A rocket accelerates by burning its onboard fuel, so its mass decreases with tim

Question

A rocket accelerates by burning its onboard fuel, so its mass decreases with time. Suppose the initial mass of the rocket at liftoff (including its fuel) is m, the fuel is consumed at rate r, and the exhaust gases are ejected with constant velocity ve (relative to the rocket). A model for the velocity of the rocket at time t is given by the equation v(t) = gt ve ln m rt m where g is the acceleration due to gravity and t is not too large. If g = 9.8 m/s2, m = 30,000 kg, r = 170 kg/s, and ve = 3,000 m/s, find the height of the rocket one minute after liftoff. (Round your answer to the nearest whole meter.)

Explanation / Answer

given  v(t) = gt ve[ln(m rt)/m]

v(t) = 9.8t 3000[ln((30000 170t)/30000)]

v(t) = 9.8t 3000[ln(1 (17t/3000)]

1 minute =60 sec

height of the rocket one minute after liftoff ,h= [0 to 60] 9.8t 3000[ln(1 (17t/3000)] dt

h=[0 to 60]4.9t2 3000 [0 to 60] [ln(1 (17t/3000)] dt

h=4.9(602-02) 3000 [0 to 60] [ln(1 (17t/3000)] dt

h =-17640 3000 [0 to 60] [ln(1 (17t/3000)] dt

for integra part u =ln(1 (17t/3000) =>du =(1/(1 (17t/3000))(-17/3000)dt , dv=dt =>v =t

u dv =uv - vdu

h =-17640 3000[[0 to 60] [ln(1 (17t/3000)]t -[0 to 60]t(1/(1 (17t/3000))(-17/3000)dt ]

h =-17640 3000[ [ln(1 (17*60/3000)]*60-[ln(1 (17*0/3000)]*0+ (17/3000)[0 to 60](t/(1 (17t/3000))dt ]

h =-17640 3000[ (60ln(0.66))-0 + (17/3000)[0 to 60](t/(1 (17t/3000))dt ]

h =-17640 3000[ (60ln(0.66))+ (17/3000)[0 to 60](t/(1 (17t/3000))dt ]

let 1 (17t/3000) = p =>t=(3000/17)(1-p) =>dt =-(3000/17)dp

t =0 =>p=1 ,t=60 =>p=0.66

h =-17640 3000[ (60ln(0.66))+ (17/3000)[1 to 0.66](((3000/17)(1-p)/p)(-(3000/17)dp) ]

h =-17640 3000[ (60ln(0.66))+ (17/3000)[0.66 to 1](((3000/17)(1-p)/p)((3000/17)dp) ]

h =-17640 3000[ (60ln(0.66))+(3000/17)[0.66 to 1](1-p)/p dp ]

h =-17640 3000[ (60ln(0.66))+(3000/17)[0.66 to 1](1/p)-1 dp ]

h =-17640 3000[ (60ln(0.66))+(3000/17)[0.66 to 1] lnp -p ]

h =-17640 3000[(60ln(0.66)) +(3000/17)[(ln1 -1)-(ln0.66 -0.66)] ]

h =-17640 3000[(60ln(0.66)) -(3000/17)[0.34+ln0.66] ]

h =17174m

h =172 km approximately

height of the rocket one minute after liftoff =172 .17 km

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