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.
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.