Question 3.12: A classic example of an accelerating control volume is a roc...

The appropriate control volume in Fig. E3.12 encloses the rocket, cuts through the exit jet, and accelerates upward at rocket speed V(t). The z momentum equation (3.49)
\sum{F}=\int_{CV}^{}{a_{rel}dm }=\frac{d}{dt}\left(\int_{CV}^{}{V\rho dv} \right)+\int_{CS}^{}{V\rho (V_{r}\cdot n )dA}
becomes
\sum{F_{x} }-\int{a_{rel} dm}= \frac{d}{dt}\left(\int_{CV}^{w d\dot{m} }{} \right)+(\dot{m}w )_{e}
or -mg-m\frac{dV}{dt}=0+\dot{m}(-V_{e})  with   m=m(t)=M_{0}-\dot{m}t
The term a_{rel}= dV/dt of the rocket. The control volume integral vanishes because of the steady rocket flow conditions. Separate the variables and integrate, assuming V =0 at t =0:
\int^{v}_{0}dV=\dot{m}V_{e}\int^{t}_{0}\frac{dt}{M_{0}-\dot{m}t }-g\int^{t}_{0}dt         or           V(t)=-V_{e} ln(1-\frac{\dot{mt} }{M_{0} } )-gt
This is a classic approximate formula in rocket dynamics. The first term is positive and, if the fuel mass burned is a large fraction of initial mass, the final rocket velocity can exceed V_{e}