A NAL is dropped from rest at a height of 40 m after striking it losses 25% of its energy to what height does it rebound and what is the velocity at the highest point.
v at ground = sqrt(2gh)
(1/2) m v^2 at ground = (1/2)m(2gh)
keep (3/4) of that so
Ke = (3/8)m(2gh)
that will be spent going back up to Hend
m g Hend = (3/8) (m g) (2h)
so
Hend = 6/8 h = (3/4) h
= 30 m
of course :) If it keeps 3/4 of the energy it will go 3/4 of original height.