The velocity of a skydiver falling to the ground

[Jamin2112]

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-kv²

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-kv²

==> dv/dt=g-(k/m)v²
==> dv/dt =(-k/m)(-gm/k + v²)
==> dv/(-gm/k + v²)=(-k/m)dt

Problem #3: Now it looks like I could use dx/(a²+x²)=(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!

 

[lanedance]

prob #1, the negative in the equation takes care of that & makes k>0

prob #2, if you don't include the drag the m will cancel out, however drag force is generally proportional to surface area whilst acceleration is proportinal to mass, so you can no longer cancel the m.

To see that consider the terminal velocity of a ball bearing & feather with the same mass. The surface area of the feather is much greater, so the drag force is greater and the feather will have a much lower terminal velocity.

 

[lanedance]

adding to #1 the best way to check is to draw a Free body diagram with all the forces

 

[lanedance]

for #3 how about trying the intergal
$$\int \frac{v.dv}{1-v^2}$$
and using the subtitution
$$u = 1-v^2$$
then
$$du = -2v.dv$$
and the integral becomes
$$\int \frac{-du}{2u}$$

I haven't followed this through but think it looks promising, though you'll have to add constants and stuff