# Thread: potential energy, how high to they end up, question

1. ## potential energy, how high to they end up, question

Tarzan is in the path of a pack of stampeding elephants when Jane swings in on a rope vince, careing Tarzan to safety. The length of the vince is 25m and Jane starts her swing with the rope horizontal. If Jane’s mass is 54kg and Tarzan’s is 82 kg, to hwat hieght above the ground will the pair swing after she rescues him? Assume rope is vertical when she grabs him.
Initially $U_1=mgh=54kg(9.81m/s^2)25m=13243.5 J$
All this potential energy is converted to kinetic energy when Jane grabs Tarzan. So $K_2=U_1=13243.5J=\frac{1}{2}m_Jv^2 \Longleftrightarrow v=\sqrt{\frac{2U_1}{m_J}}=\sqrt{\frac{2(13243.5J)} {54kg}}=22.147m/s$
When Jane and Tarzan come to rest $U_3=\frac{1}{2}(m_J+m_T)v_2^2=(m_T+m_J)gh_3 \Longleftrightarrow h_3 = \frac{\frac{1}{2}(m_J+m_t)^2v_2^2}{(m_J+m_T)g}=347 5m$

2. Originally Posted by superdude
Tarzan is in the path of a pack of stampeding elephants when Jane swings in on a rope vince, careing Tarzan to safety. The length of the vince is 25m and Jane starts her swing with the rope horizontal. If Jane’s mass is 54kg and Tarzan’s is 82 kg, to hwat hieght above the ground will the pair swing after she rescues him? Assume rope is vertical when she grabs him.
Initially $U_1=mgh=54kg(9.81m/s^2)25m=13243.5 J$
All this potential energy is converted to kinetic energy when Jane grabs Tarzan. So $K_2=U_1=13243.5J=\frac{1}{2}m_Jv^2 \Longleftrightarrow v=\sqrt{\frac{2U_1}{m_J}}=\sqrt{\frac{2(13243.5J)} {54kg}}=22.147m/s$
When Jane and Tarzan come to rest $U_3=\frac{1}{2}(m_J+m_T)v_2^2=(m_T+m_J)gh_3 \Longleftrightarrow h_3 = \frac{\frac{1}{2}(m_J+m_t)^2v_2^2}{(m_J+m_T)g}=347 5m$
yes it is. you forgot one important principle ... an inelastic collision occurs at the bottom of the swing.

using conservation of momentum ...

$v$ = jane's speed at the bottom of the swing

$v_f$ = jane+tarzan's speed after the pick up

$M_J v = (M_j + M_T)v_f$

$\frac{M_J v}{M_J+M_T} = v_f$

now convert the kinetic energy after the collision back to GPE

3. Ok so just before Jane grabs Tarzan her velocity is $v=\sqrt{\frac{2mgh}{m}}=\sqrt{2(9.81m/s^2)2m}$ and using conservation of momentum their velocity after the grab is $v=\frac{v_{oT}m_T+v_{oJ}m_J}{m_T+m_J}=8.79375m/s$ then using the fact that the kinetic energy after the grab will all be converted into potential energy again $k=\frac{1}{2}mv^2=U=mgh_2$ solving for $h_2$ I get 3.94m which makes a whole lot more sense than what I had before.