Yes I do, and if I'm to take your example as the basis of what you think my mistake really was, I think you're confused. Either way, my doubts have been cleared, so thank you! :smile:
So I suppose my mistake (as well as the rest of my classmates') was to assume that the graphical vector representation of the distances (on the picture) was identical to that of the forces, when in reality, it's not. Hence why the vertical distances might be the same but not the tension in these...
Homework Statement
Three cables are used to tie the balloon shown in the figure. Determine the vertical component of the force P exerted over the balloon at point A, if the tension on cable AB is 259 N.
http://img521.imageshack.us/img521/9687/physicsballoonproblem.png
(Sorry for the figure...
Thank you WJSwanson! I was lazy but you encouraged me to go through the formula! So this is what happened:
I listed my known variables as follows:
H = hypotenuse = initial velocity at 28° = final result
x0 = 0
x = 7.8 meters
vx0 = vx = Hcos(28)
a = 0
t = ?
y0 = 0
y = 0
vy0 =...
Yeah! Haha, typo, sorry! That's kind of what I was thinking but I'm unsure of how exactly to use what I have. Like I said, in my head I can see it all happening but I don't know where to start. Would I have to carry my hypotenuse variable all the way to the end of the problem and then do some...
Homework Statement
An athlete executing a long jump leaves the ground at a 28° angle and travels 7.80 m. What was the takeoff speed?
Xo = 0
X = 7.80m
Ay = -9.8m/s^2Homework Equations
V=Vo+at
X=Xo+VoT+(1/2)AT^2
V^2=V^2o+2A(X-Xo)The Attempt at a Solution
I made a right triangle with a 28°...