Very simple Accelerated motions

[dman5505]

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.

 

[Delphi51]

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.

 

[caretomate]

Hi, I have the same problem, did you manage to solve it?

 

[Delphi51]

Yes, it works out well. Show your work if you would like help with it.

 

[asteriatic]

Your attempt at a solution is on the right track, but there are a few errors. Firstly, the equation you used, L=v_0*t + (1/2)at^2, is not applicable in this scenario because the objects are not thrown with an initial velocity. Instead, you should use the equation L=(1/2)gt^2, which represents the distance an object falls due to gravity in a given time t.

Next, you should consider the time at which the second object is released, t_0, as the starting point for your calculations. This means that t=0 for the second object, and t=t_0 for the first object.

Using the equation L=(1/2)gt^2, we can set up two equations, one for each object, as follows:

For the second object (released at t=0): L = (1/2)gt^2

For the first object (released at t=t_0): L = (1/2)g(t-t_0)^2

Now, we can set these two equations equal to each other to find the time at which the objects have a vertical separation L:

(1/2)gt^2 = (1/2)g(t-t_0)^2

Solving for t, we get t = L/(gt_0) + t_0, which is the same as the equation given in the homework statement.

To address your confusion about the change in time for the two objects, remember that the second object is released at t=0, so its time of motion is simply t. The first object, however, is released at t=t_0, so its time of motion is t-t_0.

I hope this helps clarify the problem and your approach to solving it. Keep up the good work!