Fussy Little Dynamics Problem -_-?

Question

Problem: A sky diver jumps from a helicopter and is falling straight down at 30 m/s when her parachute opens. From then on, her downward accelereation is approximately a = g - cv2, where g = 9.81 m/s2 and c is a constant. After an initial transient period, she descends at a nearly constant velocity of 5 m/s.
a) What is the value of c, and what are its SI units?
b) What maximum deceleration is the sky diver subjected to?
c) What is her downward velocity when she has fallen 2m from the point at which her parachute opens?
OK, I worked through part a by using the chain rule to get a = (dv/ds)(ds/dt) = v(dv/ds) = g - cv2. Then I separated variables, etc, integrated both sides…And eventually ended up with: s = [-ln(9.81 - cv2)] / 2c + d, Where d was the constant of integration. I took it to be so, which I assumed to be 0 (since the initial height wasn't otherwise specified). The only thing is, that landed me with 0 = [-ln(9.81 - 900c)] / 2c

That is, for all I can see, a hopelessly retarded sticky spot. -_-
The other two parts I’m going to have an equal amount of trouble with and haven't made significant attempts at them yet. Any help would be super because I’ve got no idea where I’m going wrong! And I’ll definitely be checking for a best answer (and chances are you’ll be like the ONLY one to answer anyway….)

I wish that would have all fit up there. -_-

Sorry I spelled acceleration wrong. >_<
Also, the font screwed up when I was going on about the constant d, which I assumed to be "so"... That part shouldn't say "so", it was supposed to be s, subscript 0. Sorry.

Ahg...another formatting problem...... v2 should actually be v^2

Answer 1

the sky diver just jump-at that time he 's having a-9.81m/s2
but as soon as he oven perachute the a=g-c*v2
i think a=g-c*2v;tell me what's it?
so now a is the function of v
now veloctiy become constant after some time
it mean the resultant acceleration(a) should become 0.
a)
0=g-cv*2
or c=g/v*2=9.81/(5*2)=9.81/10=.981
unit of g=L*T^-2
and unit of v=L*T^-1
and unit =g/v*2=(L*T^-2)/(L*T^-1)
=T^-1=1/sec
b)
maximum deceleration ---
theorticaly it is,when it has the sky driver has the hightest speed.it mean at the time of opening prerachute
it a=g-c*v2=9.81- (.981)*(30)*2
=9.81(1- 6)= - 9.81*5
= -49.05 m/s^2
c)dv/dt=g-c*v2
or (dv/ds)*(ds/dt)=g-c*v2
v(dv/ds)=g-c*v2
(v/(g-c*v2))dv=ds
now integrate it
i'll tell u later i have a class
tell me i m going right or wrong

Answer 2

a) g - 25c = 0
c = 9.80665/25 = 0.39227 m^-1
b) a = 9.80665(1 - 900/25)
a = 9.80665(1 - 36)
a = 9.80665(- 35) = -343.23 m/s^2
c) vdv/(g - cv^2) = ds
(1/2c)ln(g - cv^2) = s
2cs = ln(g - cv^2) - ln(g - 900c)
2cs = ln[(g - cv^2)/(g - 900c)]
(g - cv^2) / (g - 900c) = e^(2cs)
(g - cv^2) = (g - 900c)e^(2cs)
cv^2 = g - (g - 900c)e^(2cs)
v^2 = 25[g - (g - 900g/25)e^(2cs)]/g
v^2 = 25g[1 - (1 - 900/25)e^(2cs)]/g
v^2 = 25[1 + 35e^[(2)(0.39227)(2)]
v ≈ 13.003 m/s or about 13 m/s