PHYSICS!!!! vector help plzz this is soo hard i really don't get it. pHYSICS?

Question

009 (part 1 of 1) 10 points
A person of mass 77.9 kg escapes from a burn-
ing building by jumping from a window situ-
ated 23.7 m above a catching net.
The acceleration of gravity is 9:8 m/s2 :
If air resistance exerts a force of 76.7 N on
him as he falls, determine his speed just before
he hits the net. Answer in units of m/
s.

Answer 1

You may be confused because it is a dumb problem. Air resistance isn't even approximately constant in this situation.
Still, they want an answer, so the approach that make the most sense is:

I like using work-energy on this one. Gravity does work of W1 = F*x = (mg)*(23.7m) on the person as he falls. This is positive because the force is in the same direction as motion.
W1 = (77.9kg)*(9.8m/s^2)*(23.7m) = 18093 joule
Air resistance does work of W2 = -F*x = (-76.7nt)*(23.7m) as he falls. This is negative because the force is opposite the motion.
W2 = (-76.7nt)*(23.7m) = -1815 joule
The net work W = W1+W2 all goes to kinetic energy, so:
KE = (m*v^2)/2 = W
where v is the speed of the person at the end of the fall. You know m and W...now solve for v and you're done.

Answer 2

The jumper's horizontal speed is irrelevant, so we can do this problem completely in the vertical direction. The jumper has zero speed downward the instant he leaves the window, but as gravity speeds him up (downward) he goes faster and faster downward. However, air resistance slows him along the way. We're given that the two forces are constant the whole time, so at any time, the net force is:
Fnet=Fgrav+Far
Fnet=mg(down)+76.7(up)
Fnet=(77.9)(9.8)(down)+76.7(up)
Fnet=736.42(down)+76.7(up)
Fnet=686.72(down)
Of course, down wins. So all along the way he feels an acceleration downward. F=ma so
a=F/m
a=(686.72)/(77.9)
a=8.815 down
Okay, now whats his final speed?
Vf^2 = V0^2 +2ad
Vf^2=0+2ad
Vf^2=2(9.8)(23.7)
Vf^2=464.52
Vf=21.55
Or you can say Vf=21.6m/s. Are you wondering why V0^2=0? Thats his initial vertical velocity, which I said was zero earlier.

Answer 3

persist with those steps: a million. comic strip a image. 2. set up a coordinate gadget. Which way is x useful? Which way is y useful? 3. comic strip all forces appearing on the elephant. be useful to get your angles ideal. 4. destroy your rigidity vectors into their x and y-aspects. that's achieved via trigonometry. If this step is complicated reference the vectors part of your textbook. 5. Sum up each and every of the forces interior the x-course making useful to be conscious useful and detrimental indicators looking on which course the forces are pointing. Set the sum of those forces equivalent to m*a_x and sparkling up. Do an identical for the y course. in case you persist with those steps faithfully, all rigidity issues exchange right into a chew of cake.