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
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.
1 answer
1 answer