1. Initial Setup and Coordinate System

Let us define the origin $(y=0)$ at the initial position of the platform. Upward direction is positive.

• Ball: Starts at $y = h$ with initial velocity $0$. Acceleration $-g$.
• Platform: Starts at $y = 0$ with constant velocity $u$.

2. Time of Collision

The collision occurs when the position of the ball equals the position of the platform.

Ball Position: $y_B = h – \frac{1}{2}gt^2$

Platform Position: $y_P = ut$

Equating them:

$$ ut = h – \frac{1}{2}gt^2 $$ $$ \frac{1}{2}gt^2 + ut – h = 0 $$

Using the quadratic formula for time $t$:

$$ t = \frac{-u + \sqrt{u^2 + 2gh}}{g} $$

3. Velocity Analysis and Rebound

Velocity just before collision:

• Platform velocity: $\vec{v}_P = u \, \hat{j}$
• Ball velocity: $\vec{v}_B = (-gt) \, \hat{j}$ (downwards)

Elastic Collision with Massive Platform:

For an elastic collision with a massive moving wall, the relative speed of separation equals the relative speed of approach.

$$ v_{sep} = v_{app} $$ $$ v’_B – u = u – v_B $$

Where $v’_B$ is the ball’s velocity after collision (upwards) and scalar values are used for speeds.

$$ v’_B – u = u – (-gt) \quad (\text{Note: } v_B \text{ is negative vectorially, speed is } gt) $$ $$ v’_B – u = u + gt $$ $$ v’_B = 2u + gt $$

So, the ball rebounds upwards with velocity $v’ = 2u + gt$.

4. Calculating Rebound Height

We need to find the maximum height reached by the ball above its initial position ($h$).

Let the collision height be $H_{coll} = ut$.

The additional height gained after collision ($H_{add}$) is determined by the rebound velocity $v’$:

$$ H_{add} = \frac{(v’)^2}{2g} = \frac{(2u + gt)^2}{2g} $$

The total max height from the ground is:

$$ H_{max} = H_{coll} + H_{add} = ut + \frac{(2u + gt)^2}{2g} $$

We require the height above the initial position ($\Delta H$):

$$ \Delta H = H_{max} – h $$

From the collision time equation, we know $h = ut + \frac{1}{2}gt^2$. Substituting this for $h$:

$$ \Delta H = \left( ut + \frac{(2u + gt)^2}{2g} \right) – \left( ut + \frac{1}{2}gt^2 \right) $$ $$ \Delta H = \frac{(2u + gt)^2}{2g} – \frac{g^2t^2}{2g} $$ $$ \Delta H = \frac{4u^2 + 4ugt + g^2t^2 – g^2t^2}{2g} $$ $$ \Delta H = \frac{4u^2 + 4ugt}{2g} = \frac{2u^2}{g} + 2ut $$

Now, substitute the value of $t = \frac{-u + \sqrt{u^2 + 2gh}}{g}$:

$$ \Delta H = \frac{2u^2}{g} + 2u \left( \frac{-u + \sqrt{u^2 + 2gh}}{g} \right) $$ $$ \Delta H = \frac{2u^2}{g} – \frac{2u^2}{g} + \frac{2u\sqrt{u^2 + 2gh}}{g} $$ $$ \Delta H = \frac{2u\sqrt{u^2 + 2gh}}{g} $$