Originally posted by: IGBT
..you think he had any 2nd thoughts on the way down??..sees the ground getting larger and larger..I wonder if he achieved terminal velocity??...
v=0
a=9.8 m/s
y= 393.192 meters[Assume "floor to floor" heights are approx. 15ft = 4.572meters -->> 4.572m/flx86fl]
v(f)^2 = v(i)^2 + 2a[y(f)-y(i)]
v(f)^2 = 2(9.8)(393.192)
1 mile = 1.609344kilometers
v(f) = 87.79 m/s = .08779km/s = 316.044km/h= 196mph
The "victim," lets call him Bob, will have reached terminal velocity, due to air resitance and such, before the moment of impact, so he will ahve been traveling at approx. 135mph for some time.
EDIT:
135mph = ~85km/h = 85000m/h = 8500 = 23.61m/s
v(f) = v(i) + at
23.61= (9.8)t
2.41=t
EDITx2:
y=.5(9.8)(2.41)^2
y= 28.5 meters
393.192m -28.5m = 364.692 meters
Finally
y=364.692m
t=?
v(i)= 135mph ~ 60.3504 m/s
v(f)= 135mph ~ 60.3504 m/s
a= 0
y=V(i)t
364.692= 60.3504t
t= 6 seconds
These caluclations lead me to believe that poor Bob was traveling at 135mph for approximately 6 seconds during which he covered 364.692 meters, or approximately 79.8 stories.
yeah, great chocie there Bob.