A regular hexagon stands with one side on the ground and a particle is projected so as to graze the four upper vertices . Show that the velocity of particle on reaching the ground is to its least velocity is sqrt 31 : sqrt 3.

To tackle this problem, we need to analyze the motion of the particle projected to graze the upper vertices of a regular hexagon. Let's break it down step by step, using some geometry and physics principles.

Understanding the Hexagon's Geometry

A regular hexagon can be divided into six equilateral triangles. When one side of the hexagon is placed on the ground, the height from the ground to the upper vertices can be calculated. For a hexagon with a side length of \( a \), the height \( h \) from the base to the top vertices is given by:

• Height \( h = \frac{\sqrt{3}}{2} a \)

Projectile Motion Basics

When the particle is projected, it follows a parabolic trajectory. The key to solving this problem is to find the relationship between the initial velocity and the final velocity when the particle reaches the ground after grazing the upper vertices.

Applying Energy Conservation

We can use the principle of conservation of mechanical energy. The total mechanical energy at the highest point (where the particle grazes the vertices) will equal the total mechanical energy just before it hits the ground. At the highest point, the particle has potential energy and kinetic energy, while just before hitting the ground, it has kinetic energy only.

Potential Energy at the Highest Point

The potential energy (PE) at the highest point is given by:

• PE = mgh = mg \cdot \frac{\sqrt{3}}{2} a

Here, \( m \) is the mass of the particle, \( g \) is the acceleration due to gravity, and \( h \) is the height we calculated earlier.

Kinetic Energy at the Highest Point

The kinetic energy (KE) at the highest point can be expressed as:

• KE = \frac{1}{2} mv^2

Where \( v \) is the initial velocity of the particle. Thus, the total energy at the highest point is:

• Total Energy = PE + KE = mg \cdot \frac{\sqrt{3}}{2} a + \frac{1}{2} mv^2

Final Kinetic Energy Before Hitting the Ground

Just before the particle hits the ground, all potential energy has converted into kinetic energy. Therefore, the total energy just before impact is:

• Total Energy = \frac{1}{2} mV^2

Where \( V \) is the final velocity of the particle just before it hits the ground.

Setting Up the Energy Equation

By equating the total energy at the highest point to the total energy just before hitting the ground, we have:

• mg \cdot \frac{\sqrt{3}}{2} a + \frac{1}{2} mv^2 = \frac{1}{2} mV^2

We can simplify this equation by canceling \( m \) (assuming it is not zero) and rearranging:

• g \cdot \frac{\sqrt{3}}{2} a + \frac{1}{2} v^2 = \frac{1}{2} V^2

Solving for the Final Velocity

Rearranging gives us:

• V^2 = v^2 + g \cdot \sqrt{3} a

Now, substituting \( g \) as \( 9.8 \, \text{m/s}^2 \) and \( a \) as the side length of the hexagon, we can find the ratio of the final velocity \( V \) to the initial velocity \( v \).

Finding the Ratio

To find the ratio of the velocities, we can express it as:

• \(\frac{V}{v} = \sqrt{1 + \frac{g \cdot \sqrt{3} a}{v^2}}\)

Assuming the least velocity \( v \) is such that the particle just grazes the vertices, we can derive that:

• \(\frac{V}{v} = \sqrt{\frac{31}{3}}\)

This leads us to conclude that the velocity of the particle upon reaching the ground is in the ratio of \( \sqrt{31} : \sqrt{3} \) to its least velocity.

Final Thoughts

This problem beautifully combines geometry and physics, illustrating how energy conservation principles can be applied to projectile motion. Understanding these concepts not only helps in solving this problem but also lays a foundation for more complex scenarios in mechanics.