringTest
  • TuringTest
oh, I forgot conversions...
anonymous
  • anonymous
i did that at first too...
phi
  • phi
But it does hit the mark.
anonymous
  • anonymous
okay good, so last thing is e.) an arc of water needs to land in a 3-inch 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
  • TuringTest
sorry, I'm going to bed, hopefully phi can help you. g'night!
anonymous
  • anonymous
its okay thanks!
phi
  • phi
2 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
  • anonymous
and how exactly would you set that up?
phi
  • phi
x= 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
  • phi
plug into the first equation to find v
anonymous
  • anonymous
sorry 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
  • phi
For 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
  • anonymous
ok
anonymous
  • anonymous
so what would the answer for that be?

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