Finding final velocity from a fall

[lzh]

I'm having trouble with this problem:

"A 79kg person escapes from a burning building by jumping from a window 25m above a net.
The acceleration of gravity is 9.81 m/s^2
Assuming that air resistance is a constant 88 N force on the person during the fall, find the person's velocity just before hitting the net"

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?

 

[bob1182006]

not 100% sure on this but i think:
the 79g-88N=79a is wrong the units don't 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

 

[asdfmosin]

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.

 

[lzh]

not 100% sure on this but i think:
the 79g-88N=79a is wrong the units don't 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

79g is his weight, which is in N, which cancels out

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.

but it isn't right...

 

[asdfmosin]

not 100% sure on this but i think:
the 79g-88N=79a is wrong the units don't 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'm not sure wat u mean. But the unites do cancel out propely:
(790N-88N)/79 = a.

 

[bob1182006]

well N is kg*m/s^2 79kg is just the weight you're missing the acceleration (gravity)

 

[asdfmosin]

Typo:
*79kg.

 

[lzh]

yeah g as in acceleration due to gravity

 

[bob1182006]

sry was distracted >.< i get the same answer as you.
are you providing the required number of significant digits?

 

[lzh]

rounding and sig figs doesn;t matter

 

[asdfmosin]

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.

 

[Doc Al]

I'm having trouble with this problem:

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

Your method and answer look correct to me.