So for this problem i used F=ma:
79g-88=79a
a=8.6861m/s^2
this is the correct acceleration i believe and to find final velocity i used:
Vf^2=Vi^2+2ad where d is displacement
Vf^2=2(8.6861)(25)
vf=20.85m/s
I'm using utexas's webhw service, and the answer above was not correct, can someone tell me what i'm doing wrong?

not 100% sure on this but i think:
the 79g-88N=79a is wrong the units dont cancel out properly, you know the 88N upward on the person what's the force he excerts downard?
with those two figures you can use F=ma knowing F and m to determine the actual a, and plug that into the Vf^2=Vi^2+2ad equation

I solved the problem by calculating the amount of downward force one the person; mainly his weight minus the resistance. Then I used that F in the work formula; W=FD. D would be the 25m. I got the W in Jules so I let that W = KE, which is kinetic Energy. so, W = MV^2. and solved for the desired quantity.
Same ans. So I suppose you’re ans is right.

out of curiosity; what kind of deceleration is applied form falling on the net for a safe landing. Too much would cause the person to recoil up. too little would cause he/she to hit the ground.