# Homework Help: 1-D kinematics w/ 2 objects

1. Jan 18, 2006

### scavok

I didn't have much trouble with the 1-D 1 object kinematics, but I'm having a lot of trouble with 2 objects. I'd just like some help setting the problem up:

Two stones are dropped from the edge of a 60m cliff, the second stone 1.6s after the first. How far below the top of the cliff is the second stone when the seperation between the two stones is 36m?
I thought the best way to set this up would be:

X = Displacement of first stone in terms of a
Y = Displacement of second stone in terms of a
X-Y=36m
(vt+.5at2)-(vt+.5at2)=36

(Pretend I have subscripts to indicate they aren't the same variables and don't cancel each other out right away.)

I don't know if that's right, and even if it is, I just can't think of how to factor in the 1.6s.

A little guidance on how to get started would be appreciated.

2. Jan 18, 2006

### frozen7

X = Displacement of first stone
Y = Displacement of second stone
X-Y=36m
(vt+.5at2)-(vt+.5at2)=36

the initial speed is zero for each stone, and the acceleration is same for both also since it is free fall, so the only variables is time.
I solve it by:

taking time taken by first stones = time taken by second stones + 1.6

I think the problem can be solved by using this assumption.

3. Jan 18, 2006

### scavok

I still can't get it. The correct answer should be 10.9m.

The initial speed is zero, and a=g so the equation should look like this:
(.5gt2)-(.5gt2)=36m

Where does the +1.6s come in?

(.5gt2)-(.5g(t+1.6s)2)=36m?

vv Very cool, thank you.

Last edited: Jan 18, 2006
4. Jan 18, 2006

### HallsofIvy

You need to be more precise! Let's say that the distance the first stone is below the cliff is x1 and the distance the second stone is below the cliff is x2. Let t1 be the time the first stone has been falling and t2 the time the second stone has been falling. (No, I will not "pretend you have subscripts- put them in!).

Another problem is that you have as formula (1/2)gt2+ vt but, as frozen7 said, the initial velocityis 0:
x1= (1/2)gt12 and
x2= (1/2)gt22. The distance between them is x11- x2= (1/2)gt12-(1/2)gt22.

Now, what is the relationship between t1 and t2? You appear to be replacing what I have called t2 with t1+ 1.8. Can't you see that that can't be right? The second stone doesn't start falling until 1.8 s after the first. The second stone is falling for less time than the first, not more! t2 must be less than t1. You have t2= t1- 1.8. You can replace t2 by t1- 1.8, solve for t1 and then find x2 or replace t1 by t2+ 1.8, solve for t2 and use that to find x2. The second way might be simpler since it is x2 you want to find.