Acceleration and distance problem

  • Thread starter deenuh20
  • Start date
  • #1
deenuh20
50
0

Homework Statement


In an action movie, the villain is rescued from the ocean by grabbing onto the ladder hanging from a helicopter. He is so intent on gripping the ladder that he lets go of his briefcase of counterfeit money when he is 130m above the water.


Homework Equations



Vf=Vi+ax*delta t

(Vf)squared=(Vi)squared + 2(ax)*delta x


The Attempt at a Solution



I think I'm supposed to use the first equation to figure out Vi. I calculated Vf using acceleration of 9.8 *6s, which equals 58.8 m/s. But if i throw in Vf, a, and delta t, into the first equation, Vi equals 1. I dont know if this is right so far, but I'm stuck here so any help would be appreciated. Thanks.
 

Answers and Replies

  • #2
bob1182006
492
1
where do you get 6s?
all you seem to be given is:
y=130m above the water.

Is there more to that problem? since you don't say what the question is exactly.
 
  • #3
deenuh20
50
0
Crap, my bad. This is the question:

If the briefcase hits the water 6.0s later, what was the speed at which the helicopter was ascending?
 
  • #4
bob1182006
492
1
O ok, well is the 9.8 the acceleration of the brief case? because what other force was acting on the briefcase? if the 9.8 was gravity going down, was something pulling it "up"?
 
  • #5
deenuh20
50
0
i'm not really sure. if there is anything pulling it up, it would be the helicopter. that is all the information they provided us with.
 

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