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ok im a bit rusty with my trig guys and i need help with this question. please only 3 stupid responses per 1 smart one.





A fire hose held near the ground shoots water at a speed of 6.8m/s. at what angle should the nozzle point in order that the water land 2.0 meters away.

(hint there are two angles).



so lets call the angle A
vo=6.8m/s and its components are 6.8sinA and 6.8cosA for y and x respectivley.


x= 2.0m
xo= 0m
yo=0m
y=?

so first i did this

using

x=xo+vot+0.5at*t

2= 6.8cos(A)t

so cos(A)=0.294/t

then i use the time height equation to find height at vo (which is half time.) then multiply it by 2 to get full time.

t=2vosin(A)/9.81

t=1.39sin(A)


so i have two equations put one in the other and use the trig identity

2sin(A)cos(A) = sin(2A)

i get sin(2A)= 0.848

therefore A = 29 degrees

which is wrong.

and there should be another angle i dont know how though.

Did you know without Trigonometry, there would be no engineering?

Did you know that if my auntie had balls, she'd be my uncle?

quote:

Originally posted by Paradox Lost Did you know without Trigonometry, there would be no engineering?

yea

fuck off next post i hope is an attempt to solve this .

its easy guys

...still hoping someone will chime in about the importance of lamps.

fuck its only the arabs posting.. im screwed.

I thought math was something we had going for us?

apparently not

its up to mrjivebojingles now.. you arabs are useless tits

The problem assumes the water is coming out in a straight line, but in reality, the water is coming out parabolically, and so therefore, you should be looking towards a quadratic equation for the water. It's a trick question, don't be fooled.

quote:

Originally posted by Krypton The problem assumes the water is coming out in a straight line, but in reality, the water is coming out parabolically, and so therefore, you should be looking towards a quadratic equation for the water. It's a trick question, don't be fooled.

If you read his work, you would notice that he did do a quadratic.

Do you have the answer in the back of the book?

Here's a link for equations for accelerated motion. I'm too lazy to write it out, I kinda sucked at physics.

http://id.mind.net/~zona/mstm/physi...ratedMotion.htm

quote:

Originally posted by Krypton The problem assumes the water is coming out in a straight line, but in reality, the water is coming out parabolically, and so therefore, you should be looking towards a quadratic equation for the water. It's a trick question, don't be fooled.

right fair enough but remember that velocity is instantaneous. so the initial velocity will carry an initial angle which can be thought of as linear.

the angles are like 13 and 77 in the back of the book which equal 90

my answer of 29 isnt working.

quote:

Originally posted by ziptnf If you read his work, you would notice that he did do a quadratic. Do you have the answer in the back of the book? Here's a link for equations for accelerated motion. I'm too lazy to write it out, I kinda sucked at physics. http://id.mind.net/~zona/mstm/physi...ratedMotion.htm

actually i didnt use the quadratic. since the x component neglects acceleration (gravity) your t*t is nulled out.

my 2 variables now are t and A (angle)

i made 2 equations for t so theoretically (unless my math is wrong my answer should be correct).

http://en.allexperts.com/q/Physics-...le-motion-5.htm

quote:

Originally posted by MrJiveBoJingles http://en.allexperts.com/q/Physics-...le-motion-5.htm

fuck thats almost the exact same question..and i did the same thing and they got 27 and i got 29

but the answer should be either 13 or 77


what hte hell.

quote:

Originally posted by Nrg2Nfinit fuck thats almost the exact same question..and i did the same thing and they got 27 and i got 29 but the answer should be either 13 or 77

Unless I'm misreading, those are the two angles they got:

quote:

So 2*theta = 27.6 degrees and 152.4 degrees So theta = 13.8 degrees and 76.2 degrees.

hrmm thats odd

i get 58 then divided by 2 .. i need to check my math maybe..

fuck it i had it right all the time!


2sinAcosA=.42431
sin2A=.4231
2A= 25 degrees
A= 13 degrees or 90-13 = 77 degrees


thanks mr jingles

you still struggling with math ? I remember helping you out with some calculus problems. And that was what, 2 years ago ? You still in school ?

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