Answer:
[tex]7.35 m/s^2[/tex]
Explanation:
When the skydiver is falling towards the ground, there are two forces acting on him:
- The force of gravity, downward, of magnitude [tex]mg[/tex], where
[tex]m=101 kg[/tex] is the mass of the skydiver
[tex]g=9.8 m/s^2[/tex] is the acceleration due to gravity
- The air resistance, upward, of magnitude [tex]F_r[/tex]
The net force on the skydiver must be equal to the product between mass and acceleration, so we can write:
[tex]mg-F_r = ma[/tex]
where a is the acceleration.
Here we are told to find the acceleration of the skydiver when the magnitude of the air resistance is 1/4 of his weight, which means
[tex]F_r=\frac{1}{4}mg[/tex]
Substituting into the equation and solving for a, we find the acceleration:
[tex]mg-\frac{1}{4}mg=ma\\\frac{3}{4}mg=ma\\a=\frac{3}{4}g=\frac{3}{4}(9.8)=7.35 m/s^2[/tex]
