An 80.0 kg skydiver jumps out of a balloon at an altitude of1000 m and opens the

Question

An 80.0 kg skydiver jumps out of a balloon at an altitude of1000 m and opens the parachute at analtitude of 200 m.

(a) Assuming that the total retarding force on the diver isconstant at 50.0 N with the parachute closed and constant at 3600 Nwith the parachute open, what is the speed of the diver when helands on the ground?
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(b) Do you think the sky diver will be injured? Explain.
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(c) At what height should the parachute be opened so that the finalspeed of the skydiver when he hits the ground is 5.00 m/s?
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(d) How realistic is the assumption that the total retarding forceis constant? Explain your answer.
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Explanation / Answer

mg-50=ma 80*9.8-50=80*a a=9.175m/s2 So hefalls with the resultant force F=ma=80*9.175=734N The speed at this point as v2-u2=2as v2-0=2*9.175*800 v=121.1610498m/s If 3600N acts then the resultant acceleration as 734-3600=80*a a=-35.825m/s2 v2-u2=2as v2-121.16104982=2*-35.825*200 v=18.708m/s Yes the sky diver will be injured as the speed is toohigh. 52-121.16104982=2*-35.825*h Plug in the values to get the height of it. As the atmosphere varies with height so the force constantas. Hence we get by it.

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