Question

A falling skydiver has a mass of 101 kg. What is the magnitude of the skydiver's acceleration when the upward force of air resistance has a magnitude that is equal to one-fourth of his weight?

Answer 1

Answer:

$7.35 m/s^2$

Explanation:

When the skydiver is falling towards the ground, there are two forces acting on him:

- The force of gravity, downward, of magnitude $mg$, where

$m=101 kg$ is the mass of the skydiver

$g=9.8 m/s^2$ is the acceleration due to gravity

- The air resistance, upward, of magnitude $F_r$

The net force on the skydiver must be equal to the product between mass and acceleration, so we can write:

$mg-F_r = ma$

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

$F_r=\frac{1}{4}mg$

Substituting into the equation and solving for a, we find the acceleration:

$mg-\frac{1}{4}mg=ma\\\frac{3}{4}mg=ma\\a=\frac{3}{4}g=\frac{3}{4}(9.8)=7.35 m/s^2$