Terminal velocity physics help

  • Thread starter thegreatone09
  • Start date
  • #1
Any help will be greatly appreciated.

Superwoman is hovering above the ground when a person free-falling goes by her at a terminal veloctiy of 140 km/h (39m/s). Unfortunately, the parachute doesn't open. Fortunately, Superwomanis around. If it takes her 1.9s to realize the person is in distress, what must her acceleration be if she is to catch the parachutist just before she hits the ground (so exactly at) 1000m below?
 

Answers and Replies

  • #2
FrostScYthe
80
0
0M ' *- Superwoman
| '
| * Person falling
|
|
|
-1000M_______________________________________________ GROUND

-1000 = -39m/sT (Terminal velocity is constant) T is time
T = 25.641 (Time it takes for PERSON, to hit ground)

t = 25.641 - 1.9 s (Time that Superwoman actually has)
= 23.741 s

Remember she does take her time realizing that PERSON is falling.

Then you use x(t) = Xi + Vi(t) + 0.5A(t)^2

x(t) = -1000m. We need to get to -1000 meters.
Xi = 0m. Superwoman 1000m above the -1000m therefore she's at an initial height of 0m. (just look at the picture, if it's confusing.)
Vi = 0m/s Superwoman isn't flying anywhere, that is, in the vertical direction.
t = 23.741 We know that this is the time we have to catch PERSON.
A = ? isn't that what we are finding

-1000 = 0 + 0(23.741s) + 0.5A(23.741s)^2.
-1000 = 0 + 0 + 281.871*A
-1000 = 281.871A
A = -3.548 m/s^2 or m/s/s Whichever you like

Well I could be wrong, so have someone else check it in case

os_dnw_sm0015_6.jpg
Aghhh I hate the avatars I want this pic for my avatar, except I need to get it to 50x50 :P
 
Last edited:
  • #3
Thanks!
 

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