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{\bf Question}

$(*)$ It is November 5 and a rocket of mass 50$g$ is stationary on
the ground pointing vertically to the sky.  The fuse if lit and
the motor, generating a force of 1,500$mg/s^2$ (metres grams per
(second)$^2$), burns for two seconds before it stops.  (You may
wish to take the gravitational constant as 10$m/s^2$).

\begin{enumerate}

\item{ } Write down the equation for the velocity during the first two
seconds of flight (neglect the changing mass of the rocket as fuel
is used, neglect air resistance but include gravity).

\item[ ] Find the velocity and the position for the first two
seconds.

\item
Find the velocity and position of the rocket for times later than
two seconds after liftoff (during this time only the gravitational
force acts on the rocket).

\item[ ] Determine the maximum height of the rocket.

\item[ ] Calculate the total amount of time the rocket is in the air
before it crashes back to the ground.

\end{enumerate}



\vspace{0.25in}

{\bf Answer}

$\ds 0\leq t\leq 2$

mass = 50, motor forces = 1500, gravity = 10

$\ds 50\frac{dv}{dt}=1500-50*10, \,\,\,\, \frac{dx}{dt}=v$

$\ds \frac{dv}{dt}=20, \,\,\,\, \frac{dx}{dt}=v, \,\,\,\,$ at $\ds
t=0, \,\,\,\, v=0, \,\,\,\, x=0$

$\ds \Rightarrow v=20t, \,\,\,\, x=10t^2$

hence at $\ds t=2, \,\,\,\, v=40m/s, \,\,\,\, x=40m$

$\ds 2\leq t$

mass = 50, motor force = 0, gravity = 10

$\ds 50\frac{dv}{dt}=-50*10, \,\,\,\, \frac{dx}{dt}=v$

$\ds \frac{dv}{dt}=-10, \,\,\,\, \frac{dx}{dt}=v, \,\,\,\,$ at
$\ds t=2, \,\,\,\, v=40, \,\,\,\, x=40$

$\ds \Rightarrow v=-10t+60, \,\,\,\, x=-5t^2+60t-60$

The maximum height is when $\ds v=\frac{dx}{dt}=0 \Rightarrow
-10t+60=0$

$\ds \Rightarrow t=6$ seconds

$\ds x(6)=-5(36)+60(6)-60=120$m

The rocket returns to the ground when $\ds x=0 \Rightarrow
0=-5t^2+60t-60$

$\ds \Rightarrow
t=\frac{-60-\sqrt{(60)^2-4*5*60}}{2(-5)}=6+\sqrt{24}\approx10.8$
seconds

(The positive square root gives $\ds t<6$, so is excluded for this
part of the calculation.)


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