- #1

- 986

- 9

**The velocity of a skydiver falling to the ground ....**

## Homework Statement

SORRY FOR SO MANY QUESTIONS! I'VE BEEN WORKING THROUGH EACH OF THESE UNTIL I'M COMPLETELY STUCK, SO DON'T THINK I'M JUST TRYING TO SUCK ANSWERS OUT OF YOU FOLKS!

The velocity v(t) of a skydiver falling to the ground is governed by

m(dv/dt)=mg-kv

^{2}

where m is the skydiver's mass, g is the acceleration due to gravity, k > 0 is the drag coefficient, and v(t)≥0

## Homework Equations

Going to go separable on this one

## The Attempt at a Solution

**Problem #1:**Something seems to be wrong. Since v(t)≥0, the problem is orienting positive velocity in the downwards direction. Wouldn't the drag coefficient, k, therefore be less than zero? It's pushing against the skydiver. Seems like common sense to me.

**Problem #2:**I've always been told that mass has nothing to do with the speed at which an object falls. Why is it in an equation for change in velocity, dv/dt?

m(dv/dt)=mg-kv

^{2}

==> dv/dt=g-(k/m)v

^{2}

==> dv/dt =(-k/m)(-gm/k + v

^{2})

==> dv/(-gm/k + v

^{2})=(-k/m)dt

**Problem #3:**Now it looks like I could use dx/(a

^{2}+x

^{2})=(1/a)tan

^{-1}(x/a) .... but then I'd have a= +/- √(-gm/k), which would be an imaginary number since g,m, and k are all positive!!!!!!