Rocket Propulsion Flight Time

  • Thread starter jtaz612
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  • #1
jtaz612
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The height that a model rocket launched from earth can reach can be estemated by assuming that the burn time is short compared to the total flight time, so for most of the flight the rocket is in free-fall. (This estimate neglects the burm time in calculations of both time and displacement. ) For a model rocket of specific impulse Isp = 100 s, mass ratio mo/mf = 1.2 , and initial thrust to weight ratio τ o = 5 . Estimate (a) the height the rocket can reach, and (b) the total flight time. (c) Justify the assumption used in the estimates by comparing the flight time from part (b) to the time it takes for the fuel to be spent.

Some useful equations:
The specific impulse of a rocket propellant is defined as Isp = Fth / (Rg) , where Fth is the thrust of propellant, g is the magnitude of free fall accelaration at the surface of earth, and R is the rate at which the propellant is burned.

I have not shown any work because im lost as to how to start the problem. If more of an attempt is needed let me know and i'll post what I've done so far. Thankyou.
 

Answers and Replies

  • #2
denverdoc
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well this will be a pretty crude "proof":

but total impulse, I=Isp*g*Mp where Mp=mass of propellant.
and I=deltaP=Mrocket*Vel(burnout), from mass fraction we know
(Mprop+Mrocket)/Mrocket=1.2

hence Mass Prop=.2Mrocket, subbing,

100*9.8*.2=Vb=196m/s. from kinematics, altitude,h, from burnout to apogee assuming G is constant, and no drag

2(g)h=Vb^2 =1960m, the flight time would of course be 196=a*t=20 seconds.

any thoughts re the burn time of the propellant?
 

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