Question

Refer to the textbook or other references specified in the course syllabus, read about Golden-Section Search Method for One-Dimensional Unconstrained Optimization and solve the following problem. An object can be projected upward at a specified velocity. If it is subject to linear drag, its altitude as a function of time can be computed as: 2= 20 + " (vo + mg) (1 – e-641m) mg, where z = altitude (m) above the earth's surface (defined as z = 0), zo = the initial altitude (m), m = mass (kg), c= a linear drag coefficient (kg/s), vo = initial velocity (m/s), t = time (S). Note that for this formulation, positive velocity is considered to be in the upward direction. Given the following parameter values: g = 9.81 m/s?, zo = 100 m, vo = 55 m/s, m = 80 kg, and c = 15 kg/s, the equation can be used to calculate the jumper's altitude. Determine the time and altitude of the peak elevation with the golden-section search until the approximate error falls below & = 1% with initial guesses of t1 = 2 and 1-8 s.

Answer 1

Solution. - The given function is , 7(4) = 7+ (vo 8 kg) 9- elemyt) most where, o= 100 m g9.81 m/s? Vo = 55 m/s m = 80kg C15 kg/s ti 2 s and tu = 8s or 4 € [2,8] ! On putting the values, we get, z(t) 400 + $ ( 55+ 801284) (-e67) t). 889 844 15 = 100 + 572.37 (1-e 0.1875t) - 52.32 t On applying Golden section Method: Stelo 1: to = [2,8 ] stoping tolerance E= 0.01 Step 2: Lo Here, da li se -> O Xxx x x x 4.2919 5.7086 oy, 04.618 = 0.381946 = 2.2914 X X X X X X 4.2914 5.7086 do and ta 2+2-29 14 = 4.2914 to = 8-2.2914 = 5.7086 cs CS canned w

z(+1) = 1 91.8 514 and z(t) = 177.439 . (1) ) (1) and we have to maximise the amplitude, so above below tz value of time is eliminated. x x x 1 Step 3: L2 = h Lo = 3.708 > € L = 1/2 Loe 1.4162 So, to = 5.7086 - 1.4162 = 4.2924 or z (tz) = 1 91.8471 Since, z(t.) > z.(+3) So, to maximize I beyond tz is rejected Step 4: L = 1 / 4 to 2 2.29 4 4 > € x x x to 4.2924 5.7086 4.2914 - Lo = 0.8752 = 3.4192 So, ty = 4.2924 - 0.8752 CScanned with carandrine z(tu) = 191.9944

And, to = 2+0.8752 = 2.8752 2(ts) -188.0946 - Here, z(ta) y 2 (to) 30. Beyond both is rejected x x x 2 te tz ta Step 5: La = / Lo = 1.4 16 2 > E s ak 1; 10.6409 to = 3.4172 -0.5409=208763 or (to) = 188.1029 cs. Here, z (tu) y z (to)

Step 6: 24 L5 = 15 Lo = 0.8752 > E L = 1 L = 0.33 43 3.0829 t = 3.4172 - 0.3343 z(t) = 189.97 Here, z(+₂) <z (tu) So, before tz is rejected. 3.0889 3.4172 Step & Lo= 1/5 Lo t to 2 1 1 2 0 = 0.2066 to = 3.4172 - 0.2066 = 3.2106

C or z(to) = 190.8918 z(+8) < į (tai) So, beyond to is rejected. x x x x x x Step 8 : 8. L = 1 L = 0.3343 > € Lito Lo + 0.3237 tg = 3.417 2 – 0.1 277 = 3.28 95 or z (tg) = 491.36 4 and to = 3.4172 +0.1277 23.5449 con z(to) - 192.4486 Here, e(t,0) > z(tu) go beyond tu is rejected. 191.36 Step 9: L8 = L = 0.20667 € Lg LO E 0.0189 17 1 t = 3.5 449 +0.0789 = 3.6238 or, z(t) = 192.6445 Here, (t.) (to) CSS canned with se beyond tio 3 lejected.

Step 40: Lg = I to 0.1277 ² e . Lio = I Lo= 0.0487 tiz = 3.6238 +0.0789 = 3.7027 on z(tie) = 192.7734 tis3.9029 +0.0989 = 3.4816 oh, ž (+3) = 192.8468 tu= 3.7816 +0.0789 =3-8605 or I (ty) - 192.855 tis = 3.8605+0 0789 = 3.9494 or, z(tis) = 192.7916 So, or 2(4,4) is maximum. tu = 3.8605 & 2 (tin) 192.855 CScanned with CamScanner