A rookie equilibrist

To solve this exercise, we will study each ball separately, since each one experiences a different motion:

Ball A. Upwards vertical launch.
Ball B. Downward vertical launch.

Question a)

Data

H = 12 m
v0 = 9 m/s
g = 10 m/s

Resolution

In both cases, to calculate the time that the balls stay in the air we must know at what time they touch the ground, that is, when their position y=0 m. Substituting in the equations of position for vertical motion:

Ball A

$$\begin{gathered} y_A=H+v0·t_A-\frac{1}{2}·g·t_A\Rightarrow \\ 0=12+9·t_A-\frac{10·{t_A}^2}{2}\Rightarrow \\ 0=12+9·t_A-5·{t_A}^2\Rightarrow \\ \boxed{t_A=2.69 s} \end{gathered}$$

Ball B

$$\begin{gathered} y_B=H+v0·t_B-\frac{1}{2}·g·t_B\Rightarrow \\ 0=12-9·t_B-\frac{10·{t_B}^2}{2}\Rightarrow \\ 0=12-9·t_B-5·{t_B}^2\Rightarrow \\ \boxed{t_B=0.9 s} \end{gathered}$$

Question b)

Data

H = 12 m
v0 = 9 m/s
g = 10 m/s
t_A = 2.69 s
t_B = 0.9 s

Resolution

Once we know the time that each ball takes to reach the ground, we can use that time to calculate their velocity at that moment by applying the formulas of vertical launch:

Ball A

$$\begin{gathered} v_A=v_{A0} - g · t_A \Rightarrow \\ {v}_A=9 - 10 · 2.69 \Rightarrow \\ \boxed{v_A = -17.9 m/s} \end{gathered}$$

Ball B

$$\begin{gathered} v_B=v_{B0} - g · t_B \Rightarrow \\ {v}_B=-9 - 10 · 0.9 \Rightarrow \\ \boxed{v_B = -18 m/s} \end{gathered}$$

Question c)

Data

H = 12 m
v0 = 9 m/s
g = 10 m/s
t_A = 2.69 s
t_B = 0.9 s

Resolution

Ball A reaches its maximum height when its speed is 0 m/s. In the first place we will calculate the time it takes to reach this height:

$$\begin{gathered} 0=9 - 10 · t \Rightarrow \\ \boxed{t = 0.9 s} \end{gathered}$$

Once we know the time, we can calculate the maximum height:

$$\begin{gathered} y_{\text{max}}=12+9·0.9-\frac{10·(0.9)2}{2}\Rightarrow \\ \boxed{y_{\text{max}}=16.05 m} \end{gathered}$$