Show that the distances travelled by a body freely falling from rest in the1st,2nd &3rd seconds are in the ?

Question

ratio 1:3:5.

Answer 1

distance travelled by a freely falling body (under gravity) from rest  = S  = (g/2)t²

t is the time elapsed since t = 0

S1 = (g/2)1²  = g/2

S2 = (g/2)2²  =  4g/2

S3 = (g/2)3²  =  9g/2

d1 is the distance travelled in first second = S1

d2 = distance travelled in 2nd second = S2 - S1  =   4g/2 - g/2 = 3g/2

d3 = distance travelled in 3rd second S3 - S2 = 9g/2 - 4g/2 = 5g/2

d1 : d2 :d3  =  g/2 : 3g/2 : 5g/2

d1:d2 :d3 = 1:3:5

Answer 2

Using the formula for a body falling freely starting at rest, s = 4.9t^2

s1 = 4.9

s2 = 4.9(2^2 - 1^2) = 4.9 * 3

s3 = 4.9(3^2 - 2^2) = 4.9*5

=> s1 : s2 : s3 = 1 : 3 : 5.