velocity $v_e($relative to the rocket$).$ A model for the velocity of the rocket at time $t$ is given by the equation

$v(t)=-gt-v_{e}ln(\frac{m-rt}{m})$

where $g$ is the acceleration due to gravity and $t$ is not too large. If $g=9\mathrm{.}8\frac{m}{s^2},m=31,000kg,r=160k\frac{g}{s},$ and $v_e=2,800\frac{m}{s},$ find the following.

$($a$)$ the height of the rocket $($in meters$)$ one minute after liftoff $($Round your answer to the nearest meter.$)$

$$ m

$($b$)$ the height of the rocket $($in meters$)$ after it has consumed $2,000kg$ of fuel $($Round your answer to the nearest meter.$)$

$$ m