Finding the maximum height a ball can be dropped from and still collide with a ball thrown below it I don't necessarily need a specific answer, but I could use a hint, direction, or maybe some reading material.
The question states:

A rubber ball is shot straight up from the ground with speed $V(0)$. Simultaneously, a second rubber ball at height $h$ is directly above the first ball is dropped from rest.
a) At what height above the ground do both balls collide. Your answer will be an algebraic expression in terms of $h$, $V(0)$, and $g$.
b) What is the maximum value of $h$ for which a collision occurs before the first ball falls back to the ground?
c) For what value of $h$ doe the collision occur at the instant when the first ball is at its highest point?

I have solved A and found the answer to be:
$$d = h - \frac{gh}{2V(0)}$$
EDIT: After redoing the problem, the answer comes out to be:
$$d=h-\frac{gh^2}{2V(0)^2}$$
I believe this is correct, but I am completely stumped on questions b and c.
EDIT: I'm also not sure whether or not $h$ is meant to represent the height the second ball starts at or the distance between the balls at any given moment. It seems I haven't solved the first part correctly, so I'll post my work on it:
I plugged the variables I received into the function:
$$\text{Distance} = d_0 + V_0 + \frac{1}{2}at^2$$
(where $d_0$ is the starting distance, $V_0$ is the starting velocity, $a$ is acceleration, $t$ is time), to describe the position functions for the two balls as:
$$d_0 = V(0)t - \frac{1}{2}gt^2$$
$$d_1 = h - \frac{1}{2}gt^2$$
I set $d_0 = d_1$, but I'm not sure what I should be solving for. I can solve it as:
$$V(0)t = h$$
But I can't find any kind of use for this.
 A: (a) I assume that you tackled the problem more or less like this.
Let ground level be $0$, and let the "up" direction be positive. If $U(t)$ is the displacement at time $t$ of the ball that was thrown up, then
$$U(t)=V_0t -\frac{1}{2}gt^2,$$
until the ball returns to ground. 
If $D(t)$ is the displacement at time $t$ of the ball that was dropped, then 
$$D(t)=h-\frac{1}{2}gt^2.$$
Without worrying yet about excessively large values of $t$, where the equations do not apply, we can set $U(t)=D(t)$. After simplifying, we get a very nice equation. solve for $t$ and substitute in either equation to find the collision height.
(b) The first ball returns to the ground at time $t=\dfrac{2V_0}{g}$. The problem asks us for the maximum height $h$ so that the two balls collide before the first hits the ground. There is technically no such maximum height. But we can find the $h$ so that they collide exactly when the first (and second) hits the ground. Just set the $t$ you found in part (a) equal to $\dfrac{2V_0}{g}$.
(c) The ball thrown up reaches is highest point at $t=\dfrac{V_0}{g}$. Apart from that, it is same calculation as in (b).
A: First, doublecheck your answer. Dimensionally, it doesn't make sense. $gh/v_0$ doesn't have dimensions of length.
Next, you find that they collide at some height $d$, and the problem requires that $d\ge 0$. This inequality is solved for certain values of $h$.
