Asked by Mrs Nobody
                An 80.0 kg skydiver jumps out of a balloon at an altitude of 1.00 x 10 m and opens the parachute at an altitude
of 200.0 m. omitior store it in any other
a) Assuming that the total retarding force on the diver is constant at 50 N with the parachute closed and constant
at 3.60 x 10^Nwith the parachute open, what is the speed of the diver when he lands on the ground?
b) think ? .
c) At what height should the parachute be opened so that the final speed of the skydiver when he hits the ground
is 5.00 m/s?
d) How realistic is the assumption that the total retarding force is constant?
Explain.
            
        of 200.0 m. omitior store it in any other
a) Assuming that the total retarding force on the diver is constant at 50 N with the parachute closed and constant
at 3.60 x 10^Nwith the parachute open, what is the speed of the diver when he lands on the ground?
b) think ? .
c) At what height should the parachute be opened so that the final speed of the skydiver when he hits the ground
is 5.00 m/s?
d) How realistic is the assumption that the total retarding force is constant?
Explain.
Answers
                    Answered by
            Damon
            
    1000 meters maybe ? 3600 N maybe???
If so
before chute opens:
Vi = initial speed = 0
g = -9.81 m/s^2
Force up on man = 50 N
acceleration down = 9.81-50/80 = 9.2 m/s^2
v = 0 + 9.2 t
h = 1000 - (9.2/2) t^2
200 = 1000 - 4.6 t^2
t^2 = 174
t = 13.2 seconds at open chute at 200 meters, v = 9.2 t = 121 m/s
Now 200 meters more with chute open and initial speed =121 m/s
F = m g - 3600
a = g - 360/80 = 9.81 - 45 = - 35.2
so
v = 121 - 35.2 t downward
h = 0 = 200 - (1/2)(35.5) t^2
17.6 t^2 = 200
t = 3.37 seconds
v = 121 - 35.2*3.37 = 121 - 119 = 3 m/s
You can do the rest. Check my arithmetic!
for part b, drag is sort of proportional to v^2, not constant
    
If so
before chute opens:
Vi = initial speed = 0
g = -9.81 m/s^2
Force up on man = 50 N
acceleration down = 9.81-50/80 = 9.2 m/s^2
v = 0 + 9.2 t
h = 1000 - (9.2/2) t^2
200 = 1000 - 4.6 t^2
t^2 = 174
t = 13.2 seconds at open chute at 200 meters, v = 9.2 t = 121 m/s
Now 200 meters more with chute open and initial speed =121 m/s
F = m g - 3600
a = g - 360/80 = 9.81 - 45 = - 35.2
so
v = 121 - 35.2 t downward
h = 0 = 200 - (1/2)(35.5) t^2
17.6 t^2 = 200
t = 3.37 seconds
v = 121 - 35.2*3.37 = 121 - 119 = 3 m/s
You can do the rest. Check my arithmetic!
for part b, drag is sort of proportional to v^2, not constant
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