Physics mechanics question. Answer and explain the following question?

Question

A table tennis ball is released from a height h above the table. 50% of the kinetic energy is lost at each bounce and none is lost by friction with the air. What was the height reached after the second bounce?

Answer 1

consider each bounce>

ball starts from rest, hits with v, and rises to h1

0 + mgh = 0.5 mv^2 + 0

loss 1 = 0.50[0.5mv^2] = 0.5[mgh]

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first rising KE1 = 0.5 mgh (50% remaining)



0.5mgh + 0 (ground) = 0 (new top) + mgh1

new PE = mgh1 = 0.5mgh

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o (from rest again) + 0.5 mgh = KE2 (hits again) +0

loss 2 = 0.5[KE2] = 0.25 mgh

ball will rise with remaining energy to new height

0.25 mgh + 0 (ground) = mgh2 + 0 (will come to rest)

mgh2 = 0.25 mgh

h2 = 0.25 h