A body of mass 2kg is dropped from a height of 1m. Its kinetic as it reaches the ground

To find the kinetic energy of the body as it reaches the ground, we can use the equation:

Kinetic Energy = (1/2) * mass * velocity^2

First, we need to find the velocity of the body just before it reaches the ground. We can use the equation:

Final velocity^2 = Initial velocity^2 + 2 * acceleration * distance

Since the body is dropped from rest, the initial velocity is 0 m/s. The acceleration due to gravity is approximately 9.8 m/s^2. The distance the body falls is 1m.

Plugging these values into the equation, we get:

Final velocity^2 = 0^2 + 2 * 9.8 * 1 Final velocity^2 = 0 + 19.6 Final velocity^2 = 19.6

Taking the square root of both sides, we find:

Final velocity = √19.6 Final velocity ≈ 4.43 m/s

Now that we have the final velocity, we can calculate the kinetic energy using the equation:

Kinetic Energy = (1/2) * mass * velocity^2

Plugging in the values, we get:

Kinetic Energy = (1/2) * 2 * (4.43)^2 Kinetic Energy = 0.5 * 2 * 19.6 Kinetic Energy = 19.6 Joules

Therefore, the kinetic energy of the body as it reaches the ground is approximately 19.6 Joules.