An 80.0 kg sky diver jumps out of a balloon at an altitude of1180 m and opens th

Question

An 80.0 kg sky diver jumps out of a balloon at an altitude of1180 m and opens the parachute at analtitude of 185.0 m. (a) Assumingthat the total retarding force on the diver is constant at 50.0 Nwith the parachute closed and constant at 3600 N with the parachuteopen, what is the speed of the diver when he lands on the ground?(b) Do you think the sky diver will get hurt? Explain. (c) At whatheight should the parachute be opened so that the final speed ofthe sky diver when he hits the ground is 4.80 m/s?
(d) How realistic is the assumption that the total retarding forceis constant? Explain.

Explanation / Answer

   if we apply the law of conservation energy then weget    Wnc = (KE + PEg)f -(KE + PEg)i    then we ge   (f1cos180o) d1 + (f2cos180o) d2 = [(1 / 2) mvf2 + 0] - [0 + m g yi]    vf = [2 (g yi -(f1 d1 + f2 d2) /m)]        = {2[(9.80 m /s2) (1180 m) - ((50.0 N) (800 m) + (3600 N) (185 m) /80.0 kg)]}        = ......... m / s (b)    this depends on the answer of (a) (c)    if we apply the law of conservation energy then weget    Wnc = (KE + PEg)f -(KE + PEg)i taking    d1 = 1180 m - d2    vf = 4.80 m / s    (f1 cos180o) (1180 m -d2) + (f2 cos180o) d2 =[(1 / 2) m vf2 + 0] - [0 + m gyi]    on solving we get    d2 = (m g) yi - (1180 m)f1 - (1 / 2) m vf2 / f2- f1         =......... m (d)    in reality the air drag will depend on the skydiversspeed    it will be larger than her weight only after chute isopened it will be nearly equal to his weight before he opens thechute    and again she touches down when ever she moves nearterminal speed

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