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 = 130 kg/s, and ve = 2,700 m/s, find the height of the rocket one minute after liftoff. (Round your answer to the nearest whole meter.)

Answer 1

The height of the rocket after one minute will be -7 × 10^9 m

Step-by-step explanation:

In the question expression for velocity is given. If we take the integral of that expression for time interval 0 to 60s, we will get expression for the height of rocket So,

X = ∫〖(-gt - ve ln⁡(m) - rtm)〗 dt (Over a interval 0 to 60 s)

X = -g (60^2-0) / 2 – veln(m) (60-0) - rm (60^2-0) / 2

By putting all the values we get:

X = - (9.8)(3600)/2 – (2700) (ln(30000)) (60) – (130)(30000)(3600)/2

X = -7.02 × 10^9 m

After rounding off

X = -7 × 10^9 m

The negative sign shows that direction of the rocket motion is opposite to the velocity of ejected gases.

Answer 2

the position is 7 10⁹ m

Step-by-step explanation:

Since the position is the integral of the velocity with respect to time and the equation of velocity is given in the problem

v (t) = -gt - ve ln m - rt m

v = dx / dt

dx = v dt

x = ∫ v dt

x = ∫ [-gt -ve ln m -rtm] dt

x = -g t² / 2 - veln m t - rm t² / 2

Evaluated between t = 0 and t = 60 s

x = -g ½ (60² -0) - veln m (60-0) - rm ½ (60²-0)

x = - 9.8 1800 - 2700 ln 30000 60 - 130 30000 1800

x = -17640 - 1670050 -7020000

x = 7.02 10⁹ m

x = 7 10⁹ m

This is the position of the rocket 1 minute after clearance