A rocket accelerates by burning its onboard fuel, so its mass decreases with time. Suppose the initial mass of the rocket at liftoff (including its fuel) is $m$, the fuel is consumed at rate $r$, and the exhaust gases are ejected with constant velocity $v_{e}$ (relative to the rocket). A model for the velocity of the rocket at time $t$ is given by the equation
$$v(t)=-g t-v_{e} \ln \frac{m-r t}{m}$$
where $g$ is the acceleration due to gravity and $t$ is not too large. If $g=9.8 \mathrm{~m} / \mathrm{s}^{2}, m=30,000 \mathrm{~kg}, r=160 \mathrm{~kg} / \mathrm{s}$, and
$v_{e}=3000 \mathrm{~m} / \mathrm{s}$, find the height of the rocket (a) one minute after liftoff and (b) after it has consumed $6000 \mathrm{~kg}$ of fuel.