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!

 

[Moetasim]

Your attempt at a solution is on the right track. To calculate the maximum height at which Tarzan can land, you need to solve for v(f) and then use that value to calculate the height using the equation h = (v(f)^2)/(2g). However, it is important to note that the conservation of mechanical energy only applies if there are no external forces acting on the system. In this case, there are other forces at play such as air resistance and friction, which may affect the accuracy of your calculation. Additionally, the angle of the vine and the release point of Jane may also affect the outcome. It is important to consider all these factors when solving for the maximum height.