Question

two falling sky drivers have total mass of 134 kg including parachute. the upward force of air resistance is equal to one- fourth of their weight. part A determine the acceleration of the sky drivers. express your answer to three significant figures and include the appropriate units. enter positive if the acceleration is upward and negative value if the acceleration is downward. part B after opening the parachute, the drivers descend leisurely to the groud at constant speed. what now is the force of the air resistance on the sky drivers and their parachute? express your answer to three significant figures and include the appropriate units. enter positive value f the force is upward and negative value if the force is downward.

Answer 1

A) the divers are falling together, so their combined weight is

W=mg

=134*9.8=1313.2N

its given that the upward force is one fourth of their weight

so R=w/4

=1313/4=328.35N

using the newtons second law we have the formula

F=ma=R-mg

a=R-mg/m

=328.35-(134*9.8)/134

=-7.35m/s²

so A) -7.35m/s² ( dont forget negative sign)

B) as the drivers are descending at constant speed, the net forces cancel out

mg-resistance=0

and force of air resistace will be =weight of the divers=1313.2N

B)1313.2N