Question

An object of mass 700 kg is released from rest 1000 m above the ground and allowed to fall under the influence of gravity. Assuming the force due to air resistance is proportional to the velocity of the object with proportionality constant b=50 N-sec/m, determine the equation of motion of the object. When will the object strike the ground? (Hint: Here the exponential term is too large to ignore. Use Newton's method to approximate the time t when the object strikes the ground.] Assume that the acceleration due to gravity is 9.81 m/sec Let X(t) represent the distance the object has fallen int seconds. Determine the equation of motion of the object. X(t) = 39.295t + 1923.529 e 14 - 1923.529

Answer 1

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from may law of motion, = - Fairvensancet Egresivity F = bv laval proportional) F = mg may = – bv + mg dt =- dv. s - dt mdu - breng - = - dt (bu orng 700, 2=9.8 b=50, m= du =- dt 2) » 1 d 88.8 = - dt oo = -22 at dv Vo9.8x700 570 V8137.2 -0.71 dt side int both du = so.ndt en/v. 137.21=-0.71€ +(V4 137.2) = 0.7 137.2 -0.714 v ve at

da - = e è 0.71 - 137. dt 0.71t da int = (e on 1372) at both side; No=1000 m. x v.71t 1 = & I 137.2t 6.71 -0.71t 0.71 0 -1000 = -1 e 0.7 e = 1000 $137.2t tt e-0.71€ - 710 & 97.414 + 1 -0.71 t - 711 97.41 € 711 =97.41t = t = 37.41 20 ) Cappy. o) - 7.30 egn of motion of object tas 711 minuts -0-71E -I e +137.2t. -1000 D 0.71 -0.71€ +137.2t t1ooo Oox eo.7,