A ball is dropped from a height of 1 m and loses 10% of its kinetic energy when it bounces on the ground. To what height does it rise?

When a ball is dropped from a given height h, it has potential energy proportional to that height: 
E^(pot) = mgh (considering that the potential energy on the ground, at h = 0, is zero.) 
As the ball reaches the ground, all of its potential energy transfers to the kinetic energy, which is proportional to the square of the ball's speed just before it hits the ground: 
E^k = mv^2/2 . Since the energy is conserved during the ball's fall, 
E^k = E^(pot) = mv^2/2 = mgh .
If during the collision the ball loses 10% of its kinetic energy, it will bounce back up with the kinetic energy equal to 90% of the original energy, or
0.9mv^2/2 = 0.9 mgh .
As the ball flies up, its kinetic energy is again converted to potential energy. At the maximum height, the kinetic energy is zero and the potential energy equals the kinetic energy with which the ball left the ground, or 0.9mgh.
This means the maximum height to which the ball will rise is 0.9 of the original height, 0.9 m, or 90 cm.
 
 
https://www.physicsclassroom.com/class/energy/Lesson-1/Potential-Energy