find average force?

Question

50kg person initially at rest slips off ledge 5.0m above ground person falls to ground and stops by sinking .10m in mud hole. determine average force of mud in stopping the person.

Answer 1

Potential energy equal to work done on mud
Pe=W
then
mgh=Fs and finally
F=mgh/s

F=50 x 9.8 x 5.0/.10=24,500N

Answer 2

Lets find the velocity just before hitting the mud. The initial velocity is 0, the acceleration is 9.8 and the distance is 5 so we have:
v^2 = v0^2 + 2ad
v^2=0+2(9.8)(5)
v^2=98
v=9.8995 m/s
The person then goes from that speed to zero in a distance of 0.10 meters. thats enough info to find the person's average acceleration:
v^2 = v0^2 + 2ad
=> a=(v^2-v0^2)/2d
a=(0-9.8995^2)/(2*.1)
a=-490m/s2
Now F=ma so the average force is:
F=50*(-490)
F=-24500 N
Or 24500 Newtons directed upward. Notice that v^2 and v0^2 were different each time I used them. Thats because I broke the problem into two parts.