## megd 3 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

1. TuringTest

this problem is word-for word for sure? like "below floor level" and all that weird-sounding stuff?

2. megd

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

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.

4. megd

tried graphing it and it points downward which isn't right..

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.

6. megd

tried that too and it seems like it should point upward

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?

8. TuringTest

There is no way it should be pointing upward with the angle at 60 deg

9. megd

well this if for parametric equations so yes you can

10. megd

not straight upward but to some degree yes it should be

11. 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

12. TuringTest

my calculator has it pointing upwards

13. megd

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

14. megd

what did you use as the equations then?

Are you translating by 6 inches? That would be below the x axis pointed up.

16. 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...

17. megd

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..

18. megd

but if you look at the table it appears to be positive for just a little bit

19. 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

20. TuringTest

of course we will get two answers and we want the larger one

Best wishes with this. My time's up for today. Bye guys.

22. TuringTest

later :)

23. megd

but couldnt you just look at the table? i would get between 1.1 and 1.15 if i do it that way?

24. TuringTest

what table?

25. megd

hit 2nd and graph

26. megd

i would get 1.3 seconds

27. megd

sorry 1.13

28. 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

29. megd

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

30. 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.

31. phi
32. TuringTest

thanks phi, I couldn't figure how to put parametric equations in wolfram. "plot", huh?

33. phi
34. megd

oh wow yeah my graph looks nothing like that..dont know why

35. TuringTest

oh, I forgot to use the negative root to make for the larger time, because the denominator is negative...

36. megd

well i guess it sort of does match my answers, confused about the next part though

37. TuringTest

which part?

38. megd

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?

39. 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?

40. TuringTest

by r I mean the distance travelled

41. TuringTest

*should be r=(v_x)t=21.3cos(60)t=21.3cos(60)(1.125)

42. megd

okay based on that i think it will make it to the receptacle...

43. phi

double check it's 1.5 inches so 1.5/12 feet

44. TuringTest

right, because $21.3\cos(60)(1.125)=11.98$ which is in $12\pm1.5$

45. TuringTest

oh, I forgot conversions...

46. megd

i did that at first too...

47. phi

But it does hit the mark.

48. megd

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?

49. TuringTest

50. megd

its okay thanks!

51. 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

52. megd

and how exactly would you set that up?

53. 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)

54. phi

plug into the first equation to find v

55. megd

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

56. 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)

57. megd

ok

58. megd

so what would the answer for that be?