Conservation of Mechanical Energy in rope swing

[physicsma1391]

Homework Statement

Tarzan and Jane, whose total mass is 130.0 kg, start their swing on a 5.0 m long vine when the vine is at an angle of 30 degrees with the horizontal. At the bottom of the arc, Jane, whose mass is 50.0 kg, releases the vine. What is the maximum height at which Tarzan can land on a branch after his swing continues? (Hint: Treat Tarzan's and Jane's energies as separates quanitities.)

Homework Equations

How to I calculate this?

The Attempt at a Solution

cos60=x/5
x=2.5
h=5-2.5=2.5

ME(i)=ME(f)
GPE=KE(f1)+KE(f2)
mgh=.5mv(f)^2 + .5 mv(f)^2
(130)(9.81)(2.5)=.5(50)v(f)^2 + .5(80)v(f)^2
3188.25=25v(f)^2 + 40v(f)^2

not sure if this is correct or where to go from here??

 

[Runelord]

Try this out.

When the swing starts, there is only potential energy.

gh(m1+m2)

At the bottom, there is only kinetic

1/2(m1+m2)v^2. Therefore, equate these 2 quantities and solve for v. Now, Jane (m2) releases off. So the energy on the bottom becomes:

1/2(m1+m2)v^2 - 1/2(m2)v^2 = Total

Finally, this total goes into how high he can go alone.

Total = m1 g h, solve for h. Good luck!