(adsbygoogle = window.adsbygoogle || []).push({}); 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.

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# Homework Help: Very simple Accelerated motions

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