Let's solve this problem step-by-step. The objective is to find the maximum speed at which the first ball can be thrown upward such that it does not collide with the second ball before the first ball returns to its starting height.

Problem Analysis

1. Ball 1 is thrown upwards with an initial speed $v_0$.
2. Ball 2 is dropped from rest from a height of 12 m above Ball 1's starting position.
3. We need to ensure that Ball 1 reaches back to its starting height without colliding with Ball 2.

Assumptions and Equations

1. Gravitational Acceleration: $g = 9.8 \, \text{m/s}^2$ (downward).
2. For Ball 1 (thrown upward):
   • Initial position: $y_1 = 0$.
   • Initial velocity: $v_1 = v_0$ (upwards).
   • Position as a function of time: $y_1(t) = v_0 t - \frac{1}{2} g t^2.$
3. For Ball 2 (dropped from rest):
   • Initial position: $y_2 = 12 \, \text{m}$.
   • Initial velocity: $v_2 = 0$.
   • Position as a function of time: $y_2(t) = 12 - \frac{1}{2} g t^2.$
4. Collision Condition: A collision occurs if $y_1(t) = y_2(t)$ at any time $t$ before Ball 1 returns to the ground.

Find Time of Return for Ball 1

The first ball returns to its starting height when $y_1 = 0$: $v_0 t - \frac{1}{2} g t^2 = 0.$ Factoring $t$, we get: $t (v_0 - \frac{1}{2} g t) = 0.$ This yields two solutions: $t = 0$ (initial position) and $t = \frac{2v_0}{g}$ (time to return).

Collision Analysis

To avoid collision, Ball 1 must never reach Ball 2's height during its ascent and descent. We will find the maximum $v_0$ such that their paths don't intersect:

1. Set $y_1(t) = y_2(t)$ and solve for $t$: $v_0 t - \frac{1}{2} g t^2 = 12 - \frac{1}{2} g t^2.$ Simplifying: $v_0 t = 12.$ Solving for $t$: $t = \frac{12}{v_0}.$

2. The critical condition is that the collision time $\frac{12}{v_0}$ must be greater than the time it takes for Ball 1 to reach the top of its trajectory (when its velocity is zero), which is $\frac{v_0}{g}$. So we set: $\frac{12}{v_0} > \frac{v_0}{g}.$

3. Multiply both sides by $v_0 g$: $12g > v_0^2.$

4. Solving for $v_0$: $v_0 < \sqrt{12g}.$

5. Plug in $g = 9.8 \, \text{m/s}^2$: $v_0 < \sqrt{12 \times 9.8} = \sqrt{117.6} \approx 10.84 \, \text{m/s}.$

Final Answer

The maximum speed at which the first ball can be thrown without colliding with the second ball is:

$\boxed{v_0 \approx 10.84 \, \text{m/s}}$

Would you like further details on the calculations, or have any questions?

Follow-Up Questions:

1. What would happen if the balls had a certain diameter?
2. How does the initial separation distance affect the result?
3. What if Ball 2 were thrown downward instead of being dropped?
4. How would air resistance impact the motion of both balls?
5. What if the acceleration due to gravity were different, e.g., on another planet?

Tip: When analyzing motion problems involving collisions, carefully compare the positions of both objects over time to identify critical interaction points.