Question

A skydiver weighing 800 N opens his parachute at the appropriate height from the ground. The air exerts a significant upward drag force, which is initially greater than the weight of the diver. This helps to slow the skydiver down. What is the diver's acceleration (both direction and magnitude in m/s²) if the magnitude of the drag force is 1165 N?

Answer 1

Answer:

4.4758 m/s² upwards

Explanation:

a = Acceleration

g = Acceleration due to gravity = 9.81 m/s²

Downward direction is taken as negative

$F_{net}=1165-800=365\ N$

Net force on the skydiver is 365 N

$m=\frac{F}{g}\\\Rightarrow m=\frac{-800}{-9.81}\\\Rightarrow m=81.549\ kg$

The mass of the person is 81.549 kg

$a=\frac{F_{net}}{m}\\\Rightarrow a=\frac{365}{81.549}\\\Rightarrow a=4.4758\ m/s^2$

Acceleration of the skydiver is 4.4758 m/s² upwards