a ball of mass .1kg initially at rest is dropped from height of 1 meter.ball hits the ground and bounces off the ground.upon impact with the ground its velocity reduces by 20%.the height to which the ball will rise is?

To find the height to which the ball will rise, we can use the principle of conservation of energy.

First, let's calculate the initial potential energy of the ball when it is dropped from a height of 1 meter. The formula for potential energy is given by:

Potential Energy = mass * gravity * height

Where:
mass = 0.1 kg (given)
gravity = 9.8 m/s^2 (approximate value)
height = 1 meter (given)

Potential Energy = 0.1 kg * 9.8 m/s^2 * 1 meter
Potential Energy = 0.98 Joules

Since the ball bounces off the ground and loses 20% of its velocity, it also loses 20% of its kinetic energy. However, it retains its entire potential energy. Therefore, at the highest point of its trajectory, all the energy is in the form of potential energy.

At the highest point, the potential energy is equal to the height multiplied by the gravitational potential energy.

Potential Energy = mass * gravity * height

Therefore, we can rearrange the equation to find the height:

height = Potential Energy / (mass * gravity)

height = 0.98 Joules / (0.1 kg * 9.8 m/s^2)
height = 0.98 m

So, the height to which the ball will rise is 0.98 meters.