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this problem is word-for word for sure? like "below floor level" and all that weird-sounding stuff?
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.
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.
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?
There is no way it should be pointing upward with the angle at 60 deg
well this if for parametric equations so yes you can
not straight upward but to some degree yes it should be
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
my calculator has it pointing upwards
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
what did you use as the equations then?
Are you translating by 6 inches? That would be below the x axis pointed up.
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...
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..
but if you look at the table it appears to be positive for just a little bit
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
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.
but couldnt you just look at the table? i would get between 1.1 and 1.15 if i do it that way?
hit 2nd and graph
i would get 1.3 seconds
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
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
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.
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
thanks phi, I couldn't figure how to put parametric equations in wolfram. "plot", huh?
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
oh wow yeah my graph looks nothing like that..dont know why
oh, I forgot to use the negative root to make for the larger time, because the denominator is negative...
well i guess it sort of does match my answers, confused about the next part though
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?
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?
by r I mean the distance travelled
*should be r=(v_x)t=21.3cos(60)t=21.3cos(60)(1.125)
okay based on that i think it will make it to the receptacle...
double check it's 1.5 inches so 1.5/12 feet
right, because \[21.3\cos(60)(1.125)=11.98\] which is in \[12\pm1.5\]
oh, I forgot conversions...
i did that at first too...
But it does hit the mark.
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?
sorry, I'm going to bed, hopefully phi can help you. g'night!
its okay thanks!
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
and how exactly would you set that up?
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)
plug into the first equation to find v
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
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)
so what would the answer for that be?