Uniformly accelerated motion

  • Thread starter sunbunny
  • Start date
  • #1
55
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Hey I was wondering if anyone can point me in the right direction of where to begin in this problem:

An apartment-dweller is looking out of the window if her living room. The glass area of the window measures 2.00m from the top and bottom. A freely-falling computer monitor suddenly drops past her window, taking exactly 0.200s to pass the window. How far above the top of the window was the monitor relseased (from rest) by the frusterated computer user om the higher floor?

Thanks a lot :)
 

Answers and Replies

  • #2
163
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I would first calculate the velocity of the computer as it reached the top of the window frame. With that velocity and the fact the computer started at 0 velocity you can calculate the distance that it fell

[tex]y=y_0+v_0t+\frac {1} {2} at^2[/tex]
[tex]v^2 = v_0^2 + 2a(y-y_0)[/tex]
 
Last edited:
  • #3
55
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thank you, i'll give that a try
 
  • #4
1,860
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You know how long it took to go past it her window, you know how fast it is accelerating, you know how tall her window is, so why not find how quickly it is moving when it first enters the window frame? Hope that points you in the right direction. The rest is trivial.
 

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