If body travels half of its path in the last second of its fall from rest, find the time and height of its fall.

Options

A. 0.59 s, 57 m
B. 3.41 s, 57 m
C. 5.9 s, 5.7 m
D. 5.9 s, 34.1 m

Explanation

If the body falls a height h in time t, from `2^(nd)` equation of motion we have
` h = 1//2 " gt"^(2)` …(A)
[ u = 0 as body starts from rest]
Now the distance fallen in (t – 1) s will be
` h=1//2 g (t - 1)^(2)` …(B)
So from eq. (A) & (B) distance fallen in the last second
` h- h' = (1//2) "gt"^(2) - (1//2) g (t - 1)^(2)`
` 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) "gt"^(2) = g (2t -1)`
[ as from eq .(A) h = (1/2)` "gt"^(2)` ]
or ` 4^(2) - 4t + 2 = 0 `
or ` t = [ 4 pm sqrt( (4^(2) - 4 xx 2))]//2 `
or ` t = 2 pm sqrt(2) ` or t = 0.59 or 3.41 s
0.59 s is physically unacceptable as it gives the total time t taken by the body to reach ground is lesser than one sec while according to the given problem time of motion must be greater than 1 s.
So t = 3.41 s &
` h = (1//2) xx (9.8) xx (3.41)^(2) = 57 m`
Hence correct answer is (B)