Please help me with the question below.. I’m totally confused!

A player throws a ball of mass 0.6 kg into the air, to a height of 4.0 m above the ground. The ball then falls to the ground. During the impact, 22% of the ball’s energy is lost. Calculate the height to which the ball rises after the bounce. Your assistance will be appreciated..

**Mustafa posted**

Answer:

m= 0.6 kg

h = 4.0 m

Energy = mgh

During impact 22% energy is lost. Therefore, energy remaining = 88%

Therefore the energy with which the ball rebounds = 24 x 88/100 J

This is equal to the potential energy at the highest point after bouncing

If the new height is h’

mg h’ = 88% mgh

h’ = 0.88 x h = 0.88 x 4.0 =**3.52 m**

If you have further doubts, please ask

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