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## Homework Statement

An 80 kg skydiver jumps from an altitude of 1000m and opens his chute at 200m

(a) The total retarding force on the diver is a constant 50.0N without his chute open, it is constant 3600N with his chute open. What will be the skydiver's speed when he reaches the ground?

## Homework Equations

(sum)F=ma

P.E.=mgy

K.E.=(1/2)mv^2

E(total)=P.E. + K.E.

## The Attempt at a Solution

Well, frankly I've gone through several methods and it's all become a big mess. It seems like a straight forward problem but I think I've gotten tunnel vision from it and am missing something obvious. I have a hunch I should be using PE and KE rather than the following methods, but this was all I could come up with.

I got a velocity (at 200m)=122 m/s doing this:

(sum)F=ma

750N=80a

a=9.375 m/s/s

--> 200=1000+V(initial)t+.5at^2

=200=1000+.5*9.375*t^2

t=13s (appox.)

V(@200)=9.375*13=122 m/s

(sum)F=ma

-3500N=80a

a=-35 m/s/s

0=200+122t=.5*(-35)*t^2

t= 8.34 (or -1.3... obviously time can't be negative)

V(final)=V(@200)-35*8.34

V(final)=-170 m/s.... this does not seem right at all to me.

Any suggestions? I had tried using PE and KE to solve for it, but got lost along the way.