What is the magnitude of the acceleration of twofalling sky divers (mass 147 kg

Question

What is the magnitude of the acceleration of twofalling sky divers (mass 147 kg including parachute) when theupward force of air resistance is equal to one fourth of theirweight?
After popping open theparachute, the divers descend leisurely to the ground at constantspeed. What now is the force of air resistance on the sky diversand their parachute? Please explain the concepts used toexplain.. After popping open theparachute, the divers descend leisurely to the ground at constantspeed. What now is the force of air resistance on the sky diversand their parachute? Please explain the concepts used toexplain.. Please explain the concepts used toexplain..

Explanation / Answer

Weight is the name of the force that causes thegravitational acceleration. FWeight   = mg   (m = mass)       = (147)(9.8)       = 1440.6 N pulling him down
The force of the air resistance pushing him upis 1/4 of FWeight So, Fair =(1/4)1440.6      = 360.15N up The resultant force is 1440.6 down - 360.15 up = 1080.45N down Newtons second law says Fres =ma a = Fres/m    = 1080.45N/147kg    = 7.35m/s2down
For the second part, constant speed mean a = 0,Thus from newton II Fres = 0. In other words the upward forces must balancethe downward. FWeight   =1440.6 N pulling him down as we calculated earlier, so theupward force (air resistance) must be 1440.6 N aswell. FWeight   =1440.6 N pulling him down as we calculated earlier, so theupward force (air resistance) must be 1440.6 N aswell.

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