Answer
The height of the rocket after one minute is 512375.89 m
Work Step by Step
The height of the rocket as a function of time is just the integral of the velocity function so, we integrate the velocity given
$$v(t)=-gt- v_{e}ln{ \frac {m-rt}{m}}$$
The height is
$$h(t)= \int{v(t)} dt=\int{-gt- v_{e}ln{ \frac {m-rt}{m}}dt}$$
$$h(t)= \int{v(t)} dt=-\int{gt}dt-\int{ v_{e}ln{ \frac {m-rt}{m}dt}}$$
the first integral is easy it's just
$$\int{gt} dt =\frac{1}{2}gt^{2}+c_{1}$$
The second integral could be easier with changing variables
$$y= \frac {m-rt}{m}$$
thus
$$dy=-\frac {r}{m}dt$$
or
$$dt=-\frac {m}{r}dy$$
then the second integral becomes
$$-\frac {m}{r} \int\ln{y} dy$$
which can be solved by integration by parts
$$u=\ln{y}$$ and $$dk=dy$$
then applying the integration by part formula
$$A=uk−∫kdu$$
the second integral is
$$A=y\ln{y} -\int{dy} = y\ln{y}-y +C_{2}$$
So the height function is
$$h(t)=-\frac{1}{2}gt^{2}-v_{e}\frac {m}{r}( y\ln{y}-y)+C_{1}+C_{2}$$
Changing the new variable back and defining $$C=C_{1}+C_{2}$$
$$h(t)=-\frac{1}{2}gt^{2}-v_{e}\frac {m}{r}( \frac {m-rt}{m}(\ln{(\frac {m-rt}{m})}-1))+C$$
which equal
$$h(t)=-\frac{1}{2}gt^{2}-v_{e} \frac {m-rt}{r}(\ln{(\frac {m-rt}{m})}-1)+C$$
since the rocket start the motion at t=0 from the ground then h(0)=0
and
$$h(0)=v_{e} \frac {m}{r}(\ln({1})-1)+C=-v_{e} \frac {m}{r}C=0$$
then C=0
and the height function becomes
$$h(t)=-\frac{1}{2}gt^{2}-v_{e} \frac {m-rt}{r}(\ln{(\frac {m-rt}{m})}-1)$$
Now we find the height after one minute using the constants given in the problem
$$ g=9.8 m/s^{2} ,m=30000 kg , r=160 kg/s ,v_{e}=3000 m/s$$ and $t=60 s$
$h(60)= -\frac{1}{2}9.8*60^{2}-3000 \frac {30000-160*60}{160}(\ln{(\frac {30000-160*60}{30000})}-1)=512375.89 m$
