Derive an expression for the height attained by a freely falling body after 'n' number of rebounds from the floor.

Explanation

Consider a small sphere fall freely at a height .n. onto the floor . It strikes the floor with a velocity of .u.
So taht `u_(1) = sqrt(2 gh) . . . (1)`
At the time of collision between the sphere and the floor, the initial and final velocities of floor are zero i.e., `u_(2) = 0 and v_(2) = 0 `
Let `.v_(1).` bet he final velocity of the sphere after the first collision
` :. e = (v_(2) v_(1))/(u_(1) - u_(2)) = (0 - v_(1))/( sqrt(2gh)-0) = (- v_(1))/( sqrt(2 gh))`
`v_(1) = - e sqrt(2 gh)` . . . (2)
(-) sign indicates that the sphere redounds.
`:. ` The height `(h_(1))` attained byt he sphere after first rebounds .
`:.` The height `(h_(1))` attained by the sphere after first reboundis
`h_(1) = (v_(1)^(2))/( 2g) = ((e sqrt(2gh))^(2))/(2g) = e^(2) h`
`h_(1) = (e^(2))^(1) h . . . . (3)`
Similarly we can that velocity attained by the sphere after second reboudn
`v_(2) = - e^(2) sqrt(2 gh) . . . (4)`
and maximum height attained by the sphere atter second rebound
`h_(2) = (e^(2))^(2) h . . . . (5)`
From the (2) and (4) equations velocity of the sphere rebounds after .n. collisions
`v_(n) = e^(n) sqrt(2gh)`
From the (3) and (5) equations, maximum height attained byt he sphaere after n rebounds `h_(n) = (e^(2))^(n)` h