Answer
(a) The two balls collide when the time is $t = \frac{H}{v_0}$.
(b) $H = \frac{v_0^2}{g}$
Work Step by Step
(a) For the first ball,
$y = v_0t - \frac{1}{2}gt^2$
For the second ball,
$y = H - \frac{1}{2}gt^2$
We can equate the two equations for $y$ to find where the two balls meet.
$v_0t - \frac{1}{2}gt^2 = H - \frac{1}{2}gt^2$
$v_0t = H$
$t = \frac{H}{v_0}$
The two balls collide when the time is $t = \frac{H}{v_0}$
(b) $y_{max} = \frac{v_0^2}{2g}$
When $t = \frac{H}{v_0},$
$y = H - \frac{1}{2}(g)(\frac{H}{v_0})^2$
We can equate the equations for $y$ and $y_{max}$ to find the value of $H$ in terms of $v_0$ and $g$.
$H - \frac{1}{2}(g)(\frac{H}{v_0})^2=\frac{v_0^2}{2g}$
Let's multiply both sides by $\frac{2v_0^2}{g}$.
$\frac{2Hv_0^2}{g} - H^2=\frac{v_0^4}{g^2}$
$H^2 - \frac{2Hv_0^2}{g} +\frac{v_0^4}{g^2} = 0$
$(H - \frac{v_0^2}{g})^2 = 0$
$H = \frac{v_0^2}{g}$
