A ball moving vertically downward with a speed of `10 m//s` collides with a platform. The platform moves with a velocity of `5 m//s` in downward direction. If `e = 0.8`, find the speed (in `m//s` ) of the ball just after collision.

Explanation

To solve the problem, we will use the concept of the coefficient of restitution (e), which relates the velocities of two objects before and after a collision. The formula for the coefficient of restitution is given by:

\[ e = \frac{v_2 - v_1}{u_1 - u_2} \]

Where:
- \(e\) is the coefficient of restitution
- \(v_2\) is the final velocity of the ball after the collision
- \(v_1\) is the final velocity of the platform after the collision (which is \(5 \, \text{m/s}\) downward)
- \(u_1\) is the initial velocity of the ball before the collision (which is \(10 \, \text{m/s}\) downward)
- \(u_2\) is the initial velocity of the platform before the collision (which is \(5 \, \text{m/s}\) downward)

### Step-by-Step Solution:

1. **Identify the velocities**:
- Initial velocity of the ball, \(u_1 = -10 \, \text{m/s}\) (downward is negative)
- Initial velocity of the platform, \(u_2 = -5 \, \text{m/s}\) (downward is negative)
- Final velocity of the platform after collision, \(v_1 = -5 \, \text{m/s}\) (it continues moving downward)

2. **Substitute values into the restitution formula**:
\[ e = 0.8 = \frac{v_2 - (-5)}{-10 - (-5)} \]

3. **Simplify the equation**:
\[ 0.8 = \frac{v_2 + 5}{-10 + 5} \]
\[ 0.8 = \frac{v_2 + 5}{-5} \]

4. **Multiply both sides by -5**:
\[ -4 = v_2 + 5 \]

5. **Solve for \(v_2\)**:
\[ v_2 = -4 - 5 \]
\[ v_2 = -9 \, \text{m/s} \]

6. **Interpret the result**:
The negative sign indicates that the ball is moving downward after the collision. Therefore, the speed of the ball just after the collision is \(9 \, \text{m/s}\).

### Final Answer:
The speed of the ball just after the collision is \(9 \, \text{m/s}\) downward.