Kinematics - When do they meet?

[Sean1218]

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?

 

[jrab]

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

 

[Sean1218]

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

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

 

[jrab]

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

 

[tiny-tim]

Hi Sean!

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)

 

[Sean1218]

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.

 

[tiny-tim]

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

 

[Sean1218]

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.

 

[tiny-tim]

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?

 

[Sean1218]

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

 

[tiny-tim]

… 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?

 

[Sean1218]

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

 

[tiny-tim]

So the relative acceleration is … ?

 

[Sean1218]

relative to each other, 2g I guess

 

[tiny-tim]

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: …

 

[Sean1218]

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

 

[Sean1218]

Anyone?

 

[The legend]

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.

 

[The legend]

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.

 

[Sean1218]

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.

 

[The legend]

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?

 

[Sean1218]

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?

 

[The legend]

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