Question

Find the ratio of the distances travelled by a freely falling body in first second and third seconds of its all

Answer 1

s = ut + (1/2)at2
in the first case
now, here u = 0
t1 = 1 s
a = 9.8 m/s2
so,

s1 = (1/2)x9.8x12
or
the distance travelled after 1st second will be
s1 = 4.9 m
here, the velocity after 1st second will be 
v1 = at1
or
v1 = 9.8x1 = 9.8 m/s
now, in the second case
s2 = v1t2+ (1/2)at22
here
v1 = 9.8 m/s
a = 9.8 m/s2
t2 = 1 s
thus
s2 = 9.8x1 + (1/2)x9.8x12
or
the distance travelled after 2nd second will be
s2 = 14.7 m
the velocity after the 2nd second will be
v2 = v1 + at2
or
v2 = 9.8 + 9.8x1
thus
v2 = 19.6 m/s
similarly,

the distance travelled after 3rd second will be
s3 = v2t3 + (1/2)at32
here

v1 = 19.6 m/s
a = 9.8 m/s2
t3 = 1 s
or
the distance travelled after 3rd second will be
s3 = 19.6 m
thus, the ratio will be
s1:s2:s3 = 4.9:14.7:19.6 = 1:3:4