A regular hexagon of side 10 root 3 m is kept at horizontal surface . A particle is projected with velocity v m/s at an angle b from horizontal surface such that it will just touch the all four corners of regular hexagon.

To solve the problem of a particle projected with a certain velocity at an angle that allows it to touch all four corners of a regular hexagon, we need to analyze the geometry of the hexagon and the physics of projectile motion. Let's break this down step by step.

Understanding the Hexagon's Geometry

A regular hexagon can be divided into six equilateral triangles. Each side of the hexagon is given as 10√3 m. The distance from the center of the hexagon to any vertex (the circumradius) can be calculated as the length of a side, which is 10√3 m. The height of each triangle can be derived from the properties of equilateral triangles.

Calculating the Height of the Triangles

The height (h) of an equilateral triangle with side length 'a' can be calculated using the formula:

• h = (√3/2) * a

Substituting a = 10√3 m:

• h = (√3/2) * (10√3) = 15 m

Thus, the height of each triangle is 15 m, and the total vertical distance from the center to the top vertex is 15 m.

Projectile Motion Analysis

Now, let's consider the projectile motion of the particle. The particle is projected at an angle β with an initial velocity v m/s. The horizontal and vertical components of the velocity can be expressed as:

• Vx = v * cos(β)
• Vy = v * sin(β)

Time of Flight

The time of flight (T) until the particle reaches the maximum height can be calculated using the formula:

• T = (2 * Vy) / g

Where g is the acceleration due to gravity (approximately 9.81 m/s²). The particle must reach a height of 15 m, which can be calculated using the formula for vertical motion:

• h = Vy * T - (1/2) * g * T²

Finding the Required Velocity

To ensure that the particle touches all four corners of the hexagon, it must reach a height of at least 15 m. Setting the height equation equal to 15 m and substituting for T gives us a relationship between v, β, and g:

• 15 = v * sin(β) * (2 * v * sin(β) / g) - (1/2) * g * (2 * v * sin(β) / g)²

After simplifying, we can derive an expression for v in terms of β and g. This will give us the minimum velocity required for the particle to just touch all four corners of the hexagon.

Example Calculation

Assuming the angle β is 45 degrees (which is a common angle for maximizing range), we can substitute sin(45°) = cos(45°) = √2/2 into our equations. This will allow us to calculate the specific value of v needed to achieve the desired trajectory.

Final Thoughts

By carefully analyzing the geometry of the hexagon and applying the principles of projectile motion, we can determine the necessary conditions for the particle's trajectory. This approach not only helps in solving the problem but also deepens our understanding of the interplay between geometry and physics.