Question

3. A 180-Ib person jumps out of an airplane with initial upward velocity of 5 ft/s at a height of 3000 ft (assuming the air resistance is proportional to its falling speed). The air resists the body's motion with a force of 2 lb for each ft/s of speed. Assuming the constant gravity is 32 ft/s2. 1. find its velocity v = v(t) 2. find its terminal velocity 3. find the distance of falling x = x(t) 4. find the duration in seconds until the body hits the ground

Answer 1

* According to the question the differential equation of the given mystem visi : 180 de = 2.0. ='(1 80-X32), *10= 3000, x(10) = 5 ves good andre - 2880 Yaking Laplace transform on both sides and ming its linearity: N 90X (x"(t)) = + ( x'()) – 2 880 d (1) ie, 90 (52 X (s) - 5 X60) – '601) = 5X(5)-260) - 2850 vie, 906 $k x(5) – 3000$ -5 ) = 3 X 152,-3000 - 2880 we, xcs) (90 5+-5) = 270000 S +450 - 3000 - 2850 3000 - 2880 Sügnngih = CamScanner 270000 - 2550 go's - 90.5- s. 2880 51905-5)

= 270000' _2550(905-19.05-11) gos-1 T90S-1)S _ 2880(905-1905-0). E S (90 S- S 1500 i = 27000.0 905-1 229.500 + 2550 90 S-1 + TTS - 259200 + 2880 s(905-1) + 2880 3000 35- 2550 S-905- .2550 F ES -259200 (905-905-1) T iscgost), 259200 S AS- = 2617.50 +2530 - 25875 Taking inverse Laplace transform on both ? vides and uring its linearity : DA act) = 261750*24 () + 2880 2+($) +298750X-Cistus). CS10 (1) 2 2 6 415 4 + 8 got - 25% so a = ( 2 )

Soy uct) = x1(t) À, UCH) = 2880 – 2875 € * (Aus) Du(t) = 2880 –2875 eso © X(t) = 2617.50 +2880 € -2587500* @ when the body hits the ground, u(t) = 0 → 261750 + 2880 t -258750 e 5o = 0 Et=13,512 vecand. ® So its terminal velocity = V(13.512) Scanned 460.72 ft I see

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