One ball is dropped at a height, and another ball is thrown up with the final velocity of the first ball. Where do they meet? I've found variations to this problem on stack exchange, but in most of them the initial velocity is varied.
In this version, you first drop the ball at the top of the building, and measure its final velocity right before it hits the ground. Then, you throw the second ball up from the ground at that final velocity (or rather the negative of it since it is now pointing up).
Where (i.e. upper half, the middle, or lower half) do these balls cross?
The correct answer is apparently the top half, and my professor gave an intuitive explanation. However, I have been trying to model and solve this using basic kinematic equations and have been unsuccessful. Specifically, I don't know if there is a way to express the final velocity of the first ball as a function of the height of the building (assume its given) and acceleration (9.81).
So far, I have that the displacement of the two balls can be given by:
$x_1(t) = h + \frac{1}{2}(-9.81)t^2 \quad$ ($h$ is the height)
and
$x_2(t) = v_ft - \frac{1}{2}(-9.81)t^2 \quad$ ($v_f$ is the final velocity)
Now if I could find an explicit formula for $v_f$ then I think I could probably an expression for the height where $x_1(t)$ and $x_2(t)$ meet, and then show that this is greater than $\frac{1}{2}h$?
Any suggestions on this approach and how to proceed with $v_f$?

EDIT
I think I did it actually. Using $-h = \frac{1}{2}(-9.81)t^2$ and $v_f = -9.81t$, I found $v_f = \sqrt{2\cdot 9.81 \cdot h}$, eventually finding that the height at which $x_1(t)$ and $x_2(t)$ meet is $\frac{3}{4}h$, which is above the half way point.
If this looks wrong, please let me know.
 A: The answer uses the notation introduced in the OP and presents the complete analysis.
The position kinematics of the first ball which falls down under gravity are given as $x_1(t) = h - \frac{1}{2}gt^2$ and that of the second which rises up due to the initial velocity as $x_2(t)=v_f\cdot t-\frac{1}{2}gt^2$, where $0<v_f$ is the speed of the first ball at the final instant $0<t_{total}$ of motion under gravity before hitting the ground. The velocity kinematics of the first ball are given as $v_1(t)=-gt$ and the first equation implies that $t_{total}=\sqrt{\frac{2h}{g}}$ so that we have $v_f=|v_1(t_{total})|=\sqrt{2gh}$.

Method 1: Dynamics
Therefore, the condition of the two balls being at the same height expressed mathematicaly $x_1(t)=x_2(t)$, implies that $h-\frac{1}{2}gt^2=\sqrt{2gh}\cdot t-\frac{1}{2}gt^2$ so that the time $0<\tilde{t}$ at which the condition is true is given by $\tilde{t}=\sqrt{\frac{h}{2g}}$. Thus, the height at which the two balls have the same vertical position is given as $h-\frac{1}{2}g\tilde{t}^2=h-\frac{h}{4}=\frac{3}{4}h$.

Method 2: Energy conservation
The energy conservation for both mechanical systems corresponding to the two balls implies that $\frac{1}{2}mv_f^2=mgh$. At the moment of crossing, applying the conservation law again, since the height of both balls above the ground is identical, so must their speed be. Therefore, the time of crossing $\tilde{t}=\frac{h}{2g}$ can be obtained from $v_f-g\tilde{t}=\sqrt{2gh}=g\tilde{t}$. The height corresponding to any ball at that time, say the first ball, $x_1(\tilde{1})=h-\frac{1}{2}g\tilde{t}^2=h-\frac{h}{4}=\frac{3}{4}h$.

Intuition
The ball falling down under gravity will spend most of the time in the initial part of it's motion picking up speed, which occurs near the top of the building, while the ball rising up under initial velocity will spend most of it's time in the terminal part of it's motion losing speed, which occurs near the top of the building, as well. It therefore becomes clear that height of crossing of the two balls should be near the top of the building.
A: Starting with the basic equations for the kinematics, we have object 1 falling from h with zero starting velocity, and object 2 going up from the ground:
$$ x_1(t) = h - \frac{gt^{2}}{2}  \tag{1.1} $$
$$ x_2(t) = v_f t + \frac{gt^{2}}{2} \tag{1.2} $$
Now, recall that for the falling body 1:
$$ v = gt \tag{1.3}$$
We have to compute the final velocity, taking the final time, that is the time when the object 1 hit the ground. Using 1.1 for finding $t_f$, we have:
$$ x_1(t_f) = 0 \Rightarrow  t_f = \sqrt{\frac{2h}{g}} \tag{1.4}$$
Using 1.2 and inserting the final time 1.4 :
$$ v_f = \sqrt{2gh} \tag{1.5} $$
And finally:
$$ x_2(t) = \sqrt{2gh}t + \frac{gt^2}{2} \tag{1.6}$$
Now we have to write 1.1 and 1.2 taking respect to time.
Finding the square root of 1.2 we have:
$$ t_2 = \frac{\sqrt{2gh} + \sqrt{2gh - 2gx_2}}{g} = \sqrt{\frac{2h}{g}} + \sqrt{\frac{2(h - x_2)}{g}} \tag{1.7}$$
And for 1.1:
$$ t_1 = \sqrt{\frac{2(h - x_1)}{g}} \tag{1.8}$$
Since the two objects meet at some point in space and time, this two times are equal, $t_1 = t_2 $, and calling the position where they meet $ x = x_1 = x_2 $ we have:
$$ \sqrt{\frac{2(h - x)}{g}} = \sqrt{\frac{2h}{g}} + \sqrt{\frac{2(h - x)}{g}} \tag{1.9} $$
Taking the square:
$$ \frac{2h}{g} - \frac{2x}{g} = \frac{2h}{g} + \frac{2(h - x)}{g} + \frac{4}{g}\sqrt{h(h - x)} \tag{1.10}$$
And then:
$$ \frac{-2h}{g} = \frac{4}{g}\sqrt{h(h - x)} \tag{1.11}$$
$$ \frac{12h^2}{g^2} = \frac{16hx}{g^2} $$
In conclusion:
$$ x = \frac{3}{4}h \tag{1.12}$$
Another method:
we can use more simply the conservation of energy and the equations for the velocity:
$$ v_1 = gt \tag{2.1} $$
$$ v_2 = v_f - gt \tag{2.2}$$
For finding $v_f$ we use the conservation of energy:
$$ \frac{v_f^2}{2} = gh \tag{2.3}$$
Now using conservation of energy for object 1 falling:
$$ gh = gx + \frac{v_1^2}{2} \tag{2.4}$$
For object 2 going up:
$$ \frac{v_f}{2} =  gx + \frac{v_2^2}{2} \tag{2.5} $$
Subtracting the latest two and replacing $v_f$, calling the common velocity v:
$$ v := |v_2| = |v_1| \tag{2.6} $$
From 2.1 and 2.2 taking time and equating the two times when they meet, we find:
$$ v_1 + v_2 = v_f \tag{2.7}$$
With 2.6 and 2.7 we find that $v = \sqrt{\frac{gh}{2}}$
If we insert this velocity in  2.4 or 2.5 we find:
$$ x = \frac{3}{4}h  \tag{2.8} $$
A: The answer in your Edit seems right,
You must have also needed to change one of the minus signs in
$x_2(t) = v_ft - \frac{1}{2}(-9.81)t^2 \quad$
as if you are measuring positive to be upwards, just one of them will take account of gravity being downwards.
Then you just put your first two formulae equal to each other to get
$h=v_ft$ and since you now have $v_f$ you can find the time when they meet, after substituting the time back into either formula you can get the height.
