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−velnm−rt/m

where g is the acceleration due to gravity and t is not too large. If g=9.8 m/s^2, m = 30000 kg, r = 160 kg/s, and ve = 3000 m/s, find the height of the rocket one minute after liftoff.

Answer 1

This question is incomplete, the complete question is;

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 consumed at rate r, and the exhaust gases are ejected with constant velocity v(sub e) [relative to the rocket]. A model for the velocity of the rocket at time t is given by the equation:

v(t) = -gt - v(sub e) ln [(m-rt)/m]

where g is the acceleration due to gravity, and t is not too large. If g = 9.8 m/s^2, m = 30,000 kg, r = 160 kg/s, and v(sub e)= 3000 m/s, find the height of the rocket one minute after liftoff.

Answer:

the height of the rocket one minute after liftoff is 14.8441 km

Step-by-step explanation:

Given the data in the question;

Velocity of the rocket at time t is;

v(t) = -gt - $V_{e\\}$ ln( $\frac{m-rt}{m}$)

$\int\limits^t_0 {v(t)} \, dt$ = ₀∫⁶⁰(  -gt - $V_{e\\}$ ln( $\frac{m-rt}{m}$) )dt

= -g($[\frac{t^{2} }{2}]^{60}_0$  - $V_{e\\}$ ₀∫⁶⁰In( $\frac{m-rt}{m}$) dt

= -g($[\frac{t^{2} }{2}]^{60}_0$  - $V_{e\\}$ [ t In( $\frac{m-rt}{m}$) + $\frac{m}{r}$ - t - mr$\frac{1}{r^{2} }$ In |m-rt| $]^{60}_0$

= -g($[\frac{60^{2} }{2}]$  - $V_{e\\}$ [ 60 (In(1 -  $\frac{60r}{m}$)-1) + $\frac{m(In(m^{2})-In((m-60r)^2))) }{2r}$ ]

we input our given values; g=9.8 m/s², m=30000 kg, Ve = 3000 m/s, r = 160 kg)

= -9.8($[\frac{60^{2} }{2}]$  - 3000 [ 60 (In(1 - $\frac{60(160)}{m}$)-1) + $\frac{30,000(In(30,000^{2})-In((30,000-60(160))^2))) }{2(160)}$ ]

$\int\limits^t_0 {v(t)} \, dt$ =  14844.10 m

we convert to kilometer

$\int\limits^t_0 {v(t)} \, dt$ =  14844.10 ÷ 1000

$\int\limits^t_0 {v(t)} \, dt$ =  14.8441 km

Therefore, the height of the rocket one minute after liftoff is 14.8441 km