A ball is thrown from a height of 10m with certain initial downward velocity. It hits the ground and loses 50% of its energy during the impact and rises back to the sane height. What was its initial downward velocity of projection?

Answer 1

initial energy
= (1/2)m v^2 + m g (10)
= m(.5 v^2+98.1)

final energy = m(98.1)
so
98.1 = .5(.5v^2 +98.1)
2(98.1) = .5 v^2 + 98.1
.5 v^2 = 98.1
v = sqrt(2*98.1)
v = about 14 m/s