Question

10: A rocket accelerates upward 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 in meters per second after t seconds is given by the equation r(t)--gt-ve In (mmrt) where g is the acceleration due to gravity, and t is not too large. If g 9.8 m/s m 30, 000 kg, r160 kg/s, and ve 3000 m/s, find the height of the rocket one minute after liftoff

Answer 1

Given, v(t) = -gt - v_e ln (m - rt/m) h(t) = integral v(t) dt = rho - gt - v_e ln (m - rt/m) dt Taking constants and, we have = -g integral t dt - V_e integral ln (m - rt/m) dt = gt^2/2 - V_e (m - rt)/r (1 - ln (m - rt/m)) Put t = 0 in above in above equation, solve for C, C = V_e m (every term with (t) becomes zero) Therefore h = -gt^2/2 - v_e (m - rt)/r(1 - ln (m - rt/m)) + v_e m/r Now, Just put the given values, t = 1 min = 60 sec h = -9.8 (60)^2/2 - 300 (30000 - 160 times 160)/160 (1 - ln (30000 - 160 times 160/30000)) + 30000 times 3000/16 h = -17640 - 3825500 times 1.3856 + 5625000 h = 14865 m h = 14868 m