\(E_T = E _k_i_n + E_p_o_t = constant\)
The total energy of the rock does not change.
Let h be the ratio between the height of the rock and H, so that the height of the rock = H·h.
The energy of the rock that has been dropped from a height H just before it hits the ground with h = 0:
\(E_T= 180J + mgH \ h = 180J + mgH \ 0 = 180J + 0\)
The energy of the rock when it is released from rest from height H with h = 1 :
\(E_T = E_k_i_n + mgH = 0 + mgH = 180J\)
so: \(mgH = 180J\)
You find:
\(E_T = E_k_i_n + 180J \ h = 180\)
for \(h = \frac{1}{3}\)
\(180 = E_k_i_n + 180J \frac{1}{3} = 180J \frac{2}{3} + 180J \frac{1}{3} \\ \\ E_k_i_n = 180J \frac{2}{3} = 120J\)
