Tarzan and Jane Physics Question- how to set it up!?

Question

Tarzan and Jane, whose total mass is 137 kg, start their swing on a 4.6 m. long vine when the vine is at an angle 32.6 degrees with the horizontal. At the bottom of the arc, Jane, whose mass is 47 kg, releases the vine. Acceleration of gravity is 9.81 m/s^2. What is the maximum height at which Tarzan can land on a branch after her swing continues? (Hint: Treat Tarzan's and Jane's energies as separate quantities.) Answer in units of m.

Thank you very much for your help in advance!

Answer 1

At the beginning, Tarzan and Jane have a potential energy of
{(M(T) + M(J) }(g) (4.6) ( 1 - sin 57.4 deg) which is M(both)gh.

At the bottom, they are traveling at a velocity V where all the P.E. has been converted to Kinetic Energy:

V^2 = { 2(137)(4.6)(1-sin57.4)/137 }; (from (1/2)MV^2 = KE )

Tarzan's KE is (1/2)(90)V(T)^2 = 4.6(1-sin57.4) and equals his P.E. at the highest point he will reach, H where V(T) = Tarzan's velocity after Jane lets go the vine.

Momentum must be conserved so we get
M(T)V(T) + M(J)V(J) = M V = {M(T) + M(J)}V where V is obtained from the P.E. above. We can get V(J) in terms of V(T)

and use (1/2)M(T){ V(T)}^2 = (1/2)(M)V^2 - (1/2)M(J) {V(J)}^2

to get V(T) to use in M(T)gH = (1/2){M(T)} [V(T)]^2

Answer 2

ok then. first draw a diagram to follow the path taken, have all of your angles, length of your vine, and everything that happens in the problem. split your set up into 3 sections. section 1 is energy. section 2 is a momentum. section three is another energy problem. in your first section, use energy to obtain a final velocity, use KEf+PEf=KEi+PEi. use trig to get your initial h (use 90-given angle). use this to the moment jane is dropped. now use section 2-momentum. m(tar/jane)v(from section 1)=m(tarzan)v(tarzan). this gives you a new v-(tarzan alone). now section 3 use KEf+PEf=KEi+PEi with the vi=v(tarzan), vf=0...you should have all required to obtain your new height..

Answer 3

For this speed the version in g at diverse altitudes can not be exceeded over. Use conservation of skill as your important equation, with KE=(a million/2)mv^2 and PE=-GMm/r The preliminary r is the radius of the Earth, the best r is what's being solved for (type of, we will get there). The preliminary speed is what's given, the best speed is v_f=v_i cos(theta) set up the equation, sparkling up for r_f (this is a splash gruesome, yet not undesirable). as quickly as you have r_f, subtract r_i (the radius of the Earth) to get the altitude.