# Distance as a function of time for two falling stones

• MatinSAR
In summary, the distance between two stones in free fall changes linearly with time as the stones have the same acceleration.

#### MatinSAR

Homework Statement
Suppose you drop a stone from a very tall tower. After the first stone goes down 4 meters, you drop the second stone from the same height. Does the distance between two stones decrease or increase with time?
Relevant Equations
Constant acceleration equations for free fall.
I am aware that this question is very simple and basic.
Using ##y(t)=y_0+v_{0,y}t-\frac {1}{2}gt^2## we can find distance as a function of time:
##|y_1-y_2|=|y_0+v_{0,y}t|=-y_0- v_{0,y}t##
I assumed the downward direction to be negative. So as I wrote ##D(t)=-y_0- v_{0,y}t##. It tells that the distance increases linearly with time. But I wanted to describe it to a friend, withhout math.
Can I say that the distance increases because 1st stone had nonzero initial velocity and nonzero initial position? Is this answer enough to get the question's point?

Last edited:
They do not start at the same time. Your method is OK. But your equations are incoirrect.

MatinSAR
It's probably sufficient to say that the 1st stone always has a velocity greater than the 2nd stone at any instant in time, so will always be pulling away. BTW, can you say what happens to the time interval between the 2 stones passing any particular point on the tower? That might also help you in your explanation to your friend...

MatinSAR
Or plot the velocity (or position) of each stone vs time on the same graph........each is the same curve offset by 4 seconds horizontally. You can put in air resistance roughly!

MatinSAR
hutchphd said:
They do not start at the same time.
Yes, I know.
hutchphd said:
But your equations are incoirrect.
I assumed the origin of time when the second stone is dropped and the first stone is at ##y=y_0## with an intial velocity. Why is this wrong?
berkeman said:
It's probably sufficient to say that the 1st stone always has a velocity greater than the 2nd stone at any instant in time, so will always be pulling away. BTW, can you say what happens to the time interval between the 2 stones passing any particular point on the tower? That might also help you in your explanation to your friend...
Thank you for your help.

MatinSAR said:
I assumed the origin of time when the second stone is dropped and the first stone is at ##y=y_0## with an intial velocity. Why is this wrong?
When you are trying to write for others to read, you should define your variables first and then write your equations after.

[Failing to clearly define variables is a pet peeve for me. I learned computer science in the 70's and we had that stuff drilled in to our heads repeatedly]

MatinSAR
jbriggs444 said:
[Failing to clearly define variables is a pet peeve for me. I learned computer science in the 70's and we had that stuff drilled in to our heads repeatedly]

@MatinSAR -- That's why he wears that helmet now...

jbriggs444, SammyS and MatinSAR
jbriggs444 said:
When you are trying to write for others to read, you should define your variables first and then write your equations after.
I apologize. I thought that the question is simple and doesn't need more explanation.
We drop 2nd stone at ##t=0##. At this time 1st stone is at ##y=y_0## and has velocity of ##v_{0,y}##.
So we can write their position-time equations :
##y_1=y_0+v_{0,y}t-\frac {1}{2}gt^2##
##y_2=-\frac {1}{2}gt^2##
Now we should find ##|y_1-y_2|## as I mentioned in post #1:
"Using ##y(t)=y_0+v_{0,y}t-\frac {1}{2}gt^2## we can find distance as a function of time:
##|y_1-y_2|=|y_0+v_{0,y}t|=-y_0- v_{0,y}t##
I assumed the downward direction to be negative. So as I wrote ##D(t)=-y_0- v_{0,y}t##. It tells that the distance increases linearly with time."
berkeman said:
@MatinSAR -- That's why he wears that helmet now...
Interesting.

MatinSAR said:
Is this answer enough to get the question's point?
The question's conceptual point is that, once the two stones are in free fall, their velocities change in exactly the same way because they have the same acceleration. This means that their relative velocity at the time of release of the second stone will not change. It follows that their separation will change linearly with time as you have found. Note that this result is independent of whether the stones are given initial velocities directed up, down or sideways relative to ##g##.

MatinSAR and nasu
kuruman said:
The question's conceptual point is that, once the two stones are in free fall, their velocities change in exactly the same way because they have the same acceleration. This means that their relative velocity at the time of release of the second stone will not change. It follows that their separation will change linearly with time as you have found. Note that this result is independent of whether the stones are given initial velocities directed up, down or sideways relative to ##g##.
Thank you for your time.

berkeman and kuruman