Given:
- Distance between floor and ceiling = 5m
- Particle is projected at a certain angle with velocity 20m/s.
- Maximum horizontal range = 20√N

To Find:
The value of N.

Explanation:
To find the value of N, we need to understand the concept of projectile motion and how it relates to the given information.

Projectile Motion:
Projectile motion is the motion of an object that is thrown or projected into the air, subject only to the acceleration due to gravity and the resistance of the air. The motion can be divided into two independent components: horizontal motion and vertical motion.

Horizontal Motion:
In the absence of air resistance, the horizontal motion of a projectile is uniform and unaffected by gravity. The horizontal range (R) can be calculated using the formula:

R = V₀ * t

where V₀ is the initial horizontal velocity and t is the time of flight.

Vertical Motion:
The vertical motion of a projectile is affected by gravity. The time of flight (t) can be calculated using the formula:

t = 2 * V₀ * sin(θ) / g

where θ is the angle of projection and g is the acceleration due to gravity.

The maximum horizontal range occurs when the projectile is launched at an angle of 45 degrees. At this angle, the horizontal and vertical components of the velocity are equal, resulting in the maximum range.

Calculating Maximum Range:
Given that the maximum horizontal range is 20√N, we can write:

20√N = V₀² * sin(2θ) / g

Since the angle of projection is not given, we can assume it to be 45 degrees to maximize the range. Substituting the values, we get:

20√N = (20²) * sin(90) / g
20√N = 400 / g

Simplifying further, we have:

√N = 20 / √g
N = (20 / √g)²
N = (20²) / g

Conclusion:
The value of N is (20²) / g, where g is the acceleration due to gravity.

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