If the body travels half its total path in the last second of its fall from rest, find:
(a) The time and (b) height of its fall. Explain the physically unacceptable solution of the quadratic time equation. `(g=9.8 m//s^(2))`

Explanation

If the body falls from a heigh `h` in time `t`, from 2 nd equation of motion we have
`h = (1)/(2) g t^(2)` [`u = 0` as the body starts from rest]...(i)
Now, as the distance fallen in `(t - 1)s` will be
`h = (1)/(2) g (t - 1)^(2)`.....(ii)
So, from Eqns. (i) and (ii) distance fallen in the last second
`h - h' = (1)/(2) g t^(2) - (1)/(2) g (t - 1)^(2)`
i.e., `h - h' = (1)/(2) g (2t - 1)`
But according to given problem as `(h - h') = h//2`
i.e., `((1)/(2)) h = ((1)/(2)) g (2t - 1)`
or `((1)/(2)) g t^(2) = g (2 t - 1)` [as from Eqn. (i) `h = ((1)/(2)) g t ^(2)`
or `t^(2) - 4 t + 2 = 0`
or `t = [4+- sqrt((4^(2) - 4 xx )]//2`
or `t = 2 +- sqrt2`
or `t = 0.59 s " or " 3.14 s`
0.59 s is physically unacceptable as it gives the total time `t` taken by the body to reach ground lesser than one sec while according to the given problem time of motion must be greater than, 1 s
So, `t = 3.41 s`
and `h = (1)/(2) xx (9.8) xx (3.41)^(2) = 57m`