Answer
(a) See the picture.
(b) Above $\frac{h}{2}$.
(c) $\frac{3}{4}h$
Work Step by Step
Let's find $h$ as function of $v_0$
$a=-g$, $\Delta x=h$
$v^2=v_0^2+2a\Delta x$
$0^2=v_0^2-2gh$
$h=\frac{v_0^2}{2g}$. The ball was thrown upward.
Now, let's find the position as function of time equation for both balls. The initial time is when the second ball is thrown upward:
- Ball 1: $v_{0_1}=0$ (at maximum height), $x_{0_1}=h=\frac{v_0^2}{2g}$, $a=-g$
$x_1=x_{0_1}+v_{0_1}t+\frac{1}{2}at^2$
$x_1=\frac{v_0^2}{2g}+0t+\frac{1}{2}(-g)t^2$
$x_1=\frac{v_0^2}{2g}-\frac{g}{2}t^2$
- Ball 2: $v_{0_2}=v_0$, $x_{0_2}=0$, $a=-g$
$x_2=x_{0_2}+v_{0_2}t+\frac{1}{2}at^2$
$x_2=0+v_0t+\frac{1}{2}(-g)t^2$
$x_2=v_0t-\frac{g}{2}t^2$
(a) See the picture:
Ball 1: black line
Ball 2: red line
(b) According to the graph, the balls cross paths above $\frac{h}{2}$.
(c) When the balls cross paths: $x_2=x_1$
$v_0t-\frac{g}{2}t^2=\frac{v_0^2}{2g}-\frac{g}{2}t^2$
$v_0t=\frac{v_0^2}{2g}$
$t=\frac{v_0^2}{2gv_0}=\frac{v_0}{2g}$. Replace this result in $x_1=\frac{v_0^2}{2g}-\frac{g}{2}t^2$
$x_1=\frac{v_0^2}{2g}-\frac{g}{2}t^2$
$x_1=\frac{v_0^2}{2g}-\frac{g}{2}(\frac{v_0}{2g})^2$
$x_1=\frac{v_0^2}{2g}-\frac{v_0^2}{8g}=\frac{4v_0^2}{8g}-\frac{v_0^2}{8g}=\frac{3v_0^2}{8g}$. But, $h=\frac{v_0^2}{2g}$ -> $v_0^2=2gh$
$x_1=\frac{3(2gh)}{8g}=\frac{3(2gh)}{8g}=\frac{3}{4}h$
