A person walks 200meters at 27 degrees north of East

1point) a skydiver weighing 130 lb (including equipment) falls vertically downward from an altitude of 14,000 ft and opens the parachute after 20 s of free fall. assume that the force of air resistance, which is directed opposite to the velocity, is of magnitude 0.85|v|0.85|v| when the parachute is closed and is of magnitude 17|v|17|v| when the parachute is open, where the velocity vv is measured in ft/s. assume that acceleration due to gravity has magnitude 32 ft/s/s; remember that weight is the product of mass and gravitational acceleration. find the speed of the skydiver when the parachute opens. speed is 150.6121788 ft/s find the skydiver's height when the parachute opens. height is 2338.98591 ft find the limiting velocity vlvl after the parachute opens. (think about whether this should be positive or negative.) vl=vl= 7.647 ft/s

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At a certain instant after jumping from the airplane a, a skydiver b is in the position shown and has reached a terminal (constant) speed vb = 52 m/s. the airplane has the same constant speed va = 52 m/s, and after a period of level flight is just beginning to follow the circular path shown of radius ρa = 2330 m. (a) determine the velocity and acceleration of the airplane relative to the skydiver. (b) determine the time rate of change of the speed vr of the airplane and the radius of curvature ρr of its path, both as observed by the nonrotating skydiver.