Question

Suppose an object is dropped from an airplane with initial position, y(0) = 1200 ft. Assuming linear air resistance pv ft/sa, the formulas for velocity and position are, v(t) = (vo – Vr) e-pt +ve and 1 Yo + vrt +- (vo – v-) (1 - e-pt) Use p= 0.31 , g = 32 ft/s2 to answer the following. What is the position of the object at t = 4 sec.? (Give two decimal places.) When will the object reach a position of 500 ft ? (Give three decimal places.)

Answer 1

(Ve-V3) (1-2 st) 3 y (0)=1200 ft v(t) = (vo - ks) est + V g (t) = Yo + k (t) P = 0.31, g=32432 at t=0 v=0 v/o) = (vo-vre + Vo = 0 Vo-Vert Vy =D Vo=0 also at t=0 & acceleration - dv dv = vsest at t=0 → Vy (3) le-9) - 9 = Vo = -330 V = 103.23 et/ at = 9 at t= heels y (4) - 1200 +(-103.23/4) + + (+103.23) (1-2 31 0.31 (6) Ty(4) = 1023172 ft ylt) - 500 - 1200 - 103:23() + 103:23 (1-2-013134) solving by trial and ears, we get It =9.855 seconds) 0.31