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.
Jul-07-2009 01:10
Paradox Lost
In This Twilight
Registered: Aug 2007
Location: San Francisco
Did you know without Trigonometry, there would be no engineering?
its up to mrjivebojingles now.. you arabs are useless tits
Jul-07-2009 01:18
Krypton
83.798 g/6.022x10^23
Registered: Nov 2003
Location: Texas
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.
Jul-07-2009 01:22
ziptnf
Programming your future
Registered: Jun 2008
Location: Louisville, KY
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.
