#### Answer

(a) $y = h - \frac{gh^2}{2v_0^2}$
(b) $h=\frac{2v_0^2}{g}$
(c) $h=\frac{v_0^2}{g}$

#### Work Step by Step

(a) We can write an expression for the vertical position of each ball and equate the expressions to find the time that they collide;
$h-\frac{1}{2}gt^2=v_0t-\frac{1}{2}gt^2$
$t = \frac{h}{v_0}$
We then find the vertical position at this time;
$y = h - \frac{1}{2}gt^2$
$y = h - \frac{1}{2}g(\frac{h}{v_0})^2$
$y = h - \frac{gh^2}{2v_0^2}$
(b) We can find $h$ when the vertical position of the collision $y = 0$;
$y = h - \frac{gh^2}{2v_0^2} = 0$
$\frac{gh^2}{2v_0^2} =h$
$h = \frac{2v_0^2}{g}$
(c) The time for the first ball to reach its highest point is $t = \frac{v_0}{g}$. We can write an expression for the vertical position of each ball at this time to find $h$;
$h-\frac{1}{2}gt^2=v_0t-\frac{1}{2}gt^2$
$h=v_0(\frac{v_0}{g})$
$h=\frac{v_0^2}{g}$