Kinematics Free falling acceleration Physics question

[Brownk]

Homework Statement

Okay so the question is..A stone falls from rest from the top of a cliff. A second stone is thrown downward from the same height 3.99999 seconds later with an initial speed of 78.3998 m/s. They hit the ground at the same time.

How long does it take the first stone to hit the ground? The acceleration of gravity is 9.8m/s^2.

and

How high is the cliff?

Homework Equations

vf= vi + a*t
vf^2 = vi^2 +2*a*d
d = vi + (1/2)*a*(t^2)
where vf= final velocity, vi= initial velocity, a = acceleration, d = distance, t = time

The Attempt at a Solution

Okay so this is my first time posting on this forum and forgive me if I do something wrong but here is what I tried to do.
For the first rock,
vf= ?, vi= 0m/s, t=?, a= 9.8m/s^2, and d=?
And the second,
vf=?, vi=78.3998m/s, t=?, a=9.8m/s^2, and d=?
Am I missing something or am I missing 3 variables for each of the rocks?? I don't really know where to go without another variable for one of the two...Thanks for any help

 

[JaredJames]

I'm not sure how to approach this one. Is that all the question gives?

If I had to guess, I'd assume that for stone 2, initial velocity = final velocity.

But I don't think that is the correct way to approach this.

By doing that, you could calculate the height of the cliff and then calculate the time it takes for the first stone to hit the ground using t_stone1 = t_{stonge 2} + 3.9999

 

[fizzynoob]

are you allowed to use conservation of energy? or is this purely kinematics?

 

[JaredJames]

It sounds like kinematics, given you have no masses to work with.

 

[Brownk]

I'm not sure how to approach this one. Is that all the question gives?

If I had to guess, I'd assume that for stone 2, initial velocity = final velocity.

But I don't think that is the correct way to approach this.

By doing that, you could calculate the height of the cliff and then calculate the time it takes for the first stone to hit the ground using t_stone1 = t_{stone 2} + 3.9999

Yeah, that's the question word for word.
Okay I'll try approaching it that way and edit in a few. Thanks, and yeah it's just kinematics.

 

[Saterial]

Try this.

1st ball:
Given:
v1 = 0
a= 9.8m/s^2
Δd = v1Δt + 1/2aΔt^2
Δd = 4.9t^2

2nd ball:
Given:
v1=78.4m/s
a=9.8m/s

Δd = v1Δt + 1/2aΔt^2
Δd = 78.3998t + 4.9t^2

Plug equations into each other and solve for t.

 

[JaredJames]

Try this.

1st ball:
Given:
v1 = 0
a= 9.8m/s^2
Δd = v1Δt + 1/2aΔt^2
Δd = 4.9t^2

2nd ball:
Given:
v1=78.4m/s
a=9.8m/s

Δd = v1Δt + 1/2aΔt^2
Δd = 78.3998t + 4.9t^2

Plug equations into each other and solve for t.

But the first balls time is t_ball2 + 3.9999. So the t values between the two equations aren't the same. So I don't think you can do it like that without include the plus 3.9999.

So first equation should be Δd = 4.9(t_ball2+3.9999)^2

 

[Brownk]

But the first balls time is t_ball2 + 3.9999. So the t values between the two equations aren't the same. So I don't think you can do it like that without include the plus 3.9999.

So first equation should be Δd = 4.9(t_ball2+3.9999)^2

If I set those 2 equations equal then don't I have two variables? Sorry I'm still confused this problem makes no sense

 

[jhae2.718]

You know the distance each ball falls is the same, so you can set the two equations equal to each other and solve for time.

The equation would be:
[tex]78.3998t_{ball2}+4.9t_{ball2}^2=4.9(t_{ball2}+3.99999)^2[/tex]

The times in each equation are different, but are related in that the second ball is released at 3.99999 s after the first ball is released. So, at that time, [tex]t_{ball1} = 3.99999\,\mathrm{s}[/tex] and [tex]t_{ball2}=0[/tex]. If we put everything in terms of [tex]t_{ball2}[/tex], then [tex]t_{ball1} = 3.99999\,\mathrm{s} + t_{ball2}[/tex], which can be verified by setting [tex]t_{ball2}=0[/tex].

Thus, we have two equations with two unknowns and we can substitute in and solve.