We can solve the problem by using conservation of energy.
Initially, the total energy of the rocket is only kinetic energy. This is equal to
[tex]E_i = K_i = \frac{1}{2} m v_i^2 = \frac{1}{2}(0.181 kg)(92 m/s)^2 = 766 J [/tex]
As the rocket goes higher, part of this kinetic energy converts into potential energy. Indeed, at the height h=164 m, the total energy of the rocket is the sum of the kinetic energy (at the new speed [tex]v_f[/tex]) and the potential energy:
[tex]E_f = K_f + E_p = K_f + mgh[/tex]
For the conservation of energy, [tex]E_i = E_f[/tex], so we can write:
[tex]K_i = K_f + mgh[/tex]
and so we can find the kinetic energy at height h=164 m:
[tex]K_f = K_i - mgh = 766 J-(0.181 kg)(9.81 m/s^2)(164 m)=475 J[/tex]
