Solve the following question, using octave programming.

Fuel HO) Fig. E4.19 Solution The control volume is sketched and includes the entire rocket. The reference frame attached to the rocket is accelerating upward at dH]dt?. Newton's second law is writ- ten as, using z upward, d - (Fi, PV dH ... - W- dt? 142 Mc.v. = Pel-V.)VA, IF:- (F): - | pv; dx + [ pv. Venda SpV_dx=0 = where dt since V, is the velocity of each mass element p dv relative to the reference frame attached to the control volume; the only vertical force is the weight W; and Mc.v. is the mass of the control volume. From continuity we see that mc.v. 150 - mt = 150 10t :: W = (150 - 10) X 9.81 The momentum equation becomes dH -(150 10t) x 9.81 1,2 (150 10t) = -m.Ve = -10 700 = - 7000 dt This is written as dH 700 9.81 15 - t The initial acceleration is found by letting t = 0: d2H 700 9.81 = 36.9 m/s2 15 Integrate the expression for dH/dt- and obtain dH = 700 In (15 t) - 9.81t + C dt The constant C = 700 in 15 since dH/dt = 0 at t = 0. Thus at t = 1 s the velocity is dH 15 = 700 In - 9.81 x 1 = 38.5 m/s dt 14 dt2 dt |t=0 Title: Momentum Equation in A Rocket System Objective: Determining the initial acceleration of the rocket and the velocity as the time varying by neglecting the drag on the rocket. The rocket shown in Fig. E4.19, with an initial mass of 150 kg, burns fuel at the rate of 10 kg/s with a constant exhaust velocity of 700 m/s. What is the initial acceleration of the rocket and the velocity after 1 s? Neglect the drag on the rocket. Fuel H(1) Figure 1.1 Figure 1 represents a rocket system. Determine the initial acceleration of the rocket and the velocity after time is : a) 1s b) 2s c) 3s d) 45 e) 5s f) 65 g) 7s h) 85 i) 9s j) 10s