Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Very simple Accelerated motions

  1. Nov 19, 2011 #1
    1. The problem statement, all variables and given/known data

    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.

    2. Relevant equations

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

    3. 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.
  2. jcsd
  3. Nov 19, 2011 #2


    User Avatar
    Homework Helper

    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.
  4. Dec 19, 2011 #3
    Hi, I have the same problem, did you manage to solve it?
  5. Dec 21, 2011 #4


    User Avatar
    Homework Helper

    Yes, it works out well. Show your work if you would like help with it.
Share this great discussion with others via Reddit, Google+, Twitter, or Facebook