Asked by nm

What is the acceleration (in meters/second^2) of a freely falling 74.0 kg sky-diver, if air resistance exerts a force of 264 N?
Answered by Henry
Fs = 74kg * 9.8Nkg = 725.2N,

a=Fn/m = (725.2 - 264) / 74 = 6.2m/s^2

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