Supersonic Free Fall. In October 2012, Felix Baumgartner jumped off a balloon from the stratospheric altitude of 39 km (above sea-level), reaching after 50 sec a supersonic speed of 377 m/sec, at an altitude of 28 km. The speed of sound at that altitude is 300 m/sec, therefore, he achieved a record speed of Mach 1.25. After falling for 260 seconds and reaching an altitude of 2.5 km, he opened the parachute, and eventually reached the ground after a total falling time of 9 minutes and 18 sec. Note that the ground at the landing site in Roswell, New Mexico was at an elevation of 1043 meters above sea level. The following data were recorded during the fall. V sound t (sec) 0 V (m/s) 0 79.17 53.19 h (km) (m/s) 38.969 315.38 jump altitude above sea level 7.619 309.71 speed slows substantially 2.567 330.30 parachute opens 1.043 336.27 ground is reached 180 260 558 The purpose of this problem is to reproduce these results with MATLAB. We recall from the problem 4 - Tutorial 11 that the air-drag force and downward vertical force are given by: VO y Fdrag Fdrag = PCA V2 ho mg ty h dy m = mg - Fdrag = mg -- dt 1-PCAV ground which may be re-written in the form: dy dt = g. (-), 2 mg Vc = V CAP terminal velocity (14) At high altitudes h, both the acceleration of gravity g and the air density p, and hence ve, depend on the height h, and can be represented by the following functions: g(h) = go R2 (Re+h)2 go = 9.80665 m/s, Re = 6356.766 km h p(h) = 1.2241 exp exp[- (1166) - G8102) + (1935)'] h 18.192 + oshs 40 km (15) vc (h) = 2mg (h) CAp(h) = height-dependent terminal velocity where h is in units of km in all expressions, and R. is the radius of the earth. The function p(h) provides a simple and accurate least-squares fit to the standard atmosphere data over the intervalo shs 40 km. Thus, Eq. (14) must be replaced by: dy = g(h). = acceleration dt va (h) (16) dh = - = speed dt where is measured upwards from the ground and related to the drop-distance y from the initial height by h = ho- y, as shown in the above figure. The negative sign in the second equation is because v represents the downward velocity and dh/dt, the upward velocity. Next, we discretize the time in small time steps in = (n-1), n = 1,2,..., and replace the time- derivatives by differences to obtain the computational algorithm: vin +1)-vin) vin) = ain) (hin) vi (hun) hin + I)-hin) = -Vin which can be rearranged as follows, where we also divided the velocity term of hin) by 1000 because hin) is in km instead of meters: initialize at: hil)= ho. (1)= M = 0 for n = 1,2,3,.... do: (17) viin vichn)) vin + 1) = vinain). a (n) = g(h(n)-(1-hm) hin+1)= hin)-vin).T/1000 In the rest of this problem assume the following parameter values: m = 118 kg 260 lbs, Baumgartner's total sveight including equipment, C = 1.0 drag coefficient during free-fall A1 = 0.7 m cross-sectional area during free fall C2 = 3.17 drag coefficient after parachute opens A = 25.1 m parachute area, 270 ft, V= initial jump velocity ho = 38.969 km initial jump altitude, h = 1.043 km landing site elevation above sea level, 26260 sec time parachute opens, T = 0.01 sec time step a. Given a MATLAB function (on Moodle) that calculates the speed of sound in air at any vector of heights h, as well as the corresponding absolute temperatures: [vs.T] = vsound(h) where h should be in km. Use the following formula for it: vath JyrTh) a. Given a MATLAB function (on Moodle) that calculates the speed of sound in air at any vector of heights h, as well as the corresponding absolute temperatures: [vs,T] - vsoundCh) where h should be in km. Use the following formula for it: Vs (h)= VyRT(h) 6 where y = 1.4 is the adiabatic expansion constant of air, Rs = 287.053 is the specific gas constant of air in J/K/kg, and T(h) is the absolute temperature (K). Plot the sound speed vs(h) and temperatures T (h) versus h in the range 0