# Kinematics - When do they meet?

• Sean1218
In summary, the problem involves two stones, one dropped off a cliff and the other thrown upwards from the base of the cliff with an initial velocity. The goal is to find at what time the two stones will meet. Using the uniform acceleration equations, the attempt to solve the problem involved writing equations for the displacement of each stone and substituting the displacement values into the equations. However, the equations could not be combined to find a solution without using specific values. Another approach suggested was to consider the relative acceleration between the two stones, which is zero, and use the equations of motion to find the relative distance in terms of the relative speed and time. Finally, it was mentioned that the direction chosen for positive and negative values of acceleration depends on the

## Homework Statement

A stone is dropped off a cliff of height h. At the same time, a second stone is throne straight upward from the base of a cliff with an initial velocity v. Assuming that the second rock is thrown hard enough at what time t will the two stones meet.

## Homework Equations

Uniform acceleration equations

## The Attempt at a Solution

I tried a few different things, and nothing worked. I don't have anything written down right now though.

I just wrote Δd = v1Δt + 1/2gΔt2, for both stones, substituting Δd for x in one of the equations, and h - x in the other.

So, I just had x = 1/2gΔt2 and h - x = v1Δt + 1/2gΔt2

I tried combining them, but I got to a point where I didn't think I could do anything else with it, and it just felt wrong.

Am I thinking about it wrong?

Last edited:
if you told me the numbers i could solve it for you :p

jrab said:
if you told me the numbers i could solve it for you :p

There are no numbers, could you explain how to solve it?

let me get some paper i gtg eat so ill tell you when i find out

Hi Sean!
Sean1218 said:
I just wrote Δd = v1Δt + gΔt2, for both stones, substituting Δd for x in one of the equations, and h - x in the other.

So, I just had x = gΔt2 and h - x = v1Δt + gΔt2

Yes, you have the correct basic idea, but

i] it's 1/2 g∆t2, isn't it?

ii] shouldn't one of those gs be minus?

(btw, you'll notice something unexpectedly disappears … why do you think that is?)

(and jrab, on this forum you mustn't give answers, you must only help)

ah haha, forgot the 1/2. That doesn't really change my result much though.

I still end up with 0 = gΔt2 + v1Δt - h, and I don't see how I can use the quadratic formula without numbers. Nothing disappeared or canceled out from what I can tell, since moving the - 1/2gΔt2 to the other side just makes it positive. And why would one of the g's be negative? I'm using v1 for both, so the equation is just + g.

But isn't your v1 up, and your g down?

Not really sure what you mean. v1 is going up, and g is down, but I can't just include negatives can I? g = -9.8, if I made g negative, at the end I'd just get a positive acceleration, but g isn't positive. g is the same for both equations, v1 = 0 for the first one, and it's positive for the second stone, but I don't see how I can declare that it'll always be positive without plugging in numbers, if you know what I mean.

Oh i see, you're using g = minus 9.8 (we don't usually do that) …

ok, but then if your x is always positive, doesn't the first g need to have a minus?

Anyway, here's another approach … what is the relative acceleration of the two stones?

The first equation is the same as the second equation though. v1 is 0 so the first term is 0, but g is still g (-9.8 if you plug it in) as far as I know.

and I wouldn't know where to start with relative acceleration

Sean1218 said:
… I wouldn't know where to start with relative acceleration

Well, what is the acceleration of the first stone, and what is the acceleration of the second stone?

it's the same for both, accel due to gravity.

So the relative acceleration is … ?

relative to each other, 2g I guess

Nooo! it's zero, isn't it?

And if the relative acceleration is zero, then the relationship between relative distance relative time and relative speed is … ?

I'm off to bed now :zzz: …

Can anyone else help with the original method I was using (after adding in the 1/2 that is)?

Anyone?

an easier way would be relative motion as tiny-tim suggested.
And accordingly you can plug in the equations of motion considering the relative acceleration as 0.
rel(d) = rel(speed) * time.

easy as that.

Sean1218 said:
So, I just had x = 1/2gΔt2 and h - x = v1Δt + 1/2gΔt2

Using g=+9.8m/s^2
The second one in there would have a= -g.

Now add them up, eliminating few stuff you can get your answer, your way.

The legend said:
Using g=+9.8m/s^2
The second one in there would have a= -g.

Now add them up, eliminating few stuff you can get your answer, your way.

Alright, thanks! So, why would one of them be -g instead of g? Accel due to gravity is always negative, so I don't really see why.

we don't usually use g to be negative, the direction decides what g is. Only the thing is g acts downwards, and if you choose up to be positive g becomes negative and if you choose down to be positive g becomes positive.

So get it?

Not really, I was working through it assuming up is positive, and down is negative, for both equations. Was there something I did that indicated otherwise?

then in your first equation x would be negative...