# Distance between two objects dropped at different times

1. Sep 27, 2010

### 3005

1. The problem statement, all variables and given/known data
An object is dropped from the roof. A second object is dropped 1.50 s later. How far apart are the objects when the second one has reached a speed of 12.0 m/s.

GIVEN
Obj 1 - vi=0 m/s , a=-9.8 m/s/s
Obj 2 - vi=0 m/s, vf=12.0 m/s, a=-9.8 m/s/s

2. Relevant equations
d=vf*t+1/2*a*t2

3. The attempt at a solution
d=12.0*1.50+1/2*-9.8*2.25
d=18-11.025
d=6.975

I'm not sure exactly how to do this when there's two objects :S.
The answer is 29.0 m, but I want to understand how to get it.

2. Sep 27, 2010

### ehild

Write the equations for the displacements of both objects. The usual equation is the distance travelled during the time since the object has been dropped: If you start a stopwatch when the first object is dropped, and the second one when the other one is dropped, the second watch will 1.5 s behind the first one.

ehild

3. Sep 27, 2010

### 3005

So would it be:
d=vf*t-1/2*a*tt
d=12*1.5-1/2*-9.8*2.25
d=18-(-11.025)
d=29

Oh I think I just messed up the sign on the equation.
I think its right now.(?)

4. Sep 27, 2010

### jhae2.718

You need to treat each object separately; i.e. you need an equation for the displacement of each object. (Technically these should be functions of time.)

There is a time difference between the two objects; if you know the difference, how can you relate the times of the two objects? This relation will allow you to solve the first two equations.

5. Sep 28, 2010

### ehild

I do not know how you arrived to the equation, but it is correct.

When the second object is dropped the first one has the velocity of 1.5 g and travelled 1/2 g 1.5^2 meters. After that, the distance travelled by it is
s1=1/2 g 2.25+1.5gt + g/2 t^2, and that of the second object is s2=1/2gt. The distance between the objects is

D=s1-s2 =1/2 g 2.25+1.5 gt.

The second object reaches 12 m/s velocity in t=12/g time. Plugging back for t you get D=29.0 m.

ehild