As the stone falls from rest, its initial velocity is equal to zero, i.e. u = 0
Using the second equation of motion for a freely falling body (stone)r the distance covered by the stone in the first three seconds (t = 3 s) Is
S =
ut+at2 ⇒
S3 =
0+g(3s)2 ⇒
S3 =
×10m∕s2×9s2 ⇒
S3 = 45 m
[For a freely falling body, a = g (acceleration due to gravity}.
Also as the motion is only in one direction, i.e vertically downward, distance travelled is equal to displacement]
Using
Sn =
u+a(2n−1) , the distance covered in the last second (n = t) is
St = 0 +
(2t−1) =
(2t−1) ⇒
St =
(2t - 1)
it is given that the distance covered by the stone in the last second is equal to the distance covered in the first three seconds of the motion.
⇒
S3 =
St ⇒ 45 m =
(2t - 1)
⇒ 2t - 1 = 9
⇒ t = 5 s
This is equal to the time for which the stone is in the air.
Hence, option 'B' is correct