a rocket accelerates by burning its onboard fuel, so its mass decreases with time. suppose the initial mass…

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\\frac{m}{m - rt} 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 = 3,100 m/s, find the height of the rocket one minute after liftoff. (round your answer to the nearest whole meter.)

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\\frac{m}{m - rt} 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 = 3,100 m/s, find the height of the rocket one minute after liftoff. (round your answer to the nearest whole meter.)

Answer

Explanation:

Step1: Convert time to seconds

1 minute = 60 s, so (t = 60\ s).

Step2: Substitute values into velocity - time formula

The velocity formula is (v(t)=-gt - v_e\ln\frac{m}{m - rt}). Substitute (g = 9.8\ m/s^2), (m = 30000\ kg), (r = 130\ kg/s), (v_e=3100\ m/s) and (t = 60\ s) into it. First, calculate (m-rt=30000 - 130\times60=30000 - 7800 = 22200\ kg). Then, (\ln\frac{m}{m - rt}=\ln\frac{30000}{22200}\approx\ln(1.35135)\approx0.301). (gt=9.8\times60 = 588\ m/s), (v_e\ln\frac{m}{m - rt}=3100\times0.301 = 933.1\ m/s). (v(t)=- 588-933.1=-1521.1\ m/s) (the negative sign just indicates the direction convention, we are interested in the magnitude - speed for integration).

Step3: Calculate height using integration

The height (h) is the integral of velocity with respect to time. Since (v(t)=-gt - v_e\ln\frac{m}{m - rt}), and we want to find the height from (t = 0) to (t = 60\ s). (h=\int_{0}^{60}\left(-gt - v_e\ln\frac{m}{m - rt}\right)dt). We know that (\int_{0}^{60}-gt\ dt=-\frac{1}{2}gt^{2}\big|{0}^{60}=-\frac{1}{2}\times9.8\times60^{2}=- \frac{1}{2}\times9.8\times3600=-17640\ m). For (\int{0}^{60}-v_e\ln\frac{m}{m - rt}dt), let (u = m-rt), (du=-r\ dt). (\int - v_e\ln\frac{m}{m - rt}dt=-v_e\int\ln\frac{m}{u}\times\left(-\frac{1}{r}\right)du=\frac{v_e}{r}\int(\ln m-\ln u)du). After integrating and substituting the limits (t = 0) and (t = 60) (and reversing the substitution for (u)), we find: (\int_{0}^{60}-v_e\ln\frac{m}{m - rt}dt\approx52980\ m). (h=-17640 + 52980=35340\ m). Rounding to the nearest whole - meter, (h = 35340\ m).

Answer:

35340