Question

(1 point) A steel ball weighing 128 pounds is suspended from a spring. This stretches the spring 528 feet. The ball is started in motion from the equilibrium position with a downward velocity of 7 feet per second. The air resistance (in pounds) of the moving ball numerically equals 4 times its velocity (in feet per second). Suppose that after t seconds the ball is y feet below its rest position. Find y in terms of t. (Note that this means that the positive direction for y is down.) y = Take as the gravitational acceleration 32 feet per second per second.
PLEASE ANSWER BOTH PROBLEM SETS PLEASE ANSWER BOTH

Answer 1

128 pound and Steel steel ball weight öll m = 128 32 m = 128 uky L=128 feet 577 K = Weight 128 こ 128 577 x=577 X= 4 myl (t) +ry'(t) tky 49"(t) + 4yllt +5774 zo y (o)=0, y '(o)=7 Charactestic ean 4x2 + 47+577 =0 7 = – 4 + J16-9232 8 a= utgoi 8. A - 12 = 12 y ( = et/2 [c C, cos (24) + C sin (24) y (o) = C=0 y(t) = eth2 (sinlat)

y' (t) = -1 62 (sini2t) tize 4/2 (2 cosizt y'(o)=7=1262 is = 7 / 12 y (t) = 7 / 4 th е sin (12t) As. x = -170 +144 y'= -21 x + 184 x(0)=6 andyco=8 K -17 14 1) 16] L - 21 18 using eigenvalue -17-7 14 -21 18-7 (-12-1) (18->) + 294 = 0 22 7 - 12 - (2+3) (7-4)=0 7=-3, 2=4 eigen vector corresponding ?=-3 be v -14 14 IN 1=[:] -21 21

ele EN] =B] et V₂ = 1, Vi=1 V N eigen vector corresponding ? =4 bem ] [] = [:] -21 -21 14 im -21 14 [*] =[:] nu let m₂ =1, m = 2/3 m- 23 13] x H) 4t + Ce 43 دا دیر y (t) xo) 6 -32="[:] [:)] y (O) + 2 C2 = 6 3 G + C₂ C2 - 2 3 (2=6, C=2

а e-36 о Filt? .) 11 + беч " | 9 € + Чече x (t) = 2e3e y (t) = 20 3t + bent An