Does Tarzan's Vine Break During His Swing?

[usc2013]

******I figured it out********
Tarzan, who weighs 678 N, swings from a cliff at the end of a convenient vine that is 25.0 m long (see the figure). From the top of the cliff to the bottom of the swing, he descends by 3.2 m. If the vine doesn't break, what is the maximum of the tension in the vine?

The vine will break if the force on it exceeds 805.4 N. Does the vine break? If yes, at what angle does it break (if no enter 180. deg)?

Homework Statement

Homework Equations

mgh=1/2mv^2
mv^2/r=centripetal force

The Attempt at a Solution

I solved for the max tension and got 852 N
I tried to draw triangles and couldn't get the right answer

My friend said the equation should be cos(angle)=2/3(1-h/r)+1/3(T/mg) but either that's wrong or I'm plugging in wrong

 

[jbsteele]

. The correct equation is: T = mg(cos(theta) + 2h/r), where T is the tension in the vine, m is the mass of Tarzan, g is the acceleration due to gravity, theta is the angle of the vine, h is the height of the swing, and r is the length of the vine. Plugging in the values, we get: T = 678 N (9.8 m/s^2) (cos(180 deg) + 2*3.2 m/25.0 m) = 805.4 N Since the tension is less than the maximum force the vine can handle (805.4 N < 852 N), the vine does not break and the angle remains at 180 deg.