**Power Due to Gravitation at the Highest Point**

When a body of mass 2 kg is projected at a velocity of 20 m/s at an angle of 60 degrees with respect to the horizontal, the motion of the body can be analyzed using the principles of projectile motion. At the highest point of the projectile's trajectory, the body momentarily comes to rest before starting to descend.

To determine the power due to gravitation at the highest point, we need to consider the work done by gravity. The work done by a force is given by the equation:

Work = Force × Distance × cos(theta)

where theta is the angle between the force and the displacement.

In this case, the force acting on the body is the force of gravity, which is given by the equation:

Force of gravity = mass × acceleration due to gravity

The distance over which the force of gravity acts is the vertical distance traveled by the body from its initial position to the highest point.

**Work Done by Gravity**

To calculate the distance traveled vertically, we need to consider the vertical component of the projectile's initial velocity. The initial velocity can be resolved into two components: one in the horizontal direction and the other in the vertical direction. The vertical component of the initial velocity is given by:

Vertical component of velocity = initial velocity × sin(theta)

where theta is the angle of projection.

Using the given values, the vertical component of the initial velocity is:

Vertical component of velocity = 20 m/s × sin(60 degrees) = 20 m/s × √3/2 = 10√3 m/s

To find the distance traveled vertically, we can use the equation of motion:

s = ut + (1/2)at^2

where s is the distance traveled vertically, u is the initial vertical velocity, t is the time taken to reach the highest point, and a is the acceleration due to gravity (-9.8 m/s^2).

Since the body comes to rest at the highest point, the final velocity is 0 m/s. Therefore, the equation becomes:

0 = 10√3 m/s - 9.8 m/s^2 × t

Solving for t, we get:

t = (10√3 m/s) / (9.8 m/s^2) ≈ 1.07 s

Substituting the value of t back into the equation of motion, we can find the distance traveled vertically:

s = (10√3 m/s) × (1.07 s) + (1/2) × (-9.8 m/s^2) × (1.07 s)^2

s ≈ 10.64 m

**Calculating the Work Done**

Now that we have the distance traveled vertically, we can calculate the work done by gravity at the highest point using the equation:

Work = Force × Distance × cos(theta)

The force of gravity is given by:

Force of gravity = mass × acceleration due to gravity = 2 kg × (-9.8 m/s^2) = -19.6 N

Substituting the values into the equation for work, we get:

Work = -19.6 N × 10.64 m × cos(180 degrees) = -19.6 N × 10.64 m × (-1) = 209.74 J

**Calculating the Power**

Power is defined as the rate at