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**ebaines** You can use impulse and momentum principles:

$\displaystyle m \frac {dv}{dt} = -C \frac {dm}{dt} $

Rearrange and integrate:

$\displaystyle -C \frac {dm } m = dv$

$\displaystyle -C \ln (m) = v$

$\displaystyle \Delta V = C \ln (\frac {M_0}{M_1} $ where [tex] M_0 {/tex] = initial total mass of ship plus fuel, and $\displaystyle M_1$ = final mass of ship after all fuel is spent.

Now that you have a value for $\displaystyle \Delta V $, you can calculate its final kinetic energy.

To answer the second part it would be best to introduce a variable for the ratio of the ship's mass to the ship + fuel mass. Let $\displaystyle k = \frac {M_1}{M_0}$, and the KE equation becomes

$\displaystyle KE = \frac 1 2 k M_0 C^2 (\ln (\frac 1 k))^2$

Now find the value of k that maximizes KE by setting $\displaystyle \frac {KE}{dk} = 0$.