A ball of mass 0.1 kg, initially at rest, is dropped from height of 1 m. Ball hits the ground and bounces off the ground. Upon impact with the ground, its velocity reduces by 20%. The height (in m) to which the ball will rise is:

The ratio of KE after the bounce to before the bounce is

1/2 m(.8v)^2
-----------------
     1/2 mv^2
= .8^2
= .64

so, the ball will bounce .64 as high as the original height

To find the height to which the ball will rise after bouncing off the ground, we need to consider the principle of conservation of mechanical energy.

The initial mechanical energy of the ball is given by its potential energy at the topmost point, which is equal to the product of its mass (m), acceleration due to gravity (g), and the height (h) from which it was dropped:

Initial Energy = m * g * h

Since the ball hits the ground and bounces back, we can assume the loss in energy due to the impact is proportional to the square of the reduction in velocity. In this case, the velocity is reduced by 20%, so the loss in energy can be expressed as (0.2^2) * Initial Energy.

After the ball bounces off the ground, it reaches its maximum height when all its initial kinetic energy is converted back to potential energy. At this point, the velocity of the ball will be zero.

Therefore, the final mechanical energy of the ball is zero:

Final Energy = 0

Based on the conservation of mechanical energy, we can equate the initial energy with the loss in energy:

Initial Energy - Loss in Energy = Final Energy

m * g * h - (0.2^2) * m * g * h = 0

Simplifying the equation, we get:

(1 - 0.04) * m * g * h = 0

0.96 * m * g * h = 0

Since mass (m) and the acceleration due to gravity (g) are constants, we can ignore them. Therefore, we have:

0.96 * h = 0

Solving for h, we find:

h = 0 / 0.96

h = 0

This means that the ball will not rise to any height after bouncing off the ground. It will remain on the ground.