A big Diwali rocket is projected vertically upward so as to attain a maximum height of 160 m. The rocket explodes just as it reaches the top of its trajectory sending out luminous particles in all possible directions all with same speed v. The display, consisting of the luminous particles, spreads out as an expanding, brilliant sphere. The bottom of this sphere just touches the ground when its radius is 80 m. With what speed `("in" m//s)`are the luminous particles ejected by the explosion?

Explanation

To solve the problem, we need to determine the speed at which the luminous particles are ejected when the rocket explodes at its maximum height of 160 m. The bottom of the expanding sphere of luminous particles touches the ground when its radius is 80 m.

### Step-by-Step Solution:

1. **Understanding the Problem**:
- The rocket reaches a maximum height of 160 m before exploding.
- The luminous particles spread out in all directions and form a sphere with a radius of 80 m when it touches the ground.

2. **Setting Up the Scenario**:
- Let the speed of the ejected particles be \( v \).
- The time taken for the particles to reach the ground from the maximum height can be calculated using the equations of motion.

3. **Using the Equation of Motion**:
- The distance fallen by the particles from the maximum height to the ground is 160 m.
- The equation of motion for an object falling under gravity is given by:
\[ d = v_0 t + \frac{1}{2} g t^2 \]
- Here, \( d = 160 \, \text{m} \) (the height from which they fall), \( v_0 = 0 \) (initial velocity at the top), and \( g = 9.8 \, \text{m/s}^2 \).

4. **Calculating the Time of Fall**:
- Plugging in the values, we have:
\[ 160 = 0 \cdot t + \frac{1}{2} \cdot 9.8 \cdot t^2 \]
- Simplifying gives:
\[ 160 = 4.9 t^2 \]
- Rearranging for \( t^2 \):
\[ t^2 = \frac{160}{4.9} \approx 32.65 \]
- Taking the square root:
\[ t \approx 5.7 \, \text{s} \]

5. **Finding the Radius of the Sphere**:
- The radius of the sphere is given as 80 m. The particles are ejected with a speed \( v \) and travel for a time \( t \) before reaching the ground.
- The distance traveled by the particles in the horizontal direction (which is the radius of the sphere) is given by:
\[ r = v \cdot t \]
- Substituting for \( r \):
\[ 80 = v \cdot 5.7 \]

6. **Solving for Speed \( v \)**:
- Rearranging the equation gives:
\[ v = \frac{80}{5.7} \approx 14.04 \, \text{m/s} \]

### Final Answer:
The speed at which the luminous particles are ejected by the explosion is approximately **14.04 m/s**.