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Very simple Accelerated motions

  • Thread starter dman5505
  • Start date
  • #1
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Homework Statement



At t=0 an object is released from rest at the top of a tall building. At the time t_0 a second object is dropped from the same point. Ignoring air resistance, show that the time at which the objects have a vertical separation L is given by: t=L/(gt_0 )+t_0/2.

Homework Equations



L=v_0*t + (1/2)at^2

The Attempt at a Solution



With much fidgeting with the above equation, I decided that the distance must be equal to L=(1/2)gt^2-(1/2)g(t_0+t)^2. However when I simplified the equation, L=-L/(gt_0)-t_0/2 which is the equation I supposed to get multiplied by -1. Since the object is going down, maybe L is supposed to be -L?

More importantly, I was confused on how to represent the change in time for the second object and the first object.
 

Answers and Replies

  • #2
Delphi51
Homework Helper
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Welcome to PF, dman.
I don't think your expression L=v_0*t + (1/2)at^2 can be correct because it doesn't include a variable t_0, which clearly affects L.
Why not start with an expression for the position of the first object as a function of time?
Then an expression for the position of the second. It will probably have a t - t_0 in it.
Finally, subtract the two to get L.
 
  • #3
Hi, I have the same problem, did you manage to solve it?
 
  • #4
Delphi51
Homework Helper
3,407
10
Yes, it works out well. Show your work if you would like help with it.
 

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