An 88.0 kg skydiver jumps out of a balloon at an altitude of 1 180 m and opens t

Question

An 88.0 kg skydiver jumps out of a balloon at an altitude of 1 180 m and opens the parachute at an altitude of 170 m. (a) Assuming that the total retarding force on the diver is constant at 70.0 N with the parachute closed and constant at 3 600 N with the parachute open, what is the speed of the diver when he lands on the ground? ___________m/s ??? (b) At what height should the parachute be opened so that the final speed of the skydiver when he hits the ground is 5.00m/s? _____________m ???

Explanation / Answer

(a) we have force o gravitation mg=88*9.8=862.4 N total force at the first stage is 792.4 n. F=ma --> a=792.4/88=9 m/sec^2 --> v1^2=2*x*a=2*(1180-170)*9=9090--> v1=95.34 m/sec for the second stage: F=3600-mg=-2737.6 N --> a=-31.1 m/sec ^2 --> vf^2+v1^2=2xa=2*170*31=10577 --> vf^2=10577-95.34^2=1487 --> vf=38.56 m/sec (b)

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