UPSEE Solved Model Paper 7

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+

1

2

at2
⇒ S3 = 0+

1

2

g(3s)2
⇒ S3 =

1

2

×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+

1

2

a(2n−1) , the distance covered in the last second (n = t) is
St = 0 +

g

2

(2t−1) =

g

2

(2t−1)
⇒ St =

10m∕s2

2

(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 =

10m∕s2

2

(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