Initially a total mass of a rocket is M, of which kM is the mass of the fuel. Starting from rest, the rocket gives itself a constant vertical acceleration of magnitude g by ejecting fuel with constant speed u relative to itself.
a) If m denotes its remaining mass at time t, show that the rate of decrease of m with respect to t is 2mg/u and deduce that m=Me^(-2gt/u).
b) Show that the kinetic energy of the rocket when the fuel is exhausted is (1/8)Mu^2(1-k)[ln(1-k)]^2
c) Show the value of k for which energy is a maximum is 1-e^-2
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Above diagram shows the rocket at time t = 0, t = t and t = t + dt. Let the mass of the rocket be m when t = t.
In small time dt, rocket ejects fuel of mass dm. So in time t + dt, rocket's mass is m - dm.
Further, velocity of the fuel ...
This is the solution to a rocket propulsion question. As the rocket moves upward, its mass gradually decreases. This causes the kinetic energy to increase to a certain maximum value. This problem requires careful set up of equations using basic physics principles and solving them. I have done just that in this 4-page word document.