1. Problem Visualization

Two balls are involved in this problem:

• Ball 1: Mass $m_1 = m$, dropped from height $h$.
• Ball 2: Mass $m_2 = m/2$, dropped from the same height $h$ after a time interval.

For the balls to collide, Ball 1 must reach the floor, rebound elastically, and travel upwards to meet Ball 2, which is falling downwards.

2. Kinematics of the Collision

Let the collision occur at a height $y$ from the ground.
Velocity of Ball 1 ($m$):
Ball 1 has fallen distance $h$, rebounded with speed $\sqrt{2gh}$, and then risen to height $y$. Its speed $v_1$ upwards at height $y$ is determined by conservation of energy (kinetic energy at $y$ equals potential energy loss from $h$ to $y$ relative to its virtual peak or simply kinematic equations). More simply, the speed of an object at height $y$ (whether falling from $h$ or rebounding to $h$) depends only on the distance below $h$. $$ v_1 = \sqrt{2g(h-y)} \quad (\text{upwards}) $$ Velocity of Ball 2 ($m/2$):
Ball 2 has fallen directly from $h$ to $y$. Its speed is: $$ v_2 = \sqrt{2g(h-y)} \quad (\text{downwards}) $$

3. Collision Analysis

We define the upward direction as positive.

• Initial velocity of $m_1$: $u_1 = +v$
• Initial velocity of $m_2$: $u_2 = -v$

where $v = \sqrt{2g(h-y)}$.

We need to find the velocity of the lighter ball ($m_2$) after collision, denoted as $v_2’$, to determine its maximum height. Using the formula for 1D elastic collision: $$ v_2′ = \frac{2m_1 u_1 + (m_2 – m_1)u_2}{m_1 + m_2} $$ Substitute $m_1 = m$ and $m_2 = m/2$: $$ v_2′ = \frac{2(m)(v) + (m/2 – m)(-v)}{m + m/2} $$ $$ v_2′ = \frac{2mv + (-0.5m)(-v)}{1.5m} $$ $$ v_2′ = \frac{2mv + 0.5mv}{1.5m} = \frac{2.5v}{1.5} = \frac{5}{3}v $$ So, the lighter ball rebounds upwards with speed $\frac{5}{3}v$.

4. Maximizing the Height

After the collision, the lighter ball starts at height $y$ with upward velocity $v_{up} = \frac{5}{3}v$. The additional height it rises ($\Delta H$) is: $$ \Delta H = \frac{(v_{up})^2}{2g} = \frac{(5v/3)^2}{2g} = \frac{25 v^2}{18g} $$ Substituting $v^2 = 2g(h-y)$: $$ \Delta H = \frac{25}{18g} [2g(h-y)] = \frac{25}{9}(h-y) $$ The total maximum height $H_{max}$ measured from the ground is: $$ H_{max} = y + \Delta H $$ $$ H_{max} = y + \frac{25}{9}h – \frac{25}{9}y $$ $$ H_{max} = \frac{25}{9}h – \frac{16}{9}y $$

To maximize $H_{max}$, we need to minimize the term subtracted, which depends on $y$. Since the collision must occur above the ground, the minimum possible value for $y$ is approaching $0$ (collision occurs just after the first ball rebounds).
Setting $y \to 0$: $$ H_{max} = \frac{25}{9}h $$

Calculation: Given $h = 9.0$ m. $$ H_{max} = \frac{25}{9} \times 9.0 = 25 \text{ m} $$